PROBLEMS IN COMPLEX VARIABLE THEORY JAN G.KRZYZ
PROBLEMS IN COMPLEX VARIABLE THEORY JAN G.KRZYZ
MODERN ANALYTIC AND COMPUTATIONAL METHODS IN SCIENCE AND MATHEMATICS Number@@O
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Richard Bellman Editor
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PROBLEMS IN COMPLEX VARIABLE THEORY
by
JAN G. KRZYZ Maria Curie-Sklodowska University, Lublin (Poland) Institute of Mathematics, Polish Academy of Sciences
AMERICAN ELSEVIER PUBLISHING COMPANY, INC. NEW YORK
PWN-POLISH SCIENTIFIC PUBLISHERS WARSZAWA
197 1
AMERICAN ELSEVIER PUBLISHING COMPANY, INC. 52 Vanderbilt Avenue, New York, N.Y. 10017 ELSEVIER PUBLISHING COMPANY, LTD. Barking, Essex, England ELSEVIER PUBLISHING COMPANY 335 Jan Van Galenstraat, P.O. Box 211 Amsterdam, The Netherlands
International Standard Book Number 70-153071 Library of Congress Catalog Card Number 0-444-00098-4 COPYRIGHT 1971 BY PANSTWOWE WYDAWNICTWO NAUKOWE WARSZAWA (POLAND), MIODOWA 10
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, American Elsevier Publishing Company, Inc. 52 Vanderbilt Avenue, New York, N.Y. 10017.
I'RINTliIl IN (,OI.AND
To the memory of M ieczyslaw Biernacki
Contents Foreword
xiii
Notation
xv PROBLEMS
1. Complex Numbers. Linear Transformations 1.1. Sets and Sequences of Complex Numbers . 1.2. Spherical Representation . 1.3. Similarity Transformations 1.4. Linear Transformations . 1.5. Symmetry . . . . . . . 1.6. Conformal Mappings Realized by Linear Transformations 1.7. Invariant Points of Linear Transformations 1.8. Hyperbolic Geometry [3], [21] . . . . .
3 8 10 11
12 13 14 16
2. Regularity Conditions. Elementary Functions 2.1. Continuity. Differentiability . . . . . . 2.2. Harmonic Functions . . . . . . 2.3. Geometrical Interpretation of the Derivative. 2.4. Conformal Mappings Connected with w = Z2 2.5. The Mapping w = +(Z+Z-l) . . . . . . . . 2.6. The Exponential Function and the Logarithm 2.7. The Trigonometric and Hyperbolic Functions 2.8. Inverse Trigonometric and Hyperbolic Functions . 2.9. Conformal Mapping of Circular Wedges.
19 21 23 24 25 27 28 30 30
3. Complex Integration 3.1. Line Integrals. The Index [I], [10] 3.2. Cauchy's Theorem and Cauchy's Integral Formula . 3.3. Isolated Singularities 3.4. Evaluation of Residues. . . . . . . . . 3.5. The Residue Theorem . . . . . . . . . 3.6. Evaluation of Definite Integrals Involving Trigonometric Functions 3.7. Integrals over an Infinite Interval . . . . . 3.8. Integration of Many-Valued Functions [21] . 3.9. The Argument Principle. Rouche's Theorem.
33 38 41 43 45 48 49 52 54
4. Sequences and Series of Analytic Functions 4.1. Almost Uniform Convergence 4.2. Power Series . . . . . . . . . .
57 58
Ix
CONTENTS
x
4.3. 4.4. 4.5. 4.6. 4.7. 4.8.
Taylor Series .......... . Boundary Behavior of Power Series .. The Laurent Series . . . . . . . . . Summation of Series by Means of Contour Integration Integrals Containing a Complex Parameter. The Gamma Function . . . . Normal Families [I], [6], [16] . . . . . . . . . . . . . . . . . . . . .
60 65 66 69 72 75
5. Meromorphic and Entire Functions 5.1. Mittag-Leffler's Theorem [I] . 5.2. Partial Fractions Expansions of Meromorphic Functions [11] 5.3. Jensen's Formula. Nevanlinna's Characteristic [18] 5.4. Infinite Products [I], [10] . . . . . . . . . . . 5.5. Factorization of an Entire Function [I], [6], [10]. 5.6. Factorization of Elementary Functions [11] 5.7. Order of an Entire Function [6], [10], [13], [18] .
77 78 80 82 84 85 88
6. The 6.1. 6.2. 6.3. 6.4.
90 91 92 94
Maximum Principle The Maximum Principle for Analytic Functions Schwarz's Lemma [16] . . . . . . . . . . . Subordination [6], [15], [16], [22] . . . . . . The Maximum Principle for Harmonic Functions
7. Analytic Continuation. Elliptic Functions 7.1. Analytic Continuation [1], [2], [6], [10] 7.2. The Reflection Principle [I] . . . . . 7.3. The Monodromy Theorem [I], [10]. . 7.4. The Schwarz-Christoffel Formulae [I], [10] 7.5. Jacobian Elliptic Functions sn, en, dn [1], [6], [7], [10] 7.6. The Functions cr, 1;, tJ of Weierstrass [1], [6], [7], [10]. 7.7. Conformal Mappings Associated with Elliptic Functions [24] 8. The 8.1. 8.2. 8.3. 8.4. 8.5. 8.6.
Dirichlet Problem Riemann's Mapping Theorem [I], [6], [10], [16] Poisson's Formula [I], [26], [27] . . . . . . . The Dirichlet Problem [I], [14], [15], [19], [27] Harmonic Measure [I], [15], [16], [19], [25], [26], [27], Green's Function [15], [16], [17], [27j Bergman Kernel Function [6], [12], [24]
95 96 98 100 105 106 108
Ito III 113 114 116 118
9. Two-Dimensional Vector Fields 9.1. Stationary Two-Dimensional Flow of Incompressible Fluid [4], [9] 9.2. Two -Dimensional Electrostatic Field [4]. . . . . . . . .
121 124
10. Univalent Functions 10.1. Functions of Positive Real Part [26], [27] . 10.2. Starshaped and Convex Functions [16], [17] 10.3. Univalent Functions [16], [17], [20] . . . . 10.4. The Inner Radius. Circular and Steiner Symmetrization [171. [20]. 10.5. The Method or [nncr Radius Majorization [17]
127 129 130 133 136
CONTENTS
xi
SOLUTIONS 1. Complex Numbers. Linear Transformations . 2. Regularity Conditions. Elementary Functions 3. Complex Integration. . . . . . . . . . . 4. Sequences and Series of Analytic Functions 5. Meromorphic and Entire Functions . . . 6. The Maximum Principle ...... . 7. Analytic Continuation. Elliptic Functions 8. The Dirichlet Problem. . . . . 9. Two-Dimensional Vector Fields 10. Univalent Functions. Bibliography . Subject Index
141 155 168 191 212 224 229 245 256 260 277 279
Foreword This collection of exercises in analytic functions is an enlarged and revised English edition of a Polish version first published in 1962. The book is mainly intended for mathematics students who are completing a first course in complex analysis, and its subject matter roughly corresponds to the material covered by Ahlfors's book [1]. Some chapters, for example, evaluation of residues, determination of conformal mappings, and applications in the two-dimensional field theory may be, however, of interest to engineering students. Most exercises are just examples illustrating basic concepts and theorems, some are standard theorems contained in most textbooks. However, the 'author does believe that the reconstruction of certain proofs could be instructive and is possible for an average mathematics student. When the subject matter of a particular chapter is not covered by standard textbooks, the numbers in parantheses given in the contents indicate a corresponding bibliography position which may be consulted for further information. Some problems are due to the author, and some were adopted by the author from various sources. It was beyond the scope of author's possibility to trace the original sources and therefore the detailed references are omitted. The second part of the book contains solutions of problems. In most cases a complete solution is given; in some cases, where no difficulties could be expected, or when an analogous problem has been already solved in a detailed manner, only a final solution is given. The author is well aware that it was extremely hard to avoid mistakes in a book of this kind. He did his best, however, to reduce their number to a minimum. It is the author's pleasant duty to thank W. K. Hayman, Z. Lewandowski, and Q. I. Rahman, who suggested some problems included in this collection. Thanks are also due to Mrs. J. Zygmunt for her help in preparing the manuscript, as well as to M. Stark for his help and encouragement.
Lublin, July 1969
JAN
XIII
G.
KRZYZ
Notation 1. Set theory
a is an element of the set A a is not an element of A
aeA a¢A
BcA AnB
B is a subset of A
A u,B
A""-B A ""- a {a: W(a)} A
frA r)A
o
Intersection of sets A and B Union of sets A and B The complement of B with respect to A The set A with the element a removed The set of all a having a property W(a) Closure of the set A Boundary of the set A The boundary cycle of a domain A taken with positive orientation The empty set
2. Complex numbers I"CZ
Imz
:
1=1 largz IArgz
The real part of a complex number z = x+;y, i.e. the real number x The imaginary part of a complex number z = x+iy, i.e. the real number y The conjugate of z = x+iy, i.e. the complex number x-iy The absolute value of z = x+iy, i.e. yx2+y2 The argument of z =F 0, 00, i.e. any angle 0 satisfying the equations Izl cosO = rez, Izl sinO = imz. There exists a unique value of argz which satisfies - 7 t < argz ~ 7t. It is called the principal value of argument and is denoted Argz
3. Sets of complex numbers
I::, •• z 2] (Z •• Z2)
1= ••
Z2 • •••• Zn]
Closed line segment with end points z 1, Z 2 Open line segment with end points Zl' Z2 Polygonal line joining Zl' Z2' ••• , Zn in this order xv
xvi
NOTATION
[Zl' Z2) = [Zl' Z2]""'Z2, (Zl' Z2] = [Zl' Z2]""'Zl [0, + (0), (- 00, 0] Positive and negative real axis [0, +ioo), (-ioo, 0] Positive and negative imaginary axis ( - CXl, + (0) The set of all real numbers K(zo; r) The open disk with center at Zo and radius r K( 00; 1) The set of all z with Izl > 1 The circle with center at Zo and radius r C(zo; r) The upper (lower) half-plane H+ (H-)
C (C)
The finite (extended) plane
The open quadrants of C, e.g. (+; -) The convex hull of a set A dist(a; B) = inf{x: x = la-bl, b E B} dist(A; B) = inf{x: x = la-bl, a E A, bE B} (=f; =f)
=
{z: rez > 0, imz
< O}
convA
We use the same symbol C(zo; r) for frK(zo; r), as well as- for oK(zo; r); and similarly, [zo, z t1 denotes either a set, or an oriented segment. We hope that this does not cause any misunderstanding. 4. Functions and mappings 1: 1 f(A) p(a, b) <1(a, b)
p(a;f) ro(z; y, G)
g(z, zo; G) r(zo; G)
Df M(r,f) n(y, a) :::tG
r", 0 (mod G)
0,0
One-to-one correspondence The image set of the set A under the mapping f The hyperbolic distance between a, bE K(O; 1), cf. Exercise 1.8.12 The spherical distance between a, bEe The spherical derivative of f at the point a Harmonic measure of an arc y c frG at a point Z E G (G is a domain) Green's function of a domain G with the pole Zo The inner radius of a domain G at the point Zo E G The set of all values taken by f The l.u.b. of If(z) I on the circle C(O; r) The index of the point a w.r.t. a closed, regular curve y Almost uniform convergence of a sequence of functions in G The cycle is homologous to zero w.r.t. the domain G If the quotient f(t)fg(t) tends to 0 (or remains bounded)
r
as t -
'0.
we write: f(t)
o(g(t») (or f(t)
O(g(t)))
NOTATION
xvii
5. Families of function The class of functions analytic in a domain G and such that
L 2 (G)
H11'12< +00
G
[!;
S S* SC
E Eo
The class of functions I analytic in K(O; 1) such that 1(0) = 1 and re I(z) > 0 in K(O; 1) The class of functions I analytic and univalent in K(O; 1) such that 1(0) = 0,1'(0) = 1 The class of all Ie S such that I[K(O; 1)] is starshaped w.t.t. origin The class of all Ie S such that I[K(O; 1)] is a convex domain The class of all functions F analytic and univalent in K(oo; 1) whose Laurent expansion there has the form: F(z) = z+bo+b 1 z- 1 + ... The class of all FeE which do not take value 0
6. Abbreviations a.u. Almost uniform (convergence) hHyperbolic (or noneuclidean) cont. The problem is a continuation of the foregoing one
.
PROBLEMS
CHAPTER 1
Complex Numbers. Linear Transformations 1.1. SETS AND SEQUENCES OF COMPLEX NUMBERS
1.1.1. Find the real and imaginary parts of:
(~)5 1- l. ,
2 1-3i'
1.1.2. Find all complex z satisfying 1.1.3. Show that for Izl rez
-/
zZ.
> 0:
r
=
z=
( 1+i}/3)4 . 1 ·
1 ( rZ) =2 z+Z- '
1.1.4. Evaluate all complex z for which (1 +z)(1-Z)_l is (i) purely real; (ii) purely imaginary. 1.1.5. Show that unless z
=
I:' with reC > 0 such that CZ
x+iY is real and negative, there exists a unique z.
=
1.1.6. Prove the identity
IZI +zzlz+lzI-Zzlz I1l1d
=
2(lzllz+lzzIZ)
explain its geometrical meaning.
1.1.7. Show that
la+ya 2-b 2 1+/a- Va2_bz/
=
la+bl+la-bl·
1.1.8. Prove the identities: (i)
Iz 1 (1 + IZzIZ)-zz(l + IZIIZ)[Z
(ii) II/-ztZzI2+lzI-ZzIZ (iii) II -Zt z212-'lzt-'Z212
=
IZI-ZzIZII-ZIZzlz- (ZlZZ- Zl zz)Z; (l+I ZlI 2)(1+lz2I Z); =
(1-lztlz)(1-lzzI2). 3
t. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
4
1.1.9. Show that for () i= 0, =j=27t, =j=47t, ...
1+ei6 +
...
+ein6
=
sin t(n+ l)()(sin t())_1 e in6 /2
1.1.10. Suppose IZ11 = IZ21 = IZ31 = 1. Show that Z1' Z2' Z3 are vertices of an equilateral triangle, iff Z1 +Z2+Z3 = O. 1.1.U. Suppose IZkl
=
1, k
1 to 4. Show that Zk are vertices of a rectangle,
=
4
iff 2.:Zk=O. k=1
*
1.1.12. Let Z1 Z2 be the inner (scalar) product of vectors [0, ztl, [0, Z2] and Z1 XZ 2 the component of the outer (vector) product perpendicular to the z-plane. Show that
Z1Z2
=
Z1
* Z2+Z1 Xz 2 ' i.
1.1.13. Prove that the area enclosed by the triangle [Z1' Z2' Z3' Z1] whose orientation is positive, is equal tim(z1z2+z2z3+z3z1)' 1.1.14. Prove that the area enclosed by the positively oriented polygon [Z1' Z2' ... , Zn, ztl is equal tim(z1z2+z2z3+ ... +znZ1)' 1.1.15. Find the angle IX between two vectors [Z1' Z2], [Z3' Z4]' Also find the distance of Z3 from the straight line through Z1' Z2' 1.1.16. Show that, if the centers of gravity and of circum-circle of a triangle coincide, the triangle must be equilateral. 1.1.17. Discuss the curves defined by the following complex functions of a real variable (unless stated otherwise, a, b, (J) are real and positive and -00< t < +00). (i) Z = exp(a+bit); (ii) Z = (1 +it)-l; (iii) Z = aeit+a-le-it, 0 ~ t ~ 27t; (iv) Z = at +be irot ; (v) \
Z
=
(vi) Z =
(vii) (viii)
a(1-it)e it ; O+e it )2, 0 ~ t ~ 2~;
t 2 +it4, 0 ~ t < +00; = t+it-t, t > O.
Z =
Z
1.1.1S. Suppose z(t) = x(t)+iy(t) and x, yare real differentiable functions of t E (a, h). Explain the geometrical meaning of z'(t) s'(t) I iy'(t). Show that (e lt )' ie".
1.1. SETS AND SEQUENCES
5
1.1.19. Given the curve z(B) = r«(I)e i9 , where r(B) is a positive, differentiable function of (I. Find the angle a between the tangent and the radius vector. 1.1.20. Suppose z(t) = x(t)+iy(t) is a complex, differentiable function of a real parameter t E (a, b) that does not vanish in (a, b). Show that (i) ;t argz(t) = (xy'-x'y)lz(t)I- 2;
(ii)
~ Iz(t)1
=
(xx' +yy')lz(t)I- 1 •
1.1.21. Explain the geometrical meaning of the following sets of complex numbers: (i) {z: Iz-al = Iz-bl}, a =1= b; (ii) {z: Iz+el+lz-el ~ 2a}, a> 0, lei
z+i (. .){o z: <arg z-i < 4
7t }
11l
< a;
;
(iv) {z: 0 ~ reiz < I}; (v) {z: rez 2 > a}., lI. > 0; (vi)
{z: I ;~~ I < I};
(vii) {z: Izl+rez ~ I}; (viii) {z: Iz2-11 < I}; (ix) {z: re[z(z+i)(z-i)-l]
> O}.
1.1.22. Show that the set {z: arg(z-a)(z-b) = const} is an arc of an equilateral hyperbola whose center is located at 1-(a+b).
> 0 for which the set {z: Iz 2+az+bl < R}
1.1.23. Evaluate all R i~
connected. 1.1.24. Explain the geometrical meaning of the set
{z: AlzI2_Bz-Bz+C where A, C are real, A
=1=
0, IBI2
=
O},
> AC.
1.1.25. Find the radius and the center of Apollonius circle
Iz-allz-bl- 1 = k
(k
=1=
1, k
> 0).
1.1.26. Find the equation of the circle through three not collinear points :I.ZZ,ZJ
(cf.Ex. 1.1.24).
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
6
1.1.27. Suppose IZII =/: e, 1 and 0, ZI' Z2 are not collinear. Show that the circle through ZI' Z2' zi i has [zl(1+lz212)-z2(1+lzI12)] (ZIZ2-ZIZ2)-1 as center and its radius is IZI-Z2111-ZIZ21IzIZ2-ZIZ21-I. Also show that Zi i is situated on this circle. 1.1.28. Suppose mI' m2' m3 are nonnegative, mi +m2-+ m3 = 1, and ZI, Z2' Z3 are not collinear. Show that (i) the point Zo = mizi +m2z2+m3z3 belongs to the closed triangle T with vertices ZI' Z2, Z3; (ii) conversely, for any Zo E T there exists a unique system of nonnegative mI' m2' m3 with mI+m2+m3 = 1 such that Zo = mIzI+m2z2+m3z3' The numbers mJ are called barycentric coordinates of Zo w.r.t. T.
1.1.29. The intersection of all closed and convex sets containing a given set A is called the convex hull of A and is denoted conv A. Show that
n
mk
~ 0, k
=
1, "', n; ~ mk =
I}.
1.1.30. Show that conv {ZI' Z2' ... , zn} = U Tk1m , where T k1m is the closed triangle with vertices Zt, ZI, Zm and the summation ranges over all triples {k, I, m} of positive integers ~ n. n
1.1.31. Prove that the equality
L (C -
Zk)-I = 0 implies: k=l CEconv {ZI' Z2' ... , zn}.
1.1.32. Prove following theorem (due to Gauss and Lucas): all zeros of the derivative of a polynomial are contained in the convex hull of zeros of the given polynomial. 1.1.33. Show that lim n .... oo
(1+ X+iy)n = eX(cosy+isiny). n
1.1.34. Discuss the behavior of the sequence {Zn} , Zn
= (1+i)
1.1.35. Show that, if {en} and is convergent, too.
(1+ ~) ... (1+ !).
L Ibnl
both converge, then the series
L
Cnb
1.1. SETS AND SEQUENCES
7
1.1.36. Suppose rez I ~ 0 (n = 1,2, ... ) and both that also IZnlz is convergent.
L
L Zn' L z~
converge. Show
1.1.37. Prove Toeplitz's theorem: Suppose (a"k) is an infinite matrix of complex numbers (n, k = 1, 2, ... ) which satisfies: 00
(i)
L la.kl ::::;; A
for n = 1, 2, ... ;
k=l
(ii) lim ank = 0
for n = 1,2, ... ;
k-+oo 00
(iii) lim (L ank) n--+oo
=
1.
k= 1
00
Then for any positive integer n and any convergent {en} the series
L ankCk k=l
is convergent. 00
Moreover, if Zn = Lank Cto then
lim Zn exists and is equal
lim Cn. n-+oo
k=l
P1 +Pz+ ... Pn MOl." ~ > lor n = 1,2, ... Ip11+IPzl+ ... +IPnl lim(!p11+lpzl+ ... +IPnD = +00. Show that for any convergent {zn} 1138 .. • Suppose
and
1.1.39. Suppose Zn = UO+U1 + '" +Un-1 +cun is a convergent sequence and rec > -t. Show that also Wn = UO+U1 + ... +Un -1 +un is convergent and has Ihe same limit. 1.1.40. Suppose {Pn} is a sequence of positive reals monotonically incr~asing 10 infinity. Show that for any convergent series Zn with complex terms we have:
L
1.1.41. Show that f
L ,unZn
converges and ,un
lim,un(z1+zz+ ... +zn)
--+
0 then
=
O.
n-+OO
1.1.42. Suppose {un}, {v n} converge to U and v resp. Show that Wn
rOllverges to
11'1).
1 =
-
n
(U1Vn+UZVn_1 +
'"
+unv 1)
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
8
1.2. SPHERICAL REPRESENTATION
1.2.1. Suppose OXlX2X3 is the system of rectangular coordinates whose axes OXl, OX 2 coincide with real and imaginary axes Ox, Oy of the complex plane C. Suppose, moreover, that the ray emanating from the north pole N(O; 0; 1) of the unit sphere s: xi + x~ + x~ = 1 and intersecting S at A (Xl; x 2 ; X3) intersects C at the point z. The point z = x+iy is called the stereographic projection of A(x l ; X2; x 3 ) whereas A is called the spherical image of z. Show that X
z+z 1+lzI 2 '
--~-=-
1 -
X
z-z 2 = iO+lzI2)'
X3
=
IZ12-1 1+lz1 2 '
and Xl +iX2
z=----~.
l- x3
Hence the points of the sphere S (also called Riemann sphere) can be used for geometrical representation of complex numbers. 1.2.2. Find the spherical images of eilZ , -1 +i, 3-4i. 1.2.3. Describe spherical images of northern and southern hemisphere. 1.2.4. Show that any straight line in C has a circle through N as its spherical image. 1.2.5. Show that the stereo graphic projection of any circle on S not containing N is also a circle. 1.2.6. Show that the spherical images of z, Z-l are points symmetric w.r.t. the plane OXlX 2 . 1.2.7. Find the relation between the spherical images of following points: (i) z, - z; (ii) z, z; (iii) Z, Z-l. 1.2.8. If cp, () denote the geographical latitude and longitude of A respectively, show that the stereographic pr~jection of A has the representation: z = e i9tan (t7t+tcp). 1.2.9. Show that
z, ,
correspond to antipodal points of S, iff
zC =
-1.
1.2.10. Prove that the circle AlzI 2+Bz+Bz+C = 0 (A, C real) has a great circle on S as its spherical image, iff A + C = O. 1.2.11. Show that C(zo; R) is the stereographic projection of a great circle on S, iff R2 = 1+lzoI2. 1.2.12. Find the stereographic projection of the great circle joining the points ( -] 3j ,• -
4 . 12) )-3 '-]3'
(
2 • 2 • I) 3 , f .
- -f'
1.2. SPHERICAL REPRESENTATION
1.2.13. The distance o"(Zl' Z2) between two points on S whose stereographic projections are Zl, Z2 is called the spherical distance between Zl and Z2' Show that o"(Zl' Z2) = 21z1-z21 [(1 + Iz112)(1+ Iz212)]-1/2. 1.2.14. Suppose dO", ds are lengths of infinitesimal arcs on Sand C resp. corresponding to each other under stereographic projection. Suppose, moreover, the arc of length ds emanates from the point Z E C. Show that
~; =
2(1
+ IzI2)-1.
Show, moreover, that the angle between any two regular arcs in C and the angle between their spherical images are equal. 1.2.1S. Suppose the sphere S is rotated by the angle rp round the diameter whose end points have a, _a- 1 (cf. 1.2.9) as stereographic projections. Suppose, moreover, z, Care stereographic projections of points corresponding to each other under this rotation. Show that
-C-a -- = l+aC
z-a
el'P---
l+az •
1.2.16. Suppose At, A2 E S and at, a2 are stereographic projections of At lind A 2 , resp. Find the set of all points a E C such that a is the stereographic r>rojection of a point A E S equidistant from At and A 2 • 1.2.17. Find the radius of the circle on S whose stereographic projection is C(a; r).
r
1.2.18. Suppose is a regular arc on Sand 'Y is its stereographic projection. Show that the length /(r) of r is equal to
~ 1+~ZI2 ds. y
011
1.2.19. Find the stereo graphic projection of a rhumb line on S, i.e. of a line S which cuts all meridians at the same angle.
r
1.2.20. Find the length /(F) of the rhumb line joining the points whose '\crcographic projection are Zl = r 1, Z2 = r2 eif%, 0 < ex < 2". Evaluate /(F) . 1 . 1 .• f i lor =1 =~, Z2 =-(3+IJl3). )13 2
III'
1.2.21. Evaluate the area of a spherical domain D being the spherical image II regular domain ;1 in C.
10
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
1.2.22. Show that the area ITI of a spherical triangle T with angles is equal a+{3+Y-7t.
(X,
{3, y
1.2.23. Evaluate the area of the spherical triangle T whose vertices are (0; 2- 1 / 2 ; 2- 1 / 2 ), (2- 1 /2; 2- 1 /2; 0), (0; I; 0).
1.3. SIMILARITY TRANSFORMATIONS
1.3.1. Show that each similarity transformation w = az+b (a i= 0) can be composed of a translation, a rotation and a homothety with center at the origin. 1.3.2. Prove that a similarity transformation (i) carries circles into circles and (ii) parallel straight lines into parallel straight lines; (iii) leaves the ratio (Z3-Z1)!(Z3-Z2) unchanged; (iv) leaves the angle between two curves unchanged. 1.3.3. Find a similarity transformation mapping the strip
{z: 0 < re z < I} onto the strip (w: limwl
so that (z
=
-t) +-+
(w
=
< +7tl
0).
1.3.4. Find the most general similarity transformation mapping (i) the upper half-plane onto itself; (ii) the upper half-plane onto the lower one; (iii) the upper half-plane onto the right half-plane. 1.3.5. Find the similarity transformation mapping the segment [a, b] onto [A, B] so that a +-+ A, b +-+ B.
1.3.6. Find the similarity transformation mapping the triangle with vertices 0, 1, i onto the triangle with vertices 0, 2, 1+ i. 1.3.7. Find the similarity transformation mapping the strip
{x+iy: kx+b l
:(;
y :(; kx+b 2 },
onto the strip
{IV: 0:(; re w :(; I} so that (z
=
ibJl)
+-+
(II' =-, 0).
bi
< b2 ,
11
1.4. LINEAR TRANSFORMATIONS
1.3.8. Show that for any similarity transformation w = az+b (a # 0, 1) there exists a unique invariant point Zo; show that the transformation can be . composed of rotation and a homothety center at Zo. 1.3.9. Find the invariant point zo, the angle of rotation and the ratio of homothety for the transformations in (i) Exercise 1.3.6; (ii) Exercise 1.3.7. 1.3.10. Show that for any similarity' transformation w = az+b (a # 0, 1) there exists a family of logarithmic spirals il}variant under the transformation. 1.4. LINEAR TRANSFORMATIONS
1.4.1. Show that any linear transformation w = (az+b)/(cz+d) (a, b, e, d are complex constants, ad-be # 0) is composed of a translation: Zl = z+/X, an inversion: Z2 = llz1 and a similarity transformation: w = Az2 +B (some of these transformati<;ms may, however, fall out). 1.4.2. Prove that any regular arc y and its image arc under inversion w = liz intersect the radius vector at the same angles. 1.4.3. Show that the angle between any regular arcs Yl' Y2' and the angle between their image arcs under any linear transformation are equal. 1.4.4. Show that the inversion w = liz carries the circle C(a; r) into the circle with radius r/laI2-r21-1 and center a(laI 2_r2)-1 whenever lal # r. Find the image line in case lal = r. 1.4.5. Show that under inversion any circle through -1, 1 is carried into itself. 1.4.6. Show that under linear transformations circles are mapped onto circles or straight lines. 1.4.7. Prove that under inversion, and also under linear transformations the l:l'oss-ratio
remains unchanged. 1.4.8. Find the linear transformation carrying a, b, c into 0, 1, tively. 1.4.9. Find the images of the following lines under inversion: (i) the family of circles C(a; lal); (ii) the family of parallel straight lines y = x+b;
00,
respec-
f. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
12
(iii) the family of straight lines through the origin; (iv) the family of straight lines through Zo (zo i= 0); (v) the parabola y Zk
=
x 2•
1.4.10. Show that the cross-ratio (Zl' Z2, Z3' Z4) is real, iff all the four points lie on circle or on a straight line. 1.4.11. Show that all linear transformations form a group.
1.4.12. Prove that all linear transformations w = (mz+n)/(pz+q) where m, n, p, q are integers satisfying mq-np = 1, also form a group. 1.5. SYMMETRY
Two points z, z* (z i= z*) are called symmetric w.r.t. L where L is a circle, or a straight line, if every circle through z, z* intersects L at a right angle. In particular, if L = C(a; r) and z i= a, 00 then z* lies on the ray through z and with origin at a and Iz-allz*-al = r2. The point z* is called reflection of z with respect to L.
1.5.1. If z* is the reflection of z with respect to C(a; r), then z* 1.5.2. Show that C(a; r) is an Apollonius I(C-z)/(C-z*)1 = const for C E C(a; r»).
=
a+r 2/(z-a).
circle for points z, z* (Le.
1.5.3. Find the reflection of 2+i w.r.t. C(i; 3). 1.5.4. Find the reflections w.r.t. C(O; 1) of (i) the circle C(l; 1); (ii) the hyperbola x 2_y2 = 1. 1.5.5. Show that any circle orthogonal to C(O; 1) remains unchanged after a reflection W.r.t. C(O; 1). 1.5.6. Show that the symmetry W.r.t. L remaihs preserved under linear transformations: if z +-+ W, z* +-+ w*, L +-+ Ll and z, z* are symmetric w.r.t. L, then also w, w* are symmetric w.r.t. L 1 • 1.5.7. Given the point z, find its reflection z* W.r.t. the straight line through al' a2· 1.5.S. Show that the reflection W.r.t. Ll followed by the reflection w.r.t. L2 is a linear transformation. When is the resulting transformation independent of the order of reflections? 1.5.9. Prove that the points symmetric w.r.t. both circles C(a l ; rl)' C(a2; r 2 ) are roots of the equation ([2 -til ~- rHz-a l )-I-rHz (12) .. 1.
1.6. CONFORMAL MAPPINGS
13
1.5.10. Given the straight line L through ai' a2 and the circle C(a; r), find the pair of points symmetric w.r.t. both lines. 1.5.11. Find all circles orthogonal to both circles C(O; 1), C(1; 4). 1.6. CONFORMAL MAPPINGS REALIZED BY LINEAR TRANSFORMATIONS
Given three points Zk and their image points Wk (k = 1, 2, 3) under a linear transformation w, we can determine W by solving the equation (w, Wi' W2 , W3) = (z, Zl' Z2' Z3). We can also use the symmetry invariance in order to determine the linear transformation: if we know that Cw is the image of Cz and W is the image z, then also the point W* symmetric to W w.r.t. Cw must be the image of z*, z* being the reflection of z w.r.t. Cz •
1.6.1. Find the image domain of the linear transformation: W= (i) {z: rez > 0, imz > O}, W = (ii) {z: Izl < 1, imz > O}, W= (iii) {z: 0 < argz < -t7t}, W= (iv) {z: 0 < rez < 1}, W= (v) {z: 1 < Izl < 2},
given domain in z-plane under the given (z-i) (Z+i)-l;
(2z-i) (2+iz)-1; z(z-lrl; (z-l) (Z-2)-1; z(Z-l)-l.
1.6.2. Find the linear transformation carrying the circle C(O; 1) into a straight line parallel to the imaginary axis, the point z = 4 into the point W = 0 and leaving the circle C(O; 2) invariant. 1.6.3. Find the linear transformation carrying the points a, b, c, d on the real axis (a < b < c < d) into _k- 1 , -1, 1, k- 1 (0 < k < 1). Evaluate k. 1.6.4. Find all linear transformations carrying the upper half-plane and the points 0, -1 into itself. 1.6.5. Find the image domains of the unit disk and its upper half under the linear transformation W = (5-4z) (4z-2rl. 1.6.6. Find the linear transformation carrying C(O; 1) into C(l; 1) so that rhe points 0, 1 correspond to -t, 0 respectively. 1.6.7. Find the linear transformation carrying the outside of C(O; 1) into the r igh t half-plane and the points z = 1, - i, -1 into w = i, 0, - i. What are the images of concentric circles center at the origin?
t. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
14
1.6.S. Using the property of symmetry invariance find all the linear transformations carrying the unit disk onto itself. 1.6.9. Find the linear transformation carrying C(O; 1) into C'(1; 2) and the points z = -1, 0 into w = -1, i. 1.6.10. Find the linear transformation mapping the unbounded doubly connected domain in the extended plane whose boundary consists of C(5; 4), C(-5; 4) onto {w: 1 < Iwl < R}. Evaluate R. 1.6.11. Find the linear transformation mapping the ring domain with boundary C(1; l)u C(i; }l6) onto {w: 1 < Iwl < R}. Evaluate R. 1.6.12. Find the linear transformation mapping the right half-plane with removed closed disk K(h; R), h > R, onto {w: p < Iwl < I}. Show that p=
~ - V(~r-1.
1.6.13. Find a linear transformation mapping the bounded domain whose boundary consists of C(O; 2), C(1; 1) onto a strip bounded by two straight lines parallel to the imaginary axis. 1.7. INVARIANT POINTS OF LINEAR TRANSFORMATIONS
1.7.1. If w = (az+b)/(ez+d), ad-be #- 0, and IX = (alX+b)/(elX+d), then IX is called an invariant point of the given linear trans/ormation. Find the general linear transformation with two different and finite invariant points IX, fJ. 1.7.2. Show that the general linear transformation with invariant points 0, ro is a similarity w = Az (A #- 0). 1.7.3. A transformation T whose inverse T- 1 is identical with T is called an involution. Show that a linear transformation (az+b)/(ez+d) different from identity is an involution, iff a+{[ = O. 1.7.4. Show that an involution different from identity has always two different invariant points. 1.7.5. Prove that any linear transformation with two different invariant points can be written in the standard form: (W-IX)/(W-(J)
1.7.6. Show that if L1 then
=
A(z-IX)/(z-fJ).
=
(d-a)2+4be and the sign of .
A
-=
(a jd+}lIi)/(a j-d--}lt1).
}/Lf is suitably chosen,
1.7. INVARIANT POINTS
15
1.7.7. Bring the linear transformation w of Exercise 1.7.5.
=
(z+i)/(z-i) to the standard form
1.7.S. Prove that a linear transformation with only one invariant point 00 is a translation. Also prove that a linear transformation with only one finite invariant point IX (or the parabolic transformation) has the form (W-IX)-l
=
(z-IX)-l+h
(h i= 0).
1.7.9. Find the parabolic transformation mapping ceO; R) onto itself whose only invariant point is z = R. 1.7.10. If IX, {3 are invariant points and A = IAle i9 (cf. Ex. 1.7.5) then the circle Cz with diameter [IX, {3] is carried into a circle Cw with radius R -= +IIX- {3llcosOI- 1 . 1.7.11. The sequence {zn} is defined by the recurrence formula: Zn+1 =/(zn), wherefis a linear transformation with at most 2 invariant points and Zo is given. Discuss the convergence of {z,,}. 1.7.12. Find the points of accumulation of the sequence {zn}: Zo =
0,
1.7.13. If A in Exercise 1.7.5 is real, the corresponding linear transformation is called hyperbolic, if A = eiq> (with real q;) it is called elliptic. Prove that in both cases there exists a family of circles such that any' circle of the family is mapped onto itself under the transformation. 1.7.14. Prove that for any parabolic transformation with an invariant point rx there exists a family of circles tangent to each other at IX and such that each d rde of the family is mapped onto itself under the given transformation. 1.7.15. Suppose IX, {3 are invariant points of a linear transformation which is not an identity and carries a circle C into itself. Show that either IX, f3 are ,itllated on C, or are symmetric w.r.t. C. 1.7.16. Suppose a linear transformation which is neither elliptic, nor hyperholic, has two finite and different invariant points. Show that no circle can be IIlIlpped onto itself by this transformation. 1.7.17. Suppose w = (az+b)/(cz+d), ad-bc = 1 and a+d is real. Prove thai the transformation is elliptic if la+dl < 2, hyperbolic if la+dl > 2 and (lilraholic if la+dl = 2.
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
16
1.7.18. Show that the rotations of the Riemann sphere correspond to elliptic transformations in the plane after stereo graphic projection. 1.7.19. Suppose IX, P are invariant points of an elliptic transformation and ~ 2. Prove that this transformation corresponds to a rotation of the Riemann sphere followed by a translation.
IIX-PI
1.7.20. Show that the linear transformation
w = (az+b)/(-bz+a),
lal+lbl
> 0,
corresponds to a rotation of the Riemann sphere. Evaluate stereographic projections of the end-points of the diameter being the axis of rotation, as well as the angle of rotation in terms of a and b. 1.7.21. Find the linear transformation representing the rotation of the Rieround the diameter with end-points (f; f; t), mann sphere by an angle (-f; -f; -t)·
t'lt
1.7.22. Find the general linear transformation representing a rotation of the Riemann sphere by an angle 'It. Show that this is an involution. 1.7.23. Show that for any involution with two finite invariant points IX, P which is different from identity the factor A in Exercise 1.7. 5 equal~ -1. Prove that the straight line through IX, fJ is mapped onto itself. 1.7.24. Find all
str~ight
lines remaining invariant under the involution
2wz+i(w+z)-2
=
o.
1.8. HYPERBOLIC GEOMETRY
In hyperbolic (Lobachevski-Bolyai) geometry the axioms of Euclid are valid except for the parallel axiom: there are at least two different straight lines in the plane through a given point not on the straight line L which do not meet L. There is a very simple and elegant way essentially due to Poincare of satisfying the axioms of non-Euclidean geometry by a suitable choice of cert~ configurations in Euclidean space. The points in the hyperbolic plane or h-points are the points of the unit disk K(O; 1). The straight lines (hyperbolic straight lines, or h-lines) are the arcs of circles, or straight line segments orthogonal to the unit circle and interior to it. Hyperbolic motions are linear transformations mapping K(O; 1) onto itself. Two sets of h-points are congruent if there exists an h-motion carrying one set intoanother one.
I.s. HYPERBOLIC GEOMETRY
17
We can also introduce in a natural way h-distance in the hyperbolic plane which is invariant under h-motion. Complex numbers and linear transformations are very convenient tools in analytic treatment of hyperbolic geometry.
1.8.1. Prove that there exists a unique h-line through any two h-points represented by Zl' Z2 (Zl"# Zz, IZll < 1, IZzl < 1). 1.8.2. Prove that there exists a unique h-line through a given point direction ei!l (i.e. meeting C(O; 1) at ei!l).
Zl
in a given
1.8.3. The unit circle C(O; 1) is called the h-line at infinity. Two h-lines meeting at infinity (i.e. two circular arcs orthogonal to C(O; 1) intersecting each other at a point on C(O; I) are called h-parallels. Show that there are two h-parallels to a given h-line L through a given h-point Zl not on L, as well as infinitely many h-lines through Zl not meeting L. 1.8.4. Find the general form of an h-rotation (i.e. an h-motion with a unique invariant h-point zo). 1.8.5. Find a general h-translation, i.e. an h-motion with two invariant points the h-line at infinity.
011
1.8.6. Find a general h-boundary rotation, i.e. an h-motion with a unique invariant point on C(O; 1). 1.8.7. Find a general h-motion. Verify the group property for h-motions. 1.8.8. Write parametric equation of an h-segment [Zl' ZZ]h' i.e. a subarc of h-Iine with end-points Zl' Z2. 1.8.9. Suppose the h-segments [a, Z]h' [b, W]h are congruent in the sense of hyperbolic geometry. Verify that
l(z-a)/(I-az)1 = l(w-b)/(l-bw)l· 1.8.10. Suppose C and r are two regular curves situated in the unit disk and carried into each other under an h-motion. Show that
1.8.11. Consider all regular curves situated inside K(O; 1) and joining two Ji1lcd points 0, R (0 < R < I) of K(O; 1). Show that
f. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
18
has a minimum for y being the segment [0, R], the minimum being equal 1 I+R Tlog l-R = artanhR. Hint: Verify first that we can restrict ourselves to regular curves with parametric representation 0 = O(r), where 0, r are polar coordinates.
1.S.12. If C is a regular curve situated inside K(O; 1) then
C Idzi
J l-lz[2
c
is called hyperbolic (or h-) length of C. Show that the h-segment with end points Zl' Z2 is the curve with shortest h-Iength among all regular curves in K(O; 1) joining Zl to Z2' The h-Iength of [Zl' Z2]h is called hyperbolic (or h-) distance P(Zl' Z2) of points Zl' Z2' Also show that P(Zl' Z2)
=
artanhlzl-z2111-Z1Z21-1.
1.S.13. Find the h-circle with h-radius R, i.e. {z: p(zo, z)
=
R}.
Also find its h-Iength lh' 1.S.14. Verify the usual properties of a metric for P(Zl' Z2)' 1.S.lS. Show that P(Zl, Z2) = tlog (Zl , Z2' ei/l, ei"),
where e i", Zl, Z2' ei/l are successive points of a circle orthogonal to C(O; 1). 1.S.16. Suppose a regular domain D, into [j. Prove that
CC
JJD
Dc
dxdy (l-x 2_y2)2 =:'
rc
JJ
K(O; 1), is carried under h-motion d~dYJ (l_~2-172)2 .
Q
The integral on the left is called hyperbolic (or h-) area of D and will be denoted
IDlh' 1.S.17. Find l[jlh for Q
=
-, {z: Izl < R}.
1.S.lS. Consider an h-triangle T, i.e. a domain bounded by three h-segments with angles rx, (:1, y. Show that
ITlh
=
t[1t-(rx+(:1+y)].
Hint: Take the origin as one of the vertices.
1.8.19. Evaluate the h-area
or
an h-triunglc with Vl:rti~cs
ZI' =2'
=.\.
CHAPTER 2
Regularity Conditions. Elementary Functions A complex function w = fez) defined on a set of complex numbers A is actually defined ~y a pair of real-valued functions u(x, y), vex, y) of two real variables x, y (x+iy = z) with a common domain A. In a formally identical manner as in real analysis we can introduce the notions of limit, continuity and differentiability. If fez) = u(x, y)+iv(x, y) is differentiable at zo = xo+iyo, then u, v have partial derivatives at zo satisfying Cauchy-Riemann equations at this point: Ux = vy, uy = -Vx ' On the other hand, if all the four partials of first order of u, v exist in some neighborhood of zo, are continuous and satisfy Cauchy-Riemann equations at zo, then f = u+ iv is differentiable at Zo, i.e. f(zo+h)
=
f(zo) +ah +h'fJ (h) ,
where
lim 'fJ(h) = 0; h .... O
the constant a is called the derivative off at zo. The most interesting and most important case occurs when f is defined and has a derivative at every point of some domain (or open, connected set) D in the plane. Thenfis called analytic, Iw/omorphic or regular in D. Regularity has far-reaching consequences that go much beyond what one can obtain from differentiability in the real case. The theory of analytic functions has as its central theme just the investigation or these consequences. So, for example, regularity in a domain D implies the existence of derivatives of all orders at all points of D. Since the definitions of Ihe derivative in real and complex domain are formally identical, the usual I"lIlcs of differentiation as the formulas concerning the derivative of a sum, a prodnct or a quotient, as well as the chain rule, remain the same in complex case.
2.1. CONTINUITY. DIFFERENTIABILITY
2.1. t. Discuss the continuity at z = 0 offunctions defined at z :1= 0 as follows: (i)
/~~Zl; (ii) z-lrez; (iii) z-2 rez2; (iv) z-2(rez 2)2 and equal 0 at z 19
=
O.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
20
2.1.2. Suppose J is defined and uniformly continuous in K(O; I). Prove that for any sequence {zn}, Zn E K(O; I), convergent to C(lCJ = I) there exists a limit €p(C) depending only on C, and not on a particular choice of {zn}. Also prove that F: F(z) = J(z) for z E K(O; I), F(C) = €p(C) for CE C(O; I), is continuous in K(O; 1). 2.1.3. Verify that the function J:
J(O)
=
0,
J(z)
Izl- 2 (I+i)imz2
=
satisfies Cauchy-Riemann equations at z 2.1.4. Verify that J(z) 2.1.5. Suppose J(z)
=
=
0. Is
J
differentiable at z
zrez is differentiable at z
= u(x, y)+iv(x, y)
z i= 0,
for
=
°
=
O?
only.
and the limit
limreh-1 [f(zo+h)-J(zo)] h-+O
exists. Show that the partials
u x , Vy
at Zo both exist and are equal.
2.1.6. Suppose u(x, y), v(x, y) are continuous and have continuous partials of first order at Zo = xo+iyo' If J = u+iv and the limit
lim Ihl-1If(zo+h)-J(zo)1 h-+O
exists, then either f, or 1 = u- iv has a derivative at Zo. 2.1.7. Verify that the following functions fulfill the Cauchy-Riemann equations in the whole plane: (i)J(z) = Z3; (ii)J(z) = eXcosy+iexsiny. 2.1.8. Verify that J(z) 2.1.9. If J a constant.
=
=
x(x2+y2)-1_iy(x2+y2rl is analytic in C"-O.
u+iv is analytic and satisfies u 2 = v in a domain D, then J is
2.1.10. Suppose Llu
a2 u a2 u
=
ax2
+ ay2 .
If J is analytic and does not vanish in a domain D, then
2.1.11. Prove that for an analytic function
J:
2.2. HARMONIC FUNCTIONS
21
2.1.12. Write Cauchy-Riemann equations for fez)
=
U(r, O)+iV(r, 0),
where
z
=
re i8 •
Express l' in terms of partials of U, V. 2.1.13. Prove thatf(z) = z" (n is a positive integer) satisfies Cauchy-Riemann equations and f'(z) = nz,,-l. 2.2. HARMONIC FUNCTIONS
A real-valued function u of two real variables x, y (resp. of one complex variable z = x+ iy) defined in a domain D is said to be harmonic in D, if it has continuous partial derivatives of second order that satisfy in D Laplace's equation: Llu
=
u.u+u yy
=
O.
Notice that continuity of partial derivatives of second order -implies continuity and U y, as well as continuity of u. Two functions u, v harmonic in a domain D and satisfying Cauchy-Riemann equations in D: UX = V y , Uy = -Vx are called conjugate harmonic functions. Any pair of conjugate harmonic functions u,v determines an analytic function u+iv. fix
2.2.1. Find all the functions harmonic 10 C"", (- 00, 0] which are cOllstant one the rays argz = const. 2.2.2. Find all the functions harmonic in C"", 0 which are constant on the circles ceO; r). 2.2.3. Verify that the functions u = log Izl, v = argz are conjugate harmonic functions in C"", (-00,0] and Logz = log Izl+iArgz, where Argz is the principal value of argument: -7t < Argz < 7t, is analytic in C"",(- 00, 0]. 2.2.4. Verify that ~cosy, ~siny are conjugate harmonic functions in C. Also verify that the analytic function expz
=
eXcosy+iexsiny
flllfills the identity: expLogz III
C "'-, ( 00 , 0] . 2.2.5. Show that
!
Logz
= Z_l
=
Logexpz
=
z
in C"", (00, 0] and
!
expz
=
expz in C.
2.2.6. Suppose w = fez) is analytic in a domain D and feD) II (- 00,0] Show that F(z) CO" loglf(z)l+i Argf(z) is analytic in D. Evaluate P'.
=
0.
22
1. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.2.7. Suppose u is harmonic in a domain D. Verify that! = u,,-iuy is analytic in D. 2.2.8. Suppose V1' V2 are harmonic and conjugate with u in a domain D. Verify that V1-V2 is a constant in D. 2.2.9. Suppose u is harmonic in {z: Izl > O} and homogeneous of degree m, m =/: 0, i.e. for any t > 0, u(tz) = tmu(z). Verify that v = m-1(yux-xUy) is a conjugate harmonic function. 2.2.10. Find a conjugate harmonic function v for u equal: (i) x 2_y2+xy; (ii) x3+6x2y-3xy2_2y3; (iii) X(X 2+y2)-1; (iv) (X 2_y2) X
(X 2+y2)-2. Evaluate in each case a corresponding analytic function u+iv as depending on z = x+iy. X
2.2.11. Find a conjugate harmonic function v for (1 +x 2+y2) X u(x, y) = 1+2(x2_ y 2)+(x2+y2)2 .
Write u+iv as depending on z = x+iy. 2.2.12. Show that a function u harmonic in a domain D has a conjugate harmonic function v in D, iff! = ux-iuy admits a primitive in D. 2.2.13. Find a conjugate harmonic function v for
u(x, y)
=
eX(xcosy-ysinx).
2.2.14. Write Laplace's equation in polar coordinates r, O. Verify that r"cosnO, r"sinnO are harmonic for any positive integer n. 2.2.15. Discuss the existence of nonconstant harmonic functions having the form: (i) lP(xy); (ii) !p(x+f'x 2+J2); (iii) tp(x 2+y), where tp is a suitable, realvalued function of one real variable. Fisd a corresponding conjugate harmonic function in case it does exist. 2.2.16. Given a real-valued function F with continuous partial derivatives of second order in a domain D such that F;+F; > 0 in D. Suppose a < F(x, y) < b for x+iy ED and 1p is a real-valued, continuous function of t E (a, b) such
that (F",,+Fyy)(F;+F;)-l = 1pO F. Then there exists a real-valued function tp defined in (a, b) and ~uch that tp 0 F is harmonic in D. Evaluate tp as depending on 1p. 2.2.17. Find an analogue of Exercise 2.2.16 in case F is given in polar coordinates r, O.
2.3. GEOMETRICAL INTERPRETATION
23
2.2.18. Verify the existence of functions u harmonic in C'\, (- 00, 0] and constant on confocal parabolas with foci at the origin and vertices on (0, +(0). Find all these functions. 2.2.19. Find all the functions I analytic in C '\,0 such that III has a constant value on circles x 2+y2-ax = O. 2.2.20. Find all the functions I analytic in C'\, (-IX), 0] such that argl has a constant value on circles C(O; r). 2.2.21. Verify the existence of functions u(r, 0) harmonic in C '\, ( - 00, 0] having a constant value on arcs of logarithmic spirals r = ke)./J, where r, 0 are polar coordinates, A. is fixed for all the spirals and k is a parameter determining the individual arcs. 2.2.22. Find all the functions regular in C '" 0 whose absolute value is constant = a2 sin20.
on lemniscates r2
2.3. GEOMETRICAL INTERPRETATION OF THE DERIVATIVE
If I is analytic in a domain D, Zo ED and J'(zo) =F 0, then I(zo+h)
=
l(zo)+hJ'(zo)+O(h 2 )
as
h -+ O.
This means that locally I is a similarity transformation composed of a rotation hy the angle argf'(zo) , a homothety with the ratio 1f'(zo)1 followed by a translation (=0)' The angle between any regular arcs intersecting at Zo and the angle between Ihe image arcs are equal, therefore the mapping realized by an analytic functionl withf'(z) =F 0 is said to be conformal. An analytic function realizing a conformal lind homeomorphic mapping of a domain D is said to be univalent in D. 2.3.1. The linear transformation w = (z+ I)/(z-I) carries the boundary of Ihe upper half-disk of K(O; 1) into two rays emanating from the origin (why?). Iii nd the angle (l( between the image ray of (-I, I) and the positive real axis liN well as the local length distortion A. at z = -I. 2.3.2. Given a linear transformation w = (az+b)/(cz+d) with c =F 0, find I he sets of all z for which (i) the length of infinitesimal segments is preserved; (ii) the direction of infinitesimal segments is preserved. 2.3.3. Suppose z = z(t) is a differentiable, complex-valued function of a real \1Il'inhle t E (a, b) such that z'(t) =F 0 and IV =/(z) is a conformal mapping
24
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
defined in a domain D containing all the points z(t), t E Ca, b). Show that if arg[f'(z(t))z'(t)] is constant in (a, b), then the image of the arc z = z(t), a < t< b, is a straight line segment (not necessarily bounded). 2.3.4. Show that for any linear mapping w = (az+b)/(cz+d), ad-be:F 0, c :F 0, some straight line has as its image a parallel straight line. 2.3.5. Find the sets of all z where an infinitesimal segment is expanded, contracted or preserved under the given transformation: (i) w = Z2; (ii) w = z2+2z; (iii) w = Z-l. 2.3.6. Find local magnification and the angle of local rotation at Zo under the mapping w = Z3. 2.3.7: Show that the Jacobian
~~::;]
of the mapping
= -3+4
f= u+iv, where
f is analytic in a domain D, is equal 11'12. Give a geometrical interpretation. 2.3.8. Verify that if f = u+iv is analytic and !'(zo) :F 0, then the lines u = const, v = const, intersect at Zo at the right angle.
2.3.9. Find the lines u = const, v = const for the mappings (i) w = Z2; (ii) w = Logz. 2.3.10. Find the length of the image arc under the univalent mappingf: D - C of the arc given by the equation z = z(t), a:« t b. Also find the area of the image domain of Q, QeD.
:«
2.3.11. Show that under the mapping w = Z2, the image curve of C(1; 1) is the cardioid w(If!) = 2(1 +cos If!) eiq>. Find its length and the area enclosed. 2.3.12. Evaluate the integral ~~(x2+y2)dxdy, where D is a domain situated D
in {z: rez > 0, imz > O} whose boundary consists of the segment [1,2] and three arcs of hyperbolas x 2- y2 = 1, x 2- y2 = 4, xy ~ 1. Hint: Cf. Ex. 2.3.9 (i). 2.3.13. Evaluate the length of the image arc of the segment [0, i] under the mapping w = z(1_Z)-2. 2.4. CONFORMAL MAPPINGS CONNECTED WITH w
= z'-
2.4.1. Evaluate the maxiq1al error in K(i; +0) if the mapping w = Z2 of this disk is replaced by its differential at z = i. 2.4.2. Find the image domain of the square: 0 < x < 1, 0 < y < 1 and the length of the boundary of the image domain under the mapping w Z2. Z -=, x+ iy. ",c
25
1.S. THE MAPPING w = t(z+z-l)
2.4.3. Find the univalent mapping of the domain {z: rez onto K(O; 1) such that Zo = l+i corresponds to the center.
> 0, imz > O}
2.4.4. Find the univalent mapping of K(O; 1) onto the inside of w(O)
=
2(1+cosO)eif1 ,
0::::;; 0::::;; 2".
2.4.5. Find the univalent mapping of the domain situated on the right-hand side branch of the hyperbola x2_y2 = a 2 onto K(O; 1) which carries the focus of the hyperbola into w = 0 and the vertex into w = -1. 2.4.6. Find the univalent conformal mapping of the domain
< +00, 2px < y2}, p > 0 onto the unit disk such that the points z = -t and z = 0 correspond to w = 0 {z
=
x+iy: -00
<
y
and w = 1 respectively. 2.4.7. Find the univalent conformal mapping of the domain bounded by the branch of hyperbola: x2_y2 = 1, X > 0, and the rays argz = 1=+" onto the strip {w: limzl < I}. 2.4.8. Show that the mapping z = ay2w(1+w 2)-1/2 carries the unit disk {w: Iwl < I} into the domain bounded by the branches of the hyperbola
x 2 _yZ = a2 • 2.4.9. Map conformally the inside of the right half of the lemniscate < p ::::;; a, onto the unit disk.
11I'2-a 21= p2, 0
2.4.10. Map conformally the inside of lemniscate Iw 2-a 21 = p2, P > a, onlo the unit disk. I,l
2.4.11. Map conformally the strip domain between 4(x+l), y2 = 8(x+2) onto the strip {w: lim wi < I}.
the
patabolas:
2.5. THE MAPPING w = t(Z+Z-l)
2.5.1. Suppose C is an arbitrary circle through -1, 1 and Zl' Z2 do not lie = 1. Show that one of these points lies inside C and an-
tin C and satisfy ZlZ2 t II hl'r one outside C.
2,5.2. Show that the mapping w = t(Z+Z-l) carries both the inside and the oiliside of any circle C through the points z = 1=1 in a 1: 1 manner onto the ~lIlIll' domain in the w-plane. Find the image domain. /lint: Show that (w-l)j(w+ 1) = [(z-l)j(z+ lW. 2.S.3. Show that the image domain of the upper half-plane under the mapping II'
~(Z·I-z-l) is
C,,",-{(-oo, -1] u [1, +oo)}.
26
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.5.4. Show that the image domain of the unit disk under the mapping t(Z+Z-l) is C",[-I, 1].
W=
2.5.5. Suppose C is a circle through z = =f 1 and r is a circle having a common tangent with C at z = 1 and situated in the outside of C. Describe the image under the mapping w = t(Z+Z-l). curve of
r
2.5.6. Find the image curves of: (i) circles C(O; R); (ii) rays argz
=
(1
under the mapping w = t(Z+Z-l).
2.5.7. Map conformally the ellipse {w: Iw-21+lw+21 < l00/7} slit along [-2,2] onto the annulus {z: 1 < Izl < R}. Evaluate R. 2.5.8. Map conformally the outside of the unit disk onto the outside of the ellipse: {w: Iw-cl+lw+cl > 2a} (c 2 = a2 _b 2 , a, b, c > 0). (a
2.5.9. Map conformally the domain C",{K(O; 1) u [-a, -1] u [1, b]} > 1, b > 1) onto the outside of the unit disk.
2.5.10. Map conformally the outside of the unit disk slit along (- 00, -1) onto: (i) w-plane slit along the negative real axis;
(ii) the right half-plane.
2.5.11. Map conformally the domain whose boundary consists of three rays: (-00, -1], [1, +00), [2i, +ioo) and of the upper half of C(O; 1) onto a halfplane. 2.5.12. Map conformally the domain bounded by two confocal ellipses:
{w: onto an annulus {z: Rl
2V5 <
Iw~21+lw+21
< 6}
< Izl < R 2 }. Evaluate Rl IR 2 •
2.5.13. Map conformally the domain bounded by the right-hand branches of the hyperbolas u 2 coS- 2 1X-v2 sin- 2 1X = 1, u2 cos- 2 (J-v 2 sin- 2 (J = 1 (0 < IX < (J < +1t) onto the angle {z: IX < argz < (J}. Hint: Cf. Excercise 2.5.6 (ii).
2.5.14. Show that the image W of the point w under the mapping W = w3 - 3 w describes three times an ellipse with foci -2,2, when w describes once a COIlfocal ellipse.
27
2.6. THE EXPONENTIAL FUNCnON
2.5.15. Show that under the mapping W = w3 -3w the quadrant {w: rew > 0, im w < O} is carried 1: 1 onto the complementary domain of the set {W: re W ;;;::, 0, im W ;;;::, O} u [- 2, 0]. 2.5.16. Map conformally the part of the z-plane to the left of the right-hand branch of the hyperbola x 2 - y2 = 1 on a half-plane. Hint: Map the upper half of the given domain by the mapping W = Z2. 2.6. THE EXPONENTIAL FUNCTION AND THE LOGARITHM
2.6.1. Use the identity:
expz
=
e%
=
eXcosy+iexsiny,
z
=
x+iy,
to verify that (i) le%1 = eX; (ii) exp(z+2rci) = expz; (iii) exp(zl+Z2) = (expzl)(expz2)'
"-
2.6.2. Show that for any complex w =f. 0 and any real ct the equation e% has exactly one solution z satisfying ct < imz < ct+2re.
=
w
2.6.3. Find the image domain of the strip -re < imz < re under the mapping II' = e%. Also find the images of segments (xo-rei, xo+rei) and of straight lines y =Yo·
2.6.4. Find the image of the straight line y e% (m =f. 0).
=
mx+n under the mapping
II' .•. '"
2.6.5. Find the image domain of the strip mx-re llIapping w = eZ •
<
y
< mx+re under the
2.6.6. For which z is the exponential function (i) real; (ii) purely imaginary? Evaluate the real and imaginary parts of exp(2+i) up to 4 decimals. 2.6.7. Find the image domain of the square Ix-al < e, Iyl < e under the mapping w = ~ (a, e are real, 0 < e < re, z = x+iy). Evaluate the limit of I he ratio of the areas of both domains as e ~ O. 2.6.8. Show that the principal branch of the logarithm maps conformally "d I ; r), 0 < r < 1, onto a convex domain symmetric W.r. t. the straight lines: illlll' •.•~ 0, rew = ilog(1-r2). 2.6.9. Show that the function w = Log[(z-ct)j(z-P)], where Log denotes principal branch of logarithm corresponding to largzl < re, is univalent in (' '.,'[ct,/J]. Find the image domain of C",[ct, 1'1] and also the images of: I he
28
1. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
(i) circular arcs with end-points IX, (3; (ii) Apollonius circles for the points IX, (3;
(iii) the point z
=
00.
2.6.10. Show that the function
W
=
«(3-oc){LOg[(Z-IX)/(Z-(3))}_l
is univalent in C"'- [IX, (J]. Find the image domain. 2.6.11. Suppose f is analytic in a domain D and does not take real, nonpositive values in D. Show that
If(z) I = If(X)tp(y),
x+iy = zED,
implies: fez) = where a is a real constant and b, c are complex constants. 2.6.12. Find a conformal mapping of the sphere into the w-plane such that the straight lines v = const are image lines of parallels and the straight lines u = const are the image lines of meridians (u+iv = w). Express u, v in terms of geographical coordinates 0, If on the sphere. Hint: If z is the stereographic projection of a point on the sphere, then = fez) is anaiytic. exp(az 2 +bz+c)
w
1.7. THE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
2.7.1. Starting from the definition of eZ (Ex. 2.6.1)-verify that
are analytic in C and coincide with cosz and sinz resp. on the real axis. This defines the functions sin and cos for complex z. 2.7.2. Write cosz, sinz, tanz = si~z/cosz in the form u(x, y)+iv(x, y) where u, v are real-valued functions of real variables x, y (x+iy = z). Verify that u, v are conjugate harmonic functions. 2.7.3. Show that
Isinzl2 = sin2x+sinh 2y,
Icoszl 2 = cos2x+sinh 2y.
Find all zeros of sine and cosine. 2.7.4. Verify that Isinzl ~ 1 on the boundary of any square with vertices 7t(m+t)(=Fn=i), m = 0,1,2, ... 2.7.5. Verify that Icoszl ~ 1 on the boundary of any square with vertices 7tm(=Fl =j=i), m =-= 1,2, ...
2.7. TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
29
2.7.6. Show that Isinzl ~ coshR, Icoszl ~ coshR for any z ~
2.7.7. Show that for any z with limzl Itanzl ~ [1+ (sinh 0)-2]1/2,
E
.R(O; R).
t5 > 0
[cotzl ~ [I+(sinho)-2]1/2.
2.7.8. Verify the identity sin 2z+cos 2z
=
1 for all complex z.
2.7.9. Verify the identity sinz = sinz and its analogues for cos, tan, cot. 2.7.10. Find all z for which sinz, cosz, tanz are (i) real; (ii) purely imaginary. 2.7.11. Evaluate cos(5 - i), sin(1- 5i) up to 4 decimals. Show Zo
=
fTC+ilog(4+ ]115) then sinzo
=
that
if
4.
2.7.12. Verify for complex Zl' Z2 the addition formulas: COS(Zl+Z2)
=
COSZ1COsz2-sinz1 sinz2'
sin(zl+z2) = sinz1cosz2+Cosz1sinz2'
2.7.13. Write coshz +iv(x, y), x+iy = z.
=
t(ez+e- Z), sinhz
=
t(eZ-e- z)- in the form u(x, y)+
2.7.14. Express [sinhz[2, [coshz[2 as functions of x, y. 2.7.15. Find the relation between corresponding trigonometric and hyperholic functions and give a geometric interpretation. 2.7.16. Verify the identity: 'R) (1 .) 'R) (1+1.) cot (IX+If' + - I cot (IX-If'
=
2 sin2IX+sinh2f3 h2 2R . cos IX-COS f'
2.7.17. Find the image domain of the strip [rezJ < tTC under the mapping II' sinz. Find the image arcs of segments (-tTC+iyo, tTC+iyO) and of straight lincs x = xo and verify the univalence of sine in the strip considered. 2.7.18. Find the image domain of the rectangle: 0
IInder the mapping w
=
<
rez
< tTC, 0 <
< a,
sinz. Are the angles at all vertices preserved?
2.7.19. Show that cosine is a univalent function in the strip 0 t hc
imz
image domain being the right half-plane with they ray [1,
<
< fTC, removed.
rez
+ (0)
2.7.20. Map 1: 1 conform ally the strip 0 < rez < fTC onto the unit disk slit IIlong a radius.
1'/
2.7.21. Map 1: 1 conformally the domain D to the left of the parabola 4(I-x) onto the unit disk. /lint: Consider the image domain of D''-.., (- 00,0] under the mapping t = liz.
1. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
30
1.8. INVERSE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
2.8.1. Find all solutions of the equation z = cos w in terms of the logarithm. 2.8.2. Show that w = Arccosz maps 1: 1 conformally the z-plane slit along the rays (-00, -1], [1, +(0) onto the strip 0 < rew < 7t. 2.8.3. Find Arccos+(3+i). 2.8.4. Show that w = Arctanz maps 1: 1 conformally the z-plane slit along the rays [i, +ioo), [-i, -ioo) onto the strip Irewl < t7t. Also show that ..
w
1
l+iz
= -2'1 Log-1-·-· -IZ
2.8.5. Discuss the image arcs of Apollonius circles with limit points - I , 1 and the image arcs of circles through these points under the mapping w = Arctanz. 2.8.6. Evaluate (i) Arctan(1+2i); (ii) Arctaneif1 ,
-t7t < (J < t7t.
2.8.7. Find the image domain of the w = Arctanz.
unit disk under the mapping
2.8.8. Find the principal branch Artanh of the inverse of tanh (i.e. the branch whose restriction to the real axis coincides with the real function artanh) in terms of Log. Find the image domain of the unit disk under Artanh. 2.8.9. Find the image domain of the quadrant rez w = Arsinhz.
> 0, imz > 0 under
2.8.10. Show that w = arcsinez maps the z-plane slit along the rays y = - 00 < x ~ 0 (k = 0, =f 1, =f2, ... ) onto the upper half-plane.
k7t,
1.9. CONFORMAL MAPPING OF CIRCULAR WEDGES
Elementary functions such as logarithm, exponential function, linear functions, power w = z" = exp(cdogz), carry some families of rays, circular arcs, or straight lines, into a similar family, e.g., Logz carries the family of rays argz = const into the family of parallel straight lines. A suitable superposition of such transformations enables us to map any circular wedge, i.e. a simply connected domain whose boundary consists of two circular arcs (not necessarily different, either arc can be replaced by a straight line segment), onto a disk, or half-plane. In particular the mapping w = z" for Ot: suitably chosen carries an angular sector with vertex at the origin into a half-plane. On the other hand the exponential
1.9. CONFORMAL MAPPING OF CIRCULAR WEDGES
31
function transforms zero angle into a half-plane. It may even happen that a particular circular triangle with two right angles can be mapped onto a circular wedge and hence onto a half-plane. 2.9.1. Map 1: 1 conformally a circular wedge in the z-plane whose sides intersect at a, b and make an angle IX orito a half-plane. Hint: Consider first a linear mapping carrying a, b into 0, 00 respectivel.y. 2.9.2. Map 1: 1 conformally the angle D
{z: IX < argz < {3},
=
0
< {3-1X <
27t,
onto the right half-plane. 2.9.3. Map 1: 1 conformally the upper half of the unit disk onto the upper half-plane. 2.9.4. Map 1: 1 conformally the circular sector: 0 onto the unit disk. 2.9.5. Map the circular wedge {;: Izl upper half-plane. 2.9.6. Map the wedge K(-l;
Y2) ("'I
<
K(I;
I} n
y2)
< argz <
t7t, 0
{z: Iz+iv31 >
1,
2} onto the
onto K(O; 1) so that 0
2.9.7. Map the strip {z: limzl < t} onto K(O; 1) so that 0 image arcs of segments rez = Xo and straight lines imz = Yo. 2.9.8. Map the domain K(O; 1) n (C",-K(t;
< Izi <
+-+
+-+
O.
O. Find the
t)) onto K(O; 1).
2.9.9. Map the complementary domain of the set K(i; 1) n K( -i; 1) onto the outside of K(O; 1) so that 00 +-+ 00. 2.9.10. Map the circular triangle Oal a2 :
K(1; 1)
n
[C",,-:K(I-i; y2)] n {z: Iz/(z-2)1 < 2}
onto the upper half-plane. 2.9.11. Map the w-plane slit along the ray (- 00, -{-] onto the unit disk so that 0 +-+ O. 2.9.12. Map the w-plane slit along the rays (-00, -t)], [-~- +00) onto the lInit disk so that 0 +-+ o. 2.9.13. Map the upper half-plane with the points of the upper half of the unit disk removed onto the upper half-plane. 2.9.14. Map the upper half-plane slit along the segment (0, ih], h I he upper half-plane.
> 0, onto
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
32
2.9.15. Map the upper half-plane slit along the ray [ih, +ioo), h the upper half-plane.
>
0, onto
2.9.16. Show that the mapping (l_p)2
z
p
. (l_Z)2'
w=
0
<
p
< 1,
carries the unit disk Izi < 1 slit along the radius (-1, 0] onto the w-plane slit along the negative real axis.
2.9.17. Map the unit disk Izl < 1 onto the w-plane slit along the negative real axis so that (z = 0) - (w = 1). 2.9.18. Map the unit disk Izl < 1 slit along the radius (-1, 0] onto the unit disk It I < 1 so that the points t = 0, z = p (0 < p < 1) correspond to each other. 2.9.19. Map the domain K(O; 1) "" (-1, -p], 0 disk 1t 1 < 1 so that 0 - O. '
<
p
<
1, onto the unit
2.9.20. Assume that lal < 1, Ibl < 1, a i= b, and y is the circle through a, b orthogonal to C(O; 1). Let D(a, b) be the unit disk slit along the arc of y joining b to C(O; 1), situated inside C(O; 1) and not containing a. Map conformally D(a, b) onto the unit disk It I < 1 so that (C = a) - (t = 0). 2.9.2i. Evaluate
(ddC) t
for the mapping of Exercise 2.9.20. t=O
2.9.22. Show that the mapping w = pz(z+p)(l +pz)-t, p > 1, carries' 1: 1 the outside of the unit disk in the z-p1ane into the outside of a circular arc on the circle C(O; p) symmetric w.r.t. the real axis. Verify that the angle subtended by the arc is equal to 4arcsinp-l. 2.9.23. Find the 1:1 conformal mapping of the domain C""K(O; 1) onto the w-plane slit along the arc: Iwl = 1; largwl ~ s,o that the points at 00 correspond to each other.
+7t
2.9.24. Find the 1: 1 conformal mapping of the domain C""K(O; 1) onto the w-plane slit along the arc: Iwl = R, largwl ~ lI. so that the points at 00 correspond to each other.
CHAPTER 3
Complex Integration 3.1. LINE INTEGRALS. THE INDEX
Suppose F = u+iv is a complex-valued function of a real variable t E [a, b]; F is said to be integrable (e.g. in Riemann's sense), if both real-valued functions b
U,
b
b
v are integrable; then the integraliF(t)dt is defined as ~u(t)dt+ilv(t)dt. a
a
a
Most of the properties of the real integral also hold for the integral of a complexvalued function. The lihe integrals of complex-valued functions which are a very important tool in complex analysis, can be reduced to the integrals over an interval. Suppose I' is a regular curve, i.e. a curve which is represented by the equation z = z(t), a ~ t ~ b, with continuous z(t) having a piecewise continuous nonvanishing derivative z'(t). If J is a complex-valued function defined for all z b
on 1', then the line integral ~ J(z)dz may be defined as ~ G(t)dt, where G(t) 1
a
,J(z(t))z'(t) = J(z(t))[x' (t)+ iy'(t)] . The line integral does not depend on the choice of parameter; if -I' is the arc with the opposite orientation, i.e. the arc defined by the equationz = Z(-T), -b ~ T ~ -a, then ~J(z)dz = - V(z)dz.
Ir ris defined in a domain D and has a
-]I
"
primitive Fin D (i.e. F is analytic in D and F' = f) then for any regular arc I' contained in D with end points Zl' Z2 we have ~J(z)dz = F(Z2}-F(Zl}' In particular, for any closed curve C in D we ]I
have ~ J(z)dz = O. Conversely, if the line integral of J over any regular, closed c
curve in D vanishes, then J has a primitive in D. The notion of line integral can be extended on chains n
r, which are linear forms
~ k",I'"" where I'm are regular curves and k m are integers, the integral ~ being III-I
defined as
n
L m-l
r
k m~ . If all curves I'm are closed, the corresponding chain is 1m
·.. tid to be a cycle. 33
3. COMPLEX INTEGRATION
34
Similarly as in real analysis, we also consider unoriented line integrals. If f = u+iv is defined for all points of a regular curve 'Y and s is the arc-length on 'Y, then ~ uds+i ~ vds is called unoriented line integral and is denoted ~ fds, y
y
y
or ~fldzl. l'
3.1.1. Evaluate the line integral ~ re z dz for: (i) 'Y
= [0, 1+i]; (ii) 'Y = ceO; r).
l'
3.1.2. Show that, for a regular curve 'Y with the equation z
= z(t), a <;; t <;; b,
we have b
~f(z)ldzl = ~f(z(t))lz'(t)ldt. a
1
~ Iz-llldzl.
3.1.3. Evaluate the integral
C(O;l)
3.1.4. Evaluate the integral ~ Izl dz for: (i) 'Y
l'
= [-i, i];
(ii) 'Y being the left-hand half of ceO; I) joining -i to i; (iii) 'Y being the right-hand half of ceO; 1) joining - i to i. 3.1.5. Verify the inequalities: (i) (ii)
b
b
a
a
I~ f(t)dt I <;; ~ If(t)ldt,
.
a < b;
I~ f(z)dz I .<;; ~ If(z)lldzl. l'
l'
3.1.6. Show that
I ~f(z)dzl <;; ML, r
where M = sup If(z) I on 'Y and L is the
length of 'Y' 3.1.7. Show that \
J
oQ
~= z-zo
27d, where aQ is the
closed polygonal line
[zo-a-ia, zo+a-ia, zo+a+ia, zo-a+ia, zo-a-ia]. 3.1.8. Prove that the function in this domain.
(Z-ZO)-l
analytic in C"",zo has no primitive
r
3.1.9. Suppose is a contour, i.e. a regular, closed positively oriented simple (or Jordan) curve and A is the area enclosed by r. Prove that
~xdz = -i ~ ydz = ~- ~zdz = iA.
/'
/'
r
3.1. LINE INTEGRALS. THE INDEX
35
3.1.10. Evaluate ~ zz- l dz, where r
annulus {z: 1 < Izl
r is the boundary of the
upper half of the
< 2} with positive orientation.
3.1.11. Evaluate ~ Izlzdz, where r
r
is the boundary of the upper half of
K(O; 1) with positive orientation. 3.1.12. Assume u, v are real-valued functions having continuous partials of first order in some neighborhood of Zo = xo+iyo. Prove that j= u+iv is differentiable w.r.t. z at zo, iff
lim~ ~ r .... O
1tr
j(z)dz
=
O.
C(zo:r)
3.1.13. Prove the Weierstrass mean value theorem: If j is a complex-valued function continuous in a domain D which has a primitive Fin D, then for any a, bED such that [a, b] cD we have: F(b)-F(a) = (b-a)Z; with CE convr, where convr is the convex hull of the curve r with the equation w =j(a+(b-a)t), 0::( t::( 1. Hint: Consider the integral sums of ~ j(z)dz after parametrization and in[0, b]
terpret LI tk as masses. z
3.1.14. Suppose F is analytic in a convex domain G and reF'(z) > 0 for any G. Prove that F is univalent in G (Le. F(Zl) "# F(zz) for any Zl "# zz).
E
3.1.15. Prove that the polynomial P(z) 3.1;16. Prove that the function F(z)
=
a+nz+z" is univalent in K(O; 1).
=
z+e z is univalent in the left half-plane.
3.1.17. Prove that the function F(z) = Z3 +3z is univalent in the domain to the right of the right-hand branch of the hyperbola xZ_yZ = 1. 3.1.18. Suppose cp is a complex-valued function continuous on a regular curve y and F(z)
=
~ (C-z)-lcp(C)dC. y
Show that F is continuous and analytic in any domain not containing any points of y and verify that y
3.1.19. Suppose y is a closed, regular curve and a is a point not on y. Show thllt n(y, a) = -1. ~ -dz2m z-a y
3. COMPLEX INTEGRATION
36
is an integer. The function n(y, a) is called the index of the point a with respect to 1', or the winding number of I' with respect to a. Hint: Consider the function t
h(t)
= ('
~iLdT
J z(T)-a
'" where z = z(t), 0( ~ t ~ (3, is the equation of y. Verify that u(t) -a] exp(-h(t)) is a constant.
=
[z(t)-
3.1.20. Show that n(y, a) has a constant value for all a ED, if the domain D does not contain any points of y. 3.1.21. Suppose I' is situated in a disk .1. Verify that n(y, a) a EC",L1.
=
0 for any
3.1.22. Suppose I' is a regular, closed curve not meeting the negative real axis. Show that for any a E ( - 00, 0) we have n(y, a) = O. 3.1.23. Prove Jordan's theorem for curves starlike w.r.t. the origin: If I' has the equation z(O) = r(O)e i9 , 0 ~ 0 ~ 27t, r(O) = r(27t) and r(O) is a positive, continuously differentiable function of 0,' then C", {I'} = Do u D oo , where Do, Doo are disjoint domains with a common boundary y. Moreover, n(y, z) = 1 for all zEDo and n(y, z) = 0 for all zEDoo' 3.1.24. Suppose {Yk} is a sequence of closed, regular curves such that n(Yk> 0) = 1 for' all k and ~k = inf 1'1 --+ + 00 as k --+ + 00. Show that for any fixed Cerk
z there exists ko such that n(Yk> z)
=
> k o.
1 for all k
n
3.1.25. The notion of index can be extended on cycles: If r n
a does not lie on any I'm, then n(r, a) is defined as
2:
=L
k mI'm and
m=l
kmn(Ym, a).
m=l
Suppose C1 = C(O; r1), C2 = C(O; r2 },J1 < r 2 • Evaluate n(r, a) for 1a1 < r1, r1 < 1a1 < r2 , r2 < 1a1 and (i) r= C1+C2 ; (ii) r= C1-C2 ' 3.1.26. Suppose I' is a cycle, a, b are points not on curves of I' and n(y, a) = n(y, b). Show that ~ (z-a)-1(z-b)-1dz
=
o.
y
3.1.27. Prove that for any positive integers m, n and a, b such that n(y, a) =n(y, b) we have
~ (z-a)-m(z-b)-n y
=
O.
3.t. LINE INTEGRALS. THE INDEX
37
3.1.28. Suppose Wn is a polynomial of degree at most n and a E K(O; r). Prove that
rJ
Wn(z) z"+l(z-a) dz
=
o.
C(O;r)
3.1.29. If lal < r < Ibl, show that ~ (z-a)-1(z-b)-ldz
27ti(a-b)-1.
=
C(O;r)
3.1.30. Evaluate the integral ~
(z-a)-m(z-b)-ndz,
C(O;r)
where m, n are positive integers and lal
< r<
Ibl.
3.1.31. Verify that
~
(1+Z2)-1dz
=
O.
C(O;2)
3.1.32. If
r
is a cycle and W is a polynomial, show that
~W(z)dz = O. r
3.1.33. Suppose r is a cycle and a is a point not on any curve of r. Show that
~ (z-a)-ndz
=
0
for any integer
1i
~ 2.
r
3.1.34. Suppose R(z) is a rational function, a1 , ••• , ar are the zeros of the denominator, A 1, ... , Ar are coefficients of (Z-a1)-1, ... , (Z-a r)-1 in the development of R in partial fractions and r is a cycle such that all its curves omit I he points a1, ... , ar . Show that r
~ R(z)dz = 27ti r
L Ajn(r, aj). j=1
3.1.35. If lal 1= r, show that
~ Iz-al- 2Idzl
=
27tr IlaI2-r21-1,
C
=
C(O; r).
C
Hint:
Transform
Idtl-;rdzjz.
the unoriented integral into an oriented one; put
3. COMPLEX INTEGRATION
38
3.1.36. If f(z)
=
z(l- Z)-2 and 0
< r<
1, show that
21t
1 (' ji i9)~Cl_ r 27t j I (re luv - l-r2 . o
3.1.37. If r is a regular arc joining the points z and omitting the origin, show that
~~
=
=
1, z
=
reitp (0 ~ rp ~ 27t)
logr+irp+2k7ti,
r
where k is an integer depending on
r.
3.2. CAUCHY'S THEOREM AND CAUCHY'S INTEGRAL FORMULA
Cauchy's theorem is a fundamental theorem in the theory of analytic functions. There are several forms of Cauchy's theorem, the simplest one being Goursat lemma, or Cauchy's theorem for a rectangle:
If f is analytic in a domain D and the closed rectangle R: a ~ rez ~ b, c ~ imz ~ d is contained in D, then the line integral oj f over the boundary oR oj R vanishes: ~ J(z)dz = O. iJR
Cauchy's theorem for a rectangle is on the one hand quite useful in proving local properties of analytic functions via Cauchy's integral formula and on the other hand it gives rise to more general statements, e.g. to the following homological version of Cauchy's theorem.
Suppose f is analytic in a domain D and r is a cycle homologous to zero with respect to D (in notation: r '" O(mod D)), i.e. each closed curve in r is contained in D and the index n(r, a) = 0 for any a E C'",D. Then ~f(z)dz = O.
r In particular, any closed curve in a simply connected domain is a cycle homologous to zero, hence for any function analytic in a simply connected domain D
and any closed, regular curve yin D we have: ~J(z)dz
=
0 which is Cauchy's
"
theorem for simply connected domains.
From Cauchy's theorem Cauchy's integral formula can be derived. Suppose f is analytic in a domain D and 0 (mod D) is such that the curves in omit some point a ED. Then
r '"
n(r, a)f(a)
r
1 . (' J~Ldz. = -2 7tl j z-a
r
3.2. CAUCHY'S THEOREM AND INTEGRAL FORMULA
39
r
y
In particular we may assume that is a contour situated together with its inside in D and a is a point inside y. Then n(y, a) = 1 for positive orientation of y and under these assumptions f(a)
=
_1_.
r fez)
2m Jz-a
dz.
)'
Hence the existence of derivatives of all orders, as well as the representation formula for nth derivative follow:
r dz,
3.2.1. Evaluate ~ (1+z 2 r
3.2.2. Evaluate
\'
where
1
eZcosz
J (1+z2)sinz
r
is the ellipse x 2+4y2
dz, where C
1.
=
.• /-
= C(2+z; Jl2).
c 21<
3.2.3. If f is analytic in K(O; R)""-.O, show that the value of ~ f(re i6 ) dO,
0< r
o
< R, does not depend on
r.
21<
3.2.4. Evaluate ~ f(re i6 )dO in case is analytic in K(O; R) and 0 o
3.2.5. If u is harmonic in K(O; R) and 0 u(O)
=
2~
<
r
<
R.
R, show that
2"
~ u (re W) dO .
o 21<
3.2.6. Evaluate ~ log !re i6 -a! dO, r o
<
!a!:
3.2.7. Iff is analytic in K(a; R)""-.O and lim (z-a)f(z)
=
A, show that
z-+a
~
f(z)dz
=
27tiA
for any
r
E
(0, R).
C(a;r)
3.2.8. Suppose D = C '" H, H is an enumerable set of points a1 , ••• , an, ... with liman = 00 andfis analytic in D. If y is a closed, regular curve in D and A~
. lim (z-ak)f(z) exists for all k, show that U(z)dz = 27ti LAkn(y, ak) . ....Ok
)'
Verify that the sum contains a finite number of non-zero terms.
k
40
3. COMPLEX INTEGRATION
3.2.9. Suppose C is a closed, regular curve omitting the points 0, 1, -1.
Find
all possible values of ~ Z(Z~~ 1) . C
3.2.10. Evaluate
~ z4~ 1 dz,
> 1.
a
C(.:a)
3.2.11. Evaluate
~ z2~a2 dz,
a> O.
C(O:2a) (
3.2.12. Using Cauchy's formula for f"(a) evaluate 21 . \ ( zez )3 dz, where C 'Ttl J z-a c is a contour containing a inside. 3.3.13. Evaluate in a similar way the integral
_1_\ ez dz 21ti J z(1-z)3 C in case: (i) C = C(O; t); (ii) C = C(l; t). 3.2.14. Evaluate
\ J C(O:R)
fez) dz in case a, be K(O; R) for (z-a)(z-b)
f analytic
in some domain containing K(O; R). Prove Liouville's theorem: Any function analytic in C and bounded is a constant. 3.2.15. Using Cauchy's formula for the derivative evaluate \
J
dz (z-b) (z-a)m '
lal
< r < Ibl.
C(O;r)
3.2.16. Suppose C is a contour containing 0 inside and leaving z outside and analytic in ~0 and bounded outside C. Prove that
J is
fez) = ~ \ zf(C) dC. 2m ~ (z-C) C
Evaluate the right-hand side term for z situated inside C and 3.2.17. Let Pn be the polynomial j
~ k.
(Z-Zl)(Z-Z2) •••
How many values can take the integral
~ P~~Z) C
with positive orientation omitting all
Zk?
~
O.
(z-zn) with
Zj ~ Zk
for
for C being contours
3.3. ISOLATED SINGULARITIES
41 3.3. ISOLATED SINGULARITIES
If I is analytic in K(a; r)",a, then the point a is called an isolated singularity 01f. If the limit lim/(z) exists, then the singularity is called removable: we can z ...a
extend the domain of I putting I(a)
=
lim I(z) and obtain in this way always z...a
a function analytic in the whole disk K(a; r). If lim/(z)
=
00,
then the point a
z ... a
is said to be a pole. In this case there exists a finite A #0 and a positive integer n such that lim (z-a)n/(z) = A. The number n is called the order 01 the z ...a
pole. Otherwise a is called an essential isolated singularity. If I is analytic in {z: Izl > R}, then 1(1/z) has an isolated singularity at 0 and the character of singularity of/at 00 is defined to be the same as that of/(1fz) at z = O.
z_
3.3.1. Suppose I is analytic in K(a; R)",a and lim (z-a)/(z)
=
O. Prove
that for any closed, regular curve in K(a; R)""a we have ~/(z)dz
=
O.
,.
3.3.2. Show that under the assumptions of Excercise 3.3.1 we have
1m = _1_. 2m
r
I(z) dz,
where
J z-C
O
C(a;r)
3.3.3. Under assumptions of Excercise 3.3.1 prove that a finite limit lim/(z) r...a
b exists and the function:
E
K(a; R)", a,
3.3.4. Show that if I is analytic in k(O; R) and 1(0) is a removable singularity of I(z)/z.
=
0 then the point z
=
0
3.3.5. Show that exp(1/z) has z = 0 as an essential isolated singularity. 3.3.6. Show that sin(l/z) has 0 as an essential isolated singularity. 3.3.7. Prove that z
=
k1t, k
= 0, =f 1, =f2, ... , are poles of order 2 of (sinz)-2.
3.3.8. State the character of isolated singularities (also possibly at infinity) fur the following functions:
(oo)
1 ( I') --3"; z-z
.
II
l-e"
ZS
(1 -z)2
(iii)
;
z -z
(v) exp-l--;
(IV) l+e%;
1 ( VIIoo) exptan --,'
z'
oo,
'(
1
(~Il) sm cos Z
)-1
ez l+z2' 1
(vi) (ez-l)-lexp-l-;
-z
3. COMPLEX INTEGRATION
42
3.3.9. Verify that a function analytic in the open plane C and having z = as a removable singularity, is a constant.
00
3.3.10. Show that a function analytic in the extended plane except for a finite number of poles, is rational. 3.3.11. Verify Taylor's formula: iffis analytic in K(a; R) then for any positive integer n there exists a functionfn analytic in K(a; R) such that for any z E K(a; R) we have: _ I'(a) fez) -f(a)+l!(z-a)+ ...
+
f
n
+ (z-a) fnCz).
Hint: fez) = f(a) + (z-a)fl (z), where fl is analytic in K(a; R). 3.3.12. If f(a) = ['(a) = ... = j<m-l}(a) = 0 and j<m}(a) =1= 0, the point a is said to be a zero off of order m. Show that (z-a)-mf(z) has a removable singularity at a zero a of order m. 3.3.13. Prove that [f(z)r 1 has a pole of order m at a, ifffhas a zero of order m at this point. 3.3.14. Suppose f is analytic in the extended plane except for a finite number of poles and p, q are the numbers of zeros and poles counted with due multiplicity. Prove that p = q. 3.3.15. If f is analytic in K(a; R) and j
3.4. EVALUATION OF RESIDUES
43
3.4. EVALUATION OF RESIDUES
3.4.1. Suppose f is analytic in K( a; r) "" a. Show that there exists a unique complex number A such that for any closed, regular curve y in K(a; r)""a:
~ [f(z)-A(z-a)-l]dz
=
O.
y
The constant A is called the residue of f at a and is denoted res(a;f). Hint: Consider a cycle r = y-mC(a; (5) which is ,..., 0 (mod K(a; r)""a). 3.4.2. If cp is analytic in K(a; r) and fez) =
res(a;f) = AI.
L"
k=l
Ak(z-a)-k+cp(z), show that
· Iractions f· 3.4.3. Express Z3+Z2+2 Z(Z2 _1)2 as a sum 0 f partla an d evaIuate t h e reSl·d ues at 0, 1, -1. 3.4.4. If a is a pole of first order of J, prove that res(a;f) = lim hf(a+h). h-+O
3.4.5. Suppose F is analytic in C except for the set of nonpositive integers, where it has poles of first order. Suppose, moreover, F(l) = 1 and zF(z) (-1)"
= F(z+l) for all regular points. Prove that res(-n; r) = --,-. n. Tt
(e% ) at z = 0, 1; (ii) 1e %2 at z = =fi.
3.4.6. Evaluate the residues of: (i)
zz-1
+z
3.4.7. If f is odd (i.e. fez) == -f( - z», show that
res(a;f)
=
res(-a;f).
If f is even (i.e. fez) == f( -z», show that
res(a;f) = -res( -a;f). III
3.4.8. Prove de l'Hospital's rule for analytic functions: if f, g are analytic K(a;r)'\.a, limf(z) = limg(z) = 0 (or limg(z) = (0) and the finite, z~a
or infinite limit lim Z-+Q
.
z~a
f:((Z» g z
z~a
exists, then also the limit lim l«(Z»__ exists and both z-+Q
g
Z
lim i ts are equal.
3.4.9. If z. is the pole of fez) =
(Z4 +a 4 t-:l,
3.4.10. Show that the residues of fez) =
1/" .
Z,,-l
show that res(zv;f) = -izva-4.
j(z"+a") at all its poles are equal
44
3, COMPLEX INTEGRATION
3.4.11. Suppose k of a, Prove that
~
2 is an integer and f is analytic in some neighborhood
,
-k
_ j
(k-l)! '
res(a, (z-a) fez)) -
3.4.12. Suppose Zk = (k+t)7t and f is analytic in some domain containing the real axis, Prove that
3.4.13. Show that
sinll.z) res ( 0; z3 sinpz 3.4.14. Prove that: (i) res(1; eZ (z-I)-4) "
=
lI.
61f
=
(fJ2
2)
-lI.,
i e;
(eXP(aLOgZ)) 1 1 ' (1+Z2)2 = 4T(1-a)ex p (2"am) (a real);
(n) res i;
, -ie-am (iii) res(ai; (Z2+a 2)-2em •z ) = -~(am+l), a i= 0;
(iv) res(i; [O+z2)cosht7tzr1)
1 -2,; 7t1
=
(v) res(z1; (Z-Z1t"(Z-Z2)-1) = (_1)"-1(Z1-Z2)-"' n = 1,2, "" (VI') res ("I, (1+ Z 2)-")
2 = - I'2'
"+1(2n-2) n -1 ,n
=
1 , 2 ,'" ,'
(VB00) res (1' - ,Z2"(1+ Z)-,,)
=
(1)11+1 (2n)! (n-l)!(n+l)!' n -- 1 , 2 , .. "'
yz/sinyz)
=
(-I)"27t 2n 2, n
(viii) res(n 27t 2 ;
3.4.15. If f is analytic in
C"" K(O;
=
R) and r
>
1,2'00' R, then res( CX); f) is defined
as __1_. 2m
\ ~
f(z)dz,
C( 0; r)
If fez) = P(z)+Az- 1+
R, where P
(i) res ( CX); Log z-a) z-b ; (ii) res (00;
y (z--a)(z---hj) ,
IS
a polynomial and
3.5. THE RESIDUE THEOREM
45
3.4.17. Suppose f is analytic in C except for a finite number of points. Show that the sum of residues at all singularities (the point at infinity included) equals to zero. 3.4.18. Evaluate: (i) res(oo; (z+I)-3 sin2z); (ii) res ( 00 ;
Z3COS
z I 2 ).
3.4.19. Evaluate the' residues of the following functions at all isolated singularities: (i)
(Z3_ Z S)-I;
(ii) ez /z 2(z2+4); (iii) cot 2z; (iv) coez; (v) sin ~I z+
(vi) [z(1-e- hz )]-I; (vii) z"sin!, n
z
0, =FI, =F2, ...
=
3.4.20. Suppose rp is analytic in a neighborhood of a and rp'(a) i= O. Iff has a simple pole at the point rp(a) and res(rp(a);j) = A, show that res(a;fo rp)
= A/rp'(a). 3.4.21. Iff is analytic in K(a; R) "" a, then f has in K(a; R) "" a the following Laurent expansion:
L 00
fez) =
and
An(z-a)n
res(a;f) = A_I'
n=-CX)
Evaluate in this way: (i) res(o;
Z-2 e l/ z Log
(ii) res( I; Logz cos
~=~;);
Z~I)' 3.5. THE RESIDUE THEOREM
If f is analytic in a domain D except for a set of isolated singularities ak, then I'or any cycle y homologous to zero W.r.t. D and contained in D"" U {ak} we have
k
~f(z)dz = y
21ti Ln(y, ak)res(ak;f). k
!'hc sum on the right is finite since n(y, ak) i= 0 only for a finite number of ak. III most applications y is a contour with positive orientation. Then n(y, ak) I~ either 0 in case ak is situated in the unbounded component of C "" {y}, or I ill case Ok is in the bounded component.
J. COMPLEX INTEGRATION
46
The residue theorem enables us to evaluate line integrals of ~nalytic functions over closed curves by means of residues. In what follows all the contours considered as paths of integration are taken with positive orientation unless stated otherwise.
~ -1 ~Z4 ' wh~re C is the ellipse X2 -
3.5.1. Evaluate the integral
xy+ y2 +X+
C
+y=O. 3.5.2. Evaluate the integrals: (i)
~ 1~4 r '
= C(1; 1);
r (ii)
~ (Z-1)~~1+z2)' r
= C(l+i; y"2);
r
dz r· (...) Jr (z2-1)2(z-3)2 ' III
IS
h' 'd 2/3+ 2/3 - 22/ 3 . t e asterol X y ,
r
(iv) ~ , r
Z3(Z~!-2) , r= C(O; 2).
3.5.3. Show that: (i)
~
1!z3 = - t1t'i, C is the ellipse 2x 2+y2_f = 0;
C
Z2
~
(ii)
C(O;R)
exp
(2 '3) 1t'IZ
-
1 dz
= 2n+1, n < R3 < n+1, n being a positive
integer; (iii)
rJ 2 e;"
, dz = }1t'(i-1)e1t/ 2 , y is the boundary of K(O; 1) () (+;+),
Z-I
l'
3.5.4. Verify that \
J
z3 exp (1/z) d _ -~ ,
l+z
z-
31t'1.
C(O;2)
3.5.5. Let aQ be the boundary of the square Q with vertices 1t'( =t= 1 =t= i) and let w be a point inside Q. Prove that
1 ~ ze% Logw = -2' -,.--dz. 1t'I e-w oQ
3.5. THE RESIDUE THEOREM
47
3.5.6. If a, bE K(O; r) and n is a positive integer, show that
~ C(O;,)
z-a 27ti z"Log--dz = _ _ (b n + 1 _an + 1). z-b n+l
3.5.7. If a, bE K(O; r) show that
~
y (z-a) (z-b) dz =
-
-}7ti(a-b)2
C(o;,)
for the branch of square root equal to z+O(1) near infinity. 3.5.8. Under the assumptions of Excercise 3.5.7 show that: (i)
~
y (z-a)(z-b)
~
dz = 27ti. Y (z-a) (z-b)
C(O;,)
(ii)
C(O;,)
Z
dz = 7ti(a+b);
3.5.9. Evaluate the integrals: t.~
(i) (ii)
('
j
C(O;l)
dz Y 4z 2+4z+3
1 for the branch of square root equal to 1+0(1) and 2z (1+0(1)) respectively near infinity. 3.5.10. Verify that
~ C(O;2)
z"
yl+z2 dz=
(0 for n odd, (_I)k 1· 3· .... (2k-l) 27ti
2· 4 .... ·2k for n
=
2k, k
=
1,2, ...
3.5.11. Let r be the parabola x = y2 with the orientation corresponding to increasing y and let Yl+z2 be the branch of square root with positive re~l part in the right half-plane and in a neighborhood of the origin. Show that the improper line integral
3. COMPLEX INTEGRATION
48
3.6. EVALUATION OF DEFINITE INTEGRALS INVOLVING TRIGONOMETRIC FUNCTIONS
3.6.1. Let R(u, v) be a rational function of two variables u, v. Show that the 21<
definite integral ~ R(cosO, sinO)dO may be considered as a parametrized line ~
integral
o Rl (z)dz, when; Rl is a rational function of z.
C(O;l)
3.6.2. Prove that
3.6.3. Prove that
3.6.4. If n is a positive integer, show that 27t 2 \ (1+2cosO)·cosnO dO = 'It (3-V5)". J 3+2cosO V5 . o
3.6.5. Prove that: 27t
(i) \
Jo
dO (1 +acosO)2
3.6.6. Prove that 1t
\
J
xsinx dx l+a 2-2acosx
=
~lo I+a a
g
a
'
a> 1.
-1<
Also evaluate the integral for 0 < a< I. Hint: Integrate z(a-e- iz )-l over the boundary with vertices =F'It, =F'It+in.
oR.
of the rectangle R.
3.7. INTEGRALS OVER AN INFINITE INTERVAL
3.6.7. By integrating ezz- n -
1
round C(O; 1) show that
21<
(i) ~ exp(cosO)cos(nO-sinO)dO = 27f:/n! ; o 2n:
(ii) ~ exp(cosO)sin(nO-sinO)dO
=
o.
o
3.7. INTEGRALS OVER AN INFINITE INTERVAL
The theorem of residues is quite useful in evaluating a large number of real integrals. Let e.g. J be a function analytic on the real axis and in the upper halfplane except for a finite number of isolated singularities ak with imak > 0, k = 1, 2, ... , n. The residue theorem as applied to J and the upper semicircle gives us R
n
~ J(x)dx+ ~ J(z)dz
(3.7A)
-R
27f:i:L res(ak; J)
=
F(R)
k=l
where R > max{lall, la 2 1, ... , lanl} and r(R) is the upper semicircle of C(O; R). If lim ). J(z)dz = 0 then we obtain at once from (3.7A): R-+oo F(R) n
+00
~ J(x)dx -00
=
27f:i
:L res(ak;J) k=l
which gives a convenient method of evaluating certain integrals over an infinite interval.
+00
(iii)
~ o +00
(iv)
~
7f:(a+2b) 2ab 3 ( a+ b)2' a, b >0;
dx
-00
..
so
3. COMPLEX INTEGRATION
3.7.2. Verify the inequality: '11:
~ exp(-AsinO) < rr:/A,
A> O.
o
Hint: Show that sinO> 20/rr: for 0 E (0,
-trr:).
3.7.3. Use Exercise 3.7.2 and the equality e'x = cosx+isinx to show that
+<0
cos mx (...) rJ (x2+a2)2 111
d'
X
rr:
(I)
= 40 3 +am e- am ,
C,
0
m> ;
o
. +f
(VI) ~
3.7.4. If
cos ax I+X 2 +X4 dx=
rr: . (rr: a) ( a V3)' .,l3 sm 6+2 exp - - 2 - ' a>O.
f is a continuous function in the sector {z:
0
< \z-a\ ::::;; (l,
81 ::::;; arg(z-a) ::::;; 02} and lim (z-a)J(z) = b, show that
where i'r = {z: z = a+re ill , 01 ::::;; 0 ::::;; 02} for 0
< r::::;;
(l.
3.7.5. By using the identity: 2sin 2 z = re(l-e 2IZ), z is real, and by integrating fez) = z-2(1-e 2bl) over the contour consisting of [-R, -r], -rer) , [r, R], r(R) (0 < r < R) show that
~ si~X )2 dx =
+<0 (
o
; .
3.7. INTEGRALS OVER AN INFINITE INTERVAL
51
3.7.6. Prove in a similar way that +00
.
2
mx Jr x sm (x +a 2
2
o
d 2)
-
7t (2 + -20m 1) 4a 3 am e -
=
X
(a, m > 0).
3.7.7. Verify the identity sin 3 x
im[t(l-e3iX)-f(l-eiX)]
=
for real x and prove that
rJ
+00 (
)3
sinx d _ 3
-x-
X-
o
3.7.8. By integrating fez) able contour show that
=
S 7t·
Z-3(Z2+ a 2)-1 [z+i(e iZ -l)] (a> 0) over a suit-
3.7.9. By integrating fez) = eOZ (1 +ez )-l (0 with vertices =FR, =fR+27ti show that +00
~
-00
<
a
<
1) round the rectangle
eax 7t --d x = 1+ex sina7t .
Also prove that +00
(i)
~
o +00
(ii)
~
o
ta-1 l+t d t
=
7t sina7t; .
X"'-l 7t --dx(0 1+x" - nsin(m7t/n)
< m
3.7.10. Verify in a similar manner that +00
1=
l'
J
+00
eOX
l+ex+e2x dx
=
-00
rJ 1 t
0
a- 1
+t+t
2
d =
t
27t . sin+7t(I-a) (0
y3- .sma7t
3.7.11. By integrating fez) = [(I+z 2 )cosh+7tz]-1 round the square vertices =FN, =FN+2iN, where N is an integer, show that +00
~ [(1 +x2 )cosh+7tX]-ldx
o
=
log 2.
QN
with
3. COMPLEX INTEGRATION
S2
3.7.12. By integrating fez) = eaz(e-ZiZ_l)-1 (a> 0) round the rectangle with vertices 0, 7t, 7t+iR, iR sUitably indented show that +00
~
•
smay eZ"-l
tta
7t
~--=--,--dy = -4 coth2 -
o
1
2a'
3.7.13. By integrating fez) = ea'z/sinhi (a real) round the rectangle with vertices =FR, =FR+27ti suitably indented show that +00
•
\ smax d _ 7t nh 7ta J sinhx x - 2 ta T'
o
3.7.14. By integrating e"Z Icosh 7tZ (-7t < a < 7t) round the rectangle with comers at =FR, =FR+i and suitable indentations, show that
+~oo coshax d _ 1 ( o
cosh 7tX
a x - -2 cos2
)-1.
3.7.15. Prove that . (I)
+00 (
J\ o
1 1 dx - - - .- ) -
x
slUhx
x
= log2;
+00
(1'1')
~ l-cosax 1 . h dx = log cosh Ta7t (a real); XSIU x
o
(iii)
+fJ o
x·- 1 d 7t(1-a) (0 (1+x)~ x = sin7ta
< a<
2)
.
3.B. INTEGRATION OF MANY-VALUED FUNCTIONS
3.8.1. By integrating fez) = (Logz)Z(zZ+aZrl round the upper half of the annulus {z: r < \z\ < R}, 0 < r < a < R, prove that +00
~
o
(logx)Z
7t
xZ+a z
8a
-'--;;-=-----=-.;-. dx = -
3.8.2. Prove in a similar way that
I
[7tz+4(loga)z]
'
3.B. INTEGRATION OF MANY·VALUED FUNCTIONS
53
I
3.8.3. By integrating (1 +Z2)-2 exp(aLogz) round the contour of Exercise 3.8.1, show that
-1
{z: r
< Izl < R}
n {z: 0
< argz < t7t},
show that +co
\' logt ~ 1+t 4 dt
7t 2
3.8.5. Show that +co I ogx (1') ~ --;c-=--=---dx 2)2 x(x2+a o
V
+CO
8 y2
= -
'
3 3 2- 3/2 7ta- s /2 (-2Ioga-I--47t) (a> 0);
=
2
dX -- 7t Iog 2', (II") ~ log(l+x) 1 2 +x o
Hint: Integrate (1+Z2)-lLog(z+i) round the upper half of K(O; R). 1
1
(".) \' log(x+x- ) d _ 7t I 2 III j l+x2 x-log. o
3.8.6. If a > 0, prove that
Hint: Integrate [(z2+a 2) LogZ]-l round the annulus {z: r < Izl < R} slit along [-R, -r]. 3.8.7. By integrating log(z-a) (0 < a < 1) round the cycle consisting of ('(0; 1), C(a; r) and [a+r, 1], a+r < 1, show that 2",
~ logle i8 -aldO = O. o
3.8.8. By integrating (Z2_1)-1Iogz round the boundary of
{z: r
< Izl < R}n(+; +),
3. COMPLEX INTEGRATION
S4
show that
3.8.9. Prove that 1
dx
~
(i)
o 1
~
(ii)
3
X2n
Vx(I-x2)
o
dx=~. 1.4 .... . (3n-2)
}fJ
3.6· ... 3n
3.8.10. By integrating z1-'(I-z)'(I+z)3 round the boundary of
K(O; R)". {K(O; r) u K(1; r) u [0, In,
0
< r <·to
R >
t,
show that 1
~xl-P(I-x)P(1+x)-3dx o
= 2P- 3 7tp(1-p)(sinp7t)-1
(-1
<
2).
3.8.11. Prove that 1
~ (l-x n)-1/ndx = 7t[sin(7tjn)]-1 o
(n = 2, 3, ... ).
3.9. THE ARGUMENT PRINCIPLE. ROUCHE'S THEOREM
~uppose f is meromorphic in a domain D and LI, Lf c::: D, is a subdomain whose boundary consists of a finite system I' of closed, regular curves 1'1' ... , 'Yn with a positive orientation w.r.t. LI (i.e. the index n('Y, z) = 1 for each z ELI). Suppose, moreover, that 1 is continuous and does not vanish on I' and maps I' onto a cycle r. According to the argument principle, the index n(r, 0) is equal to the difference between the number ~I zeros and the number 01 poles 011 in LI, zeros and poles being counted as many times as their order indicates. ' In particular, if 1 is analytic in D, we can determine the number of zeros off The number of zeros can be also evaluated by. using Roucbe's theorem: If J, g are analytic in D and continuous in its closure 15 and 'g(z) , < '/(z) , on the boundary of D, thenf+g and f have the same number of zeros in D. Again the zeros 'are counted wi.th due multiplicity.
3.9.1. If F is analytic in K(O; 1), continuous in K(O; 1) and ,F(z) , < 1 for all z E K(O; 1), show that the equation F(z)- z = 0 has a unique root in K(O; 1).
l.9. ARGUMENT PRINCIPLE. ROUCHE'S THEOREM
55
3.9.2. Show that the polynomial P(z) in the first quadrant (+; +).
=
z8+3z 3 + 7z+5 has exactly two roots
3.9.3. Show that the polynomial P(z) in each quadrant.
=
z4+2z3 -2z+IO has exactly one root
3.9.4. If a, b are real and different from zero, show that the polynomial P(z) z211+a2z2n-l+b2 has for even n exactly n roots of positive real part and n-I roots of positive real part for odd n. =
3.9.5. Show that P(z) quadrant.
=
z5+2z 3 +2z+3 has exactly one root in the first
3.9.6. Verify that all zeros of P(z) = z5- z +16 are contained in the annulus 1 < Izl < 2 and two of them have a positive real part. 3.9.7. Show that exactly two roots of P(z) in the right half-plane. 3.9.8. Show that the polynomial P(z) strip D = {z: 0 < imz < I}.
=
= z5+5z 3 +2z2+4z+1 are situated z4+3z+3 has a unique root in the
3.9.9. Show that the polynomial P(z) = z4+iz 3 +1 has a unique root in the first quadrant and four roots in the disk K(O; f). 3.9.10. Evaluate the number q of roots situated in the unit disk for the polynomials: (i) z4-5z+l; (ii) z8-4z 5+Z 2_1. 3.9.11. Ir" a K(O; I).
> e, show that the equation az" =
ez
has exactly n solutions in
3.9.12. Show that no roots of the equation (z+ I)e-~ in the right half-plane. 3.9.13. If A > I, show that the equation z+e-~ positive real part. 3.9.14. Prove that the polynomial I+z+az", n at least one root in R(O; 2).
=
z+2 are contained
= A has one solution with
~
2, has for any complex a
3.9.15. Suppose 0 < lal < 1 and p is a positive integer. Show that the equation (;=-I)P = ae- Z has exactly p simple roots with positive real part and all of these arc located inside K(I; I). 3.9.16. Prove that all roots of the equation tanz-z = 0 are real and each ill tcrval «n - -}) 'It, (n +~-) 'It), n =1= 0, contains exactly one root All'
3. COMPLEX INTEGRATION
S6
3.9.17. If
laml < 1
(m = 1,2, ... ,
n), Ibl < 1 and n
F(z)
=
IT l-a z-a z
m ,
in=1
show that the equation F(z)
=
m
b has exactly n roots in the unit disk.
3.9.18. If F is analytic in K(O; a) and continuous in .K(O; a), IF(z) I > m on C(O; a) and IF(O) I < m, show that F has at least one zero in K(O; a). 3.9.19. If I is analytic in the annulus A z E A and Yt =
=
{z: r
I( C(O; t)), show that for any t E
< Izi < R}, I(z)
=1=
a for all
(r, R) the index n(y" a) has
the same value. 3.9.20. If I is analytic in a domain D except for one simple pole
tinuous in l5"",zo, I/(z) I = every value a with lal > 1.
and con1 on D",D, show that I takes in D exactly once Zo
CHAPTER 4
Sequences and Series of .Analytic Functions 4.1. ALMOST UNIFORM CONVERGENCE
Following Saks and Zygmund [10] we shall call a sequence of functions {In} ddi'iied in an open set G almost uniformly (a.u.) convergent on G to alunction!, if {In} tends to I uniformly on each compact subset of G. In what follows w~ use the notation: In =l GI. If all functions!.. are analytic in a domain D and!.. =l'DI, then also lis analytic in D. Moreover, for any fixed, positive integer k, JC:) =l DJC k). We can also consider a.u. convergent series of functions. The so-called M-test of Weierstrass yields a quite convenient, sufficient condition for a.u. convergence of functional series. Let L Un be a series of functions defined on an open set G. If for each compact subset F of G there exists a convergent series L Mn with positive terms such that Iun(z) I ~ Mn for all z E F, then L Un is a.u. convergent on G. 4.1.1. Show that the sequence {zexp (_-}n 2 z 2 )} is uniformly convergent on the real axis and at the same time it is not a.u. convergent in any disk K(O; r). 4.1.2. Prove that the series co
00
n=1
n=1
l>n(Z) = I: zII[(1-zII)(1- zn+1)]-1 is a.u. convergent in K(O; 1) to z(1-Z)-2 and also a.u. convergent in C,,-K(O; 1) to (1-z)-2. co
4.1.3. Prove that the ~eries
l: 3-nsinnz
is a.u. convergent in the strip
n=O
lil11zl < log3, its sum I being analytic in this domain. Evaluate /,(0). co
4.1.4. If I is analytic in K(O; 1) and 1(0) = 0, show that the series l:/(z") is !l.U. convergent in K(O; 1) and its sum is analytic. n=1 57
4. SEQUENCES AND SERIES
58
4.1.5. Prove that in any closed disk K(zo; r) leaving outside all negative in00
tegers the series
L
•
(_1)n+1(z+n)-1 is uniformly convergent and its sum is ana-
n=1
lytic in C""{-l; -2; -3; ... }. 4.1.6. Suppose y is a closed, regular curve not meeting any negative integer and I is the analytic function of Exercise 4.1.5. Evaluate ~ I(z) dz. l'
4.1.7. If Izl
<
1 and .,;(n) is the number of divisors of the integer n, show that
I z n(1-z")-1 L .,;(n)z". 00
00
=
n=1
n=1
Prove the a.u. convergence of both series in K(O; 1). 4.1.8. Prove the identity 1+ (ktl)z+ (kt2)Z2+ ...
k = 0, 1, 2, ... , Izl
= (l-z)-k-l,
< 1.
4.1.9. Find the set of points of convergence of the functional series L unCz) in case un(z) equals to: (i) zn(1 +z2n)-1; (ii) n- 2cosnz; (iii) (z+n)-2; (iv) (q"z+q-n z -1-2)-1 (0 < q < 1). 4.1.10 .. Prove that the series
1+ ,-" 00
z2(z2+1 2) ... (Z2+ n2) ~ [(n+1)!j2
.n=O
S
convergent for any z, its sum being analytic in the finite plane. 4.1.11. By using the identity
n=-oo 00
z 1= 0, =F1, =f2, ... (cf. e.g. Ex. 4.5.15) find the sum
L
(z-n)-3.
"=-00
4.1.12. If the sequence of functions {In} analytic and univalent in a domain G is a.u. convergent in G and 1= lim/n , show that I is univalent, unless it is a constant. 4.2: POWER SERIES 00
A series of functions of the form
L
an(z-a)n is called a power series. The
n=O
least upper bound R of nonnegative r for which the sequence {Ianl rn} is bounded
4.2. POWER SERIES
59
is called the radius of convergence of the given power series and the disk K(a; R) is called the disk of convergence. In case R -> 0 the power series is a.u. convergent in its disk of convergence, its sum being analytic in this disk. Outside its disk of convergence, i.e. in C~K(a; R), the power series is divergent. The radius of convergence can be evaluated by means of the Cauchy-Hadamard formula: R = (lim VianO-I. 4.2.1. If an:l= 0 (n
=
1, 2, ... ) and the limit lim an/an+!
=q
exists, show
00
that the radius of convergence of the power series
2: anz"
is equal to Iql.
n=O
4.2.2. Evaluate the radius of convergence of: •
(1)
L 00
(ii)
n=O
I
.
z";
n=1
2: 2-nz 2n ; 00
(iii)
L :~ 00
(2n)1 (nl)2 z";
L (n+an)z". 00
(iV)
n=O
n=O
4.2.3. If the radii of convergence of L anzn, L bnzn are equal R 1 , R2 resp., show that: (i) the radius of convergence R of L anbnz" satisfies R ~ RI R 2; (ii) the radius of convergence R' of
(bn :1= 0, n = 0, 1, 2, ...)
, " ban z"
L
n
satisfies R' ~ R 1 /R 2; (iii) the radius of cohvergence Ro of
2: (anb +an O
_l
bl+ ... -t-aobn)z"
satisfies Ro ~ min(RI' R 2 ). 4.2.4. If the sum of the power series L an z" is real in some interval (- ~, " > 0, show that all coefficients an are real. 4.2.5. Suppose R > 0 is the radius of convergence of the power series
2~ ~
o
Hint:
1/12 =fJ.
00
If(re fl)l 2dO = i
L lanl r2n, 2
n=O
2: anz" n=O
ILnd fez) is its sum. Prove ParsevaI's identity: 21<
00
0 < r < R.
~),
01. SEQUENCES AND SERIES
60
4.2.6. (cont.) If 1 is bounded: I/(z) I ~ M for all z
E
K(O; R), show that
L lanl2R2n ~ M2. 00
n=O
4.2.7. (cont.) If r < R and M(r) = sup I/(re i f1) I, 0 ~ () ~ 27t, show that
lanl
~
,-nM(r),
4.2.8. If the sum of a power series disk, show that an = 0(1).
n
L anz"
= 0, 1, ... is defined and bounded in the unit
00
L anz' n=O
4.2.9. If I(z) =
univalent in K(O; r) (i.e.
00
in K(O; r) .and lall ~ Zl'
Z2
E
K(O; r),
Zl
L nlanlr n- l , n=2
show that 1 is
=/: Z2 implies I(Zl) =/: I(Z2»). 00
4.2.10. By using Exercise 4.2.9 find a disk of univalence of )-, z"1 • . "--J n n=O
4.2.11. If series
L cnz" has a positive radius of convergence R, show that the power
L ~i
z" has an infinite radius of convergence and the sum
1 of the
latter series satisfies
I/(z)1
~
M(()exp(lzll()R),
w~ere
0
< () < 1.
00
4.2.12. If I(z) =
L anr' is univalent in K(O; 1) and n=O
image domain of K(O; r), 0
A(r) is the area of the
< r < 1, show that
4.3. TAYLOR SERIES
If 1 is analytic in a domain D and a is an arbitrary point of D, there exists a sequence of complex numbers {an} such that for any K(a; r) c::: D the po~er 00
series
L an(z-a)n is a.u. convergent in K(a; r) its sum being equal/(z). The power
n=O 00
series
L an(z-a)n is called Taylor series 011 center at a and the
Taylor's coeffi-
n=O
cients an can be expressed either by successive derivatives of 1 at the point a:
ao=/(a),
an =
I(n)(a) I
n.
(n=I,2, ... ),
4.3. TAYLOR SERIES
61
or by Cauchy's coefficient formula: an =
2~i ~ (c~~jn+l dC
(n = 0,1,2, ... )
C(a:r)
where r is such that [((a; r) c D. The radius of convergence R of Taylor series center a is not less than the distance between c",-.n and the point a. On the other hand, the circle of con00
vergence C(a; R) of the power series
I
an(z-a)n contains at least one singular
n=O
point. Each point on C(a; R) which is not regular, is called singular. A point .
00
bE CCa; R) is called regular if there exists a power series
L
bn(z-b)n conver-
n=O 00
gent in some disk K(b; ~), ~ > 0, such that both series have identical sums in K(a; R) n K(b;
~).
L
00
an(z-a)n,
n=O
L
bn(z-b)n
n=O
4.3.1. Evaluate four initial, non-vanishing coefficients of Taylor series with
center at the origin for the following functions and find the corresponding radius of convergence R:
z
(i)
Log(1+z)
z
(ii)
Arctanz' (iii) .) (IV
]Icosz (take the branch corresponding to the value 1
1 at the origin);
•
cosz' (v) Log (1 +e); (vi) expe. 4.3.2. By applying Weierstrass theorem on term by term differentiation of II.U.
convergent series of analytic functions prove the following theorem: 00
Suppose
L
unCz) is a.u. convergent in K(O; R) and fez) is its sum; suppose,
n~O
lIloreover, ulI(z) =, anO+alllz+ ... +anlczk+ ... in K(O; R) for n = 0,1,2, ...
4. SEQUENCES AND SERIES
62 co
Then all the series Lank are convergent and their sums are equal to correspondn=O
ing Taylor coeffiCients of f at the origin. 4.3.3. Evaluate four initial, non-vanishing Taylor coefficients at the ongm for the following functions and find the corresponding radius of convergence R:--
z (")' Z (1.) exp -1--; II sm -1--'
-z
-z
4.3.4. Evaluate nth coefficient of Taylor series at the origin and its radius of convergence for following functions:
. 2"1(Log l-z 1)2 ;
(1)
(ii) (Arctanz)2; (iii) (Arctanz) Log(l+z2): (iv) cos 2z; (v) Log(1+Yl+z 2 ).
Hint: Differentiate the given 'function;
(vi) Log( Yl+z+ Yl-z). [Take in (v), (vi) these branches of the square root which are equal 1 at the origin.] . 4.3.5. Find the radius of convergence for Taylor series with center at the origin for sin 7tZ2 jsin 7tZ . 4.3.6. Find Taylor series with center at the origin for these branches of (i) (1+ J/l+Z)1/2;
(ii) (1+ yl+Z)-1/2
which are equal 21/2, 2- 1/2 at the origin. Hint: Cf. Exercise 1.1.5.
4.3.7. Show that the coefficients Cn of Taylor series of (1- z- Z2)-1 with center at the origin satisfy: Co = C1 = 1, cn+2 = cn+1 +cn (n ~ 0). Find the explicit formula for Cn by representing the given function as a sum of partial fractions. co
Also find the radius of convergence of
L
cnz". The sequence {cn} is so called
n=O
Fibonacci sequence.
4.3.8. Prove that fez) = ,_1_ - ~ + ~ has a removable singularity at the e Z -l z 2 origin and is an odd function. Hence
4.3. TAYLOR SERIES
63 co
_1_ _ ~ ~ e%-1 z+2
=
~ (_I)k-1 ~ 2k-1
L..J
(2k)!z.
k=1
Evaluate B 1 , B 2 , ... , Bs. Find the radius of convergence of the power series on the right and show that lim IBn I = + 00. The constants Bk are called Bernoulli numbers.
V
4.3.9. Show that co
1 hi T zcot T Z -
(1')
1+~ ( l)k-1 Bk
L..J -
(2k)! z
2k-1
,
k= 1
co
(1'1') zcotz
=
22k
1- "'-, Bk z 2k, Iz I < ~ (2k)!
7t.
k=1
4.3.10. Express by means of Bernoulli numbers the nth Taylor coefficient at the origin and the radius of convergence of the corresponding Taylor series for the following functions:
sinz (1.) Log--:
z
(ii) Logcosz; (iii) tanz; (iv) (COSZ)-2; (v) tan 2 z;
tanz (VI.) L og--; z
.. ) ( VB
Z
-.-.
smz
+
4.3.11. By using the power series expansions of Log(1 z), sinz near the origin evaluate three initial, non-vanishing terms of the power series expansion of
Log sinz and compare the obtained result with Exercise 4.3.10 (i). ;>
4.3.12. Verify that
q;(z)
= -. /
V
z Arctan'" / z l-z l-z
is analytic in some neighborhood of z of its Taylor series at the origin.
V
=
O. Evaluate the radius of convergence
4. SEQUENCES AND SERIES
64
4.3.13. (cont.) Verify that rp satisfies the differential equation
2z(l-z)rp'(z)
with the initial condition rp(O)
=
o.
=
rp(z)+z
Prove the identity:
_ 2 3 2·4 5 2·4·6 7 rp(z) - z+3 z +TI z +T5.'7z +
... ,
Izl < 1.
4.3.14. Prove that . 2 _ 2 1 2 4 1 2.4 6 (Arcsmz) - z +"2·3 z +3·TIz +
... ,
Izl
< 1.
4.3.15. If f is analytic in K(O; R) and Rn is the remainder of Taylor series center at the origin, i.e.
Rn(z)
=
z"
f(z)- f(O)-z!, (0)- ... - - , j
n.
show that Izl
< r < R.
Hint: Cf. Exercise 3.1.28. 4.3.16. If f is analytic in K(O; R) and
sn(z)
=
f(O) +z!' (0) + ...
+ n.z~ j
show that Izl < r < R. 4.3.17. If f is analytic in K(a; R), show that
fez)
=
f(a)+2{~(Z-a)!,(z+a) + (z-a)3 j",(z+a) + 3 2
2
2 .3!
+ (~~~)!5
2
j<5)( z~a)+ .. }
z E K(a; R).
4.3.18. If f is analytic in K(O; R) and f(re i9 ) = P«())+iQ«()), 0 show that the coefficients an of Taylor series at the origin satisfy 2n:
an
=
_1_ \' P«())e-in9d() n:rn j o
27<
=
_1_ \' iQ«())e-in9d(). n:rn j
0
< r < R,
4.4. BOUNDARY BEHAVIOR OF POWER SERIES
65
4.3.19. If J is analytic in K(O; 1),/(0) = 1 and reJ(z) > 0 in K(O; 1), show that Taylor coefficients an of J at the origin satisfy: lanl ~ 2, n = 1, 2, ... 4.3.20. (cont.) Verify that
(1-lzi)/(l+lzi)
~
IJ(z) I ~ (l+lzi)/(I-lzi)
and show that equality can actually be attained. 4A. BOUNDARY BEHAVIOR OF POWER SERIES
4.4.1. Discuss the behavior of the following power series on the circle of convergence 00
(i) Lzn; n=1
00
(iv) _~ (-1)" L...J logn
z3n-1
.
"=2
4.4.2. Give an example of a divergent series
vergence of
L anz"
L an
such that the radius of con00
is equal to I and the finite limit lim
L
anr" exists.
' .... 1- n=O
4.4.3. If limzn = eiIJ, IZnl
< I (n = 1,2, ... ) and all Zn are situated inside a Stolz angle with vertex eiIJ, i.e. an angle made up by two chords of the unit disk with a common vertex eiIJ, show that. the sequence Prove the converse, too.
l'~i~~~i'} is
bounded.
00
4.4.4. Prove Abel's Limit Theorem: If
L ane1n6
is convergent and {zn} is
n=O
!lny sequence approaching ei6 inside a Stolz angle with vertex at ei6 , then
Hint: If Sn = aO+a 1 ei6 + ... +aneinIJ , show that
00
L Ski'
is convergent in
k=O 00
1\·(0; 1) and express
L k=O
ak~
as a Toeplitz transform of {sn}, cf. Exercise 1.1.37.
4. SEQUENCES AND SERIES
66
4.4.5. Prove that (i) sinO-tsin20+tsin30- ... = to, 101 < 1t"; (ii) cosO-tcos20+tcos30- ... = log(2costO), 01= (2k+l)1t" with k being an integer.
4.4.6. Prove that (i) sinO+t sin30+i- sin50+ ... = t1t", 0
< 0 < 1t";
(ii) cosO+tcos30+i-cos50+ ... = tlog(cottO) for 01= k1t" with k being an integer.
4.4.7. Prove that
2·4 2.4.6 1 .r 2 1- 3 +3.5-"3.5.7 + ... = y'210g(l+ .,2). Hint: Cf. Exercise 4.3.13. 4.4.8. Prove that
1 2.4·6 [{ ./;;;"\]2 1 2 1 2.4 1- 2 . 3 + 3 , 3.5 -4·"3-:-0+=··· log\.I+.,2, . Hint: Cf. Exercise 4.3.14. 4.4.9. Prove that
(1.)
Jl'2)1/2.
1 1. 3 . 5 _ ( 1+ 1+ 2 .4-2.4.6.8+ ... 2
'
(ii) !_~+ 1·3·5·7 _ ... = [2(I+V2)r1 / 2 • 2 2.4·6 2.4·6.8·10 Hint: Cf. Exercise 4.3.6. 4.4.10. Prove that
Hint: Consider Arsinhz. 4.5. THE LAURENT SERIES 00
A functional series of the form
L
An(z-a)n is called the Laurent series
n=-oo
with the center a. If the Laurent series is convergent for some zo, then its regular " 00
part, i.e. the power series
2: An(z-a)n is convergent in K(a;
n=O
Izo-a\). whereas
4.5. THE LAURENT SERIES
67 00
its singular or principal part, i.e. the senes
L: A_n (z-a)-n,
is convergent in
n=l
C"'K(a; Izo-aj) which is an immediate consequence of well-known properties of power series. Hence,. i.f the set of points of convergence contains interior points, there exists a maximal, non-empty annulus of convergence {z: r < Iz-al < R} and the sum of Laurent series is analytic in this annulus. Conversely, if f is analytic in {z: r < Iz-al < R}, where 0 ::( r < R ::( +00, then it can be expressed 00
L:
as the sum of a Laurent series
An(z-a)n which is absolutely and a.u. con-
"=-00
vergent in this annulus. The development is unique. If C = C(a; p) and r < p < R, then the coefficients An of the Laurent development are equal: An=
2~i ~f(c)(,-a)-n-ldC
(n = 0, =fl, =f2, ... ).
c
The Laurent development also holds in case r = 0, i.e. when a is an isolated singularity. Since any Laurent series with a non-empty annulus of convergence and A-l = 0 has a primitive, it follows that res(a;f) = A_i' 4.5.1. Give an example of a Laurent series with an empty annulus of convergence and a non-empty set of points of convergence. 00
L
4.5.2. Show that
anz~
has a non-empty annulus of convergence, iff
4.5.3. Find the Laurent development of (z2-l)j(z+2)(z+3) in: (i) {z: 2
< Izl <
3}; (ii) {z: Izl
>
3}.
4.5.4. Show that the Laurent coefficients An of
z-a (b-a) Log z-b (
-1 )
in
C", R(} (a+b);}lb-al)
vanish for n ;;:;:: 2. Find A 1 , A o, A-1 • 4.5.5. Find the Laurent development of (z-a)-1(z-b)-1, 0 < lal (i) lal < Izl < Ibl; (ii) Izl > b.
< Ibl,
4.5.6. Find the Laurent development of (1 + Z2)-1 (2+ Z2)-1 for (i) 1 < Izl
< V2; (ii) Izl > V2.
4.5.7. Find the Laurent expansion of Log [Z2 j(z2-l)] for Izl
>
1.
for
4. SEQUENCES AND SERIES
68
4.5.8. Express the Laurent coefficients of exp(z+z-l) at the origin in terms of (i) integrals involving trigonometric functions; (ii) sums of infinite series by using the identity exp(z+z-l)
=
e Z e l/z •
4.5.9. The Bessel function In(z) is defined as the nth coefficient of Laurent expansion at the origin (n ~ 0): <1:)
expCtz(C-C- 1))
2::
=
In (z)Cn.
n=-oo
Prove that ~
In(z)
=
. -; J cos(nO-zsm(J)d(J
00
~
1 \'
=
o
6
(-l)k(1_z )2k+n k!(n~k)!
k=O
4.5.10. Suppose f has only the following singularities in the extended plane: a simple pole at -1 with res( -1; f) = 1, a double pole at 2 with res(2; f) = 2; moraover,/(O) = -t,/(I) = f. Find the Laurent development offin the annulus 1 < Izl < 2.
4.5.n. Find the Laurent development of z-l(l-z)-l in annular neighborhoods of: (i) z = 0; (ii) z = 1; (iii) z = 00. 4.5.12. Find the Laurent development of f(z)
nulus 1 < Izl
= [
(z-1 )z(z_ 2) ] 1/2 in the an-
< 2 assuming imf(t) > O.
4.5.13. Find the Laurent development of both branches of
[(z-a)(z-b)p/2
Izl > max(lal, Ibl).
in
<1:)
4.5.14. Show that TC 2 /sin 2 TCz and h(z)
=
2::
(z-nr 2 are both analytic in
n=-oo
C""-N, where N is the set of all integers. Also show that the principal parts of both functions in the annular neighborhood of any point of N are the same.
4.5.15. (cont.) Show that g(z)
deduce that g
=
TC 2 sm 2 TCz
= - ; - - - -h(z)
is bounded in C. Hence
O.
4.5.16. Show that
L <1:)
TCcotTCZ
=
1 2z 2 - + 2 z-n Z . n=l
=~+ ~'(_1 +~) z 6 z-n n n=-oo
for z i= 0, =fl, =f2, ... (here and in the following a prime after a summation sign indicates that the term corresponding to n = 0 is omitted).
4.6. SUMMATION OF SERIES
69
4.5.17. Find the Laurent expansion of 7t cot 7tZ in the annuli: (i) 0 < Izl < 1; (ii) 1 < Izl < 2. Express the Laurent coefficients by means of <1:)
Sk = Ln- 2k
(k
= 1,2, ... ).
n=1
By using Exercise 4.3.9 (ii) find the relation between Sk and Bernoulli numbers B k . 4.6. SUMMATION OF SERIES BY MEANS OF CONTOUR INTEGRATION
4.6.1. Suppose J is meromorphic in the finite plane C and has a finite number of poles a1 , a2 , ••• , am none of which is an integer. Suppose, moreover, that N
lim zJ(z) z~oo
=
L
O. Show that the limit lim
J(n) exists and equals to
N~+oon=_N
m
- L
res(ak; 7tJ(z)cot7tz).
k=1
Hint: Integrate 7if(z) cot 7tZ over the boundary (}QN of the square QN with corners (N+t)(=fl=fi); also cf. Exercise 2.7.3. 4.6.2. Verify that
~
L..J
S=
2
(n +n+l)
-1
=
n=-oo
27t 7t y'3 3tanh-2-'
j/
4.6.3. Show that for all a outside an exceptional set (which should be determined in each case) the following formulas hold: (i)
f
(n 2 +a 2 )-1
=
~ ( : coth7ta- :2 );
n=1
1
1).
sinh tta y'I +sin tta y'i" - 2 a 3 y'i" cosh tta y' 2-cos7ta y'2 a4 '
.. ) 2:<1:) (4 _ - (7t ( II n +a4)-1 ---. n=1
... ) 2:<1:) ( III n=1
.
n2 4 n +a 4
_
7t 2ay2
- .
<1:)
(iv) 2 : (n4-a 4r1 na 1
=
2~4 -
J/i" .,
sinh tta J/I- sin tta cosh7tav'2-cos7tay2
4:
3
(cot7ta+coth7ta).
4. SEQUENCES AND SERIES
70
4.6.4. If a is not an integer, show that
L 00
(n-a)-2
'=
1t 2/sin 21ta.
"=-00
4.6.5. If none of a, b is an integer, show that
L( ) 00
n-a _1(n- b)-l -- -1t 2 cot1ta-cot1tb . 1ta-1tb
00
L
4.6.6. Evaluate --
(n 2
+1)-1.
"=1
4.6.7. Suppose J is meromorphic in the finite plane C and has a finite number of poles a 1 , a2, ... , am none of which is an integer. Suppose, moreover, that lim zJ(z) = 0. Show that the limit z-+oo
exists and is equal to m
- Lres[ak; 7tj(z)/sin1tz]. k=l Hint: Integrate 7tj(z)/sin 1tZ over the boundary aQN of the square QN with corners (N+t) (=fl=fi).
4.6.8. Prove that
.L 00
(1)
"=0
(-1)" n a
1 a
~+ 2 =-2 2
00
(ii) ~
(_1)"
~ n4-a 4 "=1
a -# =fn, =fin (n
>. 00
(iii)
..:.......J. "=-00
=
=
. + 2asm.1th 1ta·,a-# 0, =fl,.=f2z, ... ;
.
_1 _~(_1_ 2a 4 4a 3 sin1ta
+
1)
sinh1ta '
0,1,2, ... );
. 1ta 1ta 1ta. 1ta sm . /_ cosh • /_ + cos • /_ smh • /_ (-1)" _ _ 1t_ v2 v2 v2 v2 n 4+a 4 - a 3 2 1ta + .nh2 1ta sm SI
y2 .
.
y2
y2
4.6. SUMMATION OF SERIES
71
tta
7ta.
L 00
(iv)
n4
n=-oo
a #-
+a4
cos
7t
(-1)"n 2 =
ay2 .
• 2
sm
=f1 =fi 2 n (n
=
. tta
7ta
172 smh l72 ~ sm l72 cosh y2 . h2 7ta
tta
y2
y'2+ sm
0, 1, 2, ... ) in (iii) and (iv).
7t sin az 3. , where -7t < a < 7t, round the boundary sm7tZ i)QN of the square QN with corners (N+}) (=f1=fi), verify the formulas:
· . 4•6•• 9 By mtegratmg
Z
4.6.10. If x is a complex number different from an integer, -7t QN is the square with corners (N +-t) (=f 1=fi), show that
I
= N
rJ
cos az
(X 2 _Z 2 ) sin 7tZ
dz
-+
0
N
as
-+
<
a
< 7t and
+ 00 .
iJQN
Verify the formula: 00
7tcosax sin7tx
=
~+ 2x ~ (-1)" C~S7t~ . ~
x
x -n
11=1
=f~) ,
4.6.11. If a#-O is real and RN are rectangles with corners (N +-t )(=f 1 show that the integrals
IN
=
~
sin 7tZ
s~nh 7taz dz -+ 0
N -+
as
a
+00.
iJRN
By using this show that
L 00
(·1)
(-l)"n _- - -~ .1_ _ _ 1 2
sinh 7tan
27ta
a
"=1
( .1.1)
1 2 eTt_e-Tt - e 27t _e- 27t
L 00
(-l)"n
sinh(7tn/a)
for real a#-O;
"=1
+e
37t
3 -e
1
37t
•••
=
s;-.
4.6.12. Verify that the equality of Exercise 4.6.11 (i) also holds for complex " outside the imaginary axis.
4. SEQUENCES AND SERIES
72
4.6.13. If x is complex and not an integer, -7t < a < 7t, and QN is the square with comers (N+i- )(=f 1=fi), show that the integrals I
~
-
N -
zsinaz dz~ 0 (x 2 -z 2 )sin7tz
+00
N~
as
ilQN
and verify the formula •
co
•
~. s~nax = ~ (_I)n nsman .
2
L,,=1
sm 7tX
x2-n 2
4.6.14. Show that co
.
7t _I(-I)n(2n+l) (1) -(n + T1 )2 -Z 2' COS7tZ
1
Z
3
5
•
1= =fT' =fT' =fT' ... ,
n=O
I
co
(11oo)
7t cosh 7tZ
=
n=O
(-I)n(2n+l) Z2+ (n+T1)2'
l'
Z
3'
5'
1= =fTI, =fTI, =fTI, ...
4.6.15. By integrating fez) = 7tZ- 7 cot7tzcoth 7tZ round a suitable contour show that 197t7 coth7t coth27t coth37t -1-7-+~7-+ 3 7 + oo. = 56700
4.7. INTEGRALS CONTAINING A COMPLEX PARAMETER THE GAMMA FUNCTION
Suppose W(z, C) is a complex-valued function of two complex variables z, C defined and continuous on Gx {r}, where G is a domain and {r} is the set of points of a regular curve r. If W;(?, C) exists and is continuous on G X {r}, then H(z) = ~ W(z, C)dC is analytic in G; moreover, H'(z) = ~ W;(z, OdC. r
In particular
r
r
can be a segment [a, b] of the real axis. If the limit +co
lim
~ W(z, t)dt exists, it will b~ denoted ~ W(z, t)dt.
b->co [a,b]
a
+co
4.7.1. The integral ~ W(z, t)dt is said to be almost uniformly (a.u.) cona
vergent in a domain G, if for any compact subset F c G~nd any exists A such that for any b, B with A < b < B we have
B
> 0 there
73
4.7. THE GAMMA FUNCTION B
I~ W(Z, t)dt! < e
for all z in F.
b
Ifboth W(z, t), W;(z, t) are continuous on Gx [a, +(0) and H(z) =
+00
ia W(z, t)dt
is a.u. convergent in G, show that H is analytic in G. Hint: Use Weierstrass theorem on a.u. convergent series of analytic functions. +00
4.7.2. Prove that ~ e-Iltdt is a.u. convergent
m
the
right
half-plane
o
1
+ 00
{z:rez>O}and ~ e-ztdt--=Ofor all z with rez>O. By separating o z
real and imaginary parts in both terms find the values of two real integrals. +00
4.7.3. Prove that H(z) = ~ e-'t,,-ldt is analytic in the open plane and 1
+00
H'(O)
=
~ t- 1 e-:,}dt.
1
4.7.4. If +00
and
l
~
is a complex-valued function of a real variable x
1~(x)ldx
<
+00,
E ( - 00,
+ (0)
show that
-00 +00
~ (x-z)-l~(x)dx
J(z) =
-00
is analytic in the upper half-plane. 4.7.5. The gamma function as .defined by the equality +00
r(z)= ~
(Z-l
e-'dt,
rez>O,
o
ill analytic in the right half-plane and satisfies zr(z) = r(z + 1). Show that this functional equation defines a function meromorphic in C whose only singularilies are simple poles at z=O,-I,-2, ... with res(-n;r(z) = (-I)R/n ! 4.7.6. Prove that a function G meromorphic in the finite plane C satisfies = zG(z), iff G(z) = r(z)p(z), where P is meroIIwrphic in C and has the period 1. Ihe functional equation G(z+l)
4.7.7. Prove that +00
~ exp( -x")4x = oc- 1r(oc- 1 ),
oc >
o.
o
4.7.8. By integrating e-"zS-l round the boundary of D(~. R)
= {z:
~
< Izl <
R, 0
<
argz
<
n/2},
4. SEQUENCES AND SERIES
74
show that +00
~ y,-le- iY dy
• =
F(s)exp(-i-7tis),
0
<
res
< 1.
o
4.7.9. If 0
<
oc
<
1, show that
+00
(i)
~ x-a.cos x dx
=
r(1-oc)sini-oc7t;
o +00
(ii)
~ x-a.sinxdx = r(l-oc)cos-}OC7t. o
Also verify the left-hand side continuity of the integral (ii) at oc
=
1.
4.7.10. Verify that "'/2
~
sinP- 10cosq- 10dO
o
1 r(i-p)F(i-q) = -2 . ----.o~_;_~~ r(i-(p
+q))
4.7.11. Show that "'/2
(i) ~ (tanO)"'dO
= i-7t(cos-}7toc)-l,
o
-1
<
oc
<
1;
'
"'/2
(ii)
~ ysinOdO = (27t)3 / 2[F(i-)r 2.
o
Hint: Use the formula F(z)F(I-z) = 7t/sin7tz. 4.7.12. Express the following elliptic integrals by means of gamma function: "'/2
(i) ~ (l-tsin20)-1/2dO; o "'/0
(ii) ~ (1-i-sin 20) 1/2dO. o
Hint: Put sinO
=
V1- Yu.
4.7.13. Prove that r{! )F{!} ... r( n:l )
Hint: Show that n = n-1
k=1
n-1 (
k=l
k
I1 sin ~n = n . 2
IT
1- n•
= n- 1/2 (27t)(n-1)/2.
2k .) 1- exp ~ and hence deduce the formula n
4.8. NORMAL FAMILIES
75
4.7.14. Prove that 1
~ logr(t)dt
=
tlog27t.
o
4.7.15. Evaluate the integral a+ 1
/(a)= ~ logr(t)dt,
a>O.
a
4.8. NORMAL FAMILIES
A family ff of functions I analytic in a domain D is said to be normal, if every sequence {J,,} of functions In E ff contains a subsequence {Ink} which either converges a.u. in D, or diverges to 00 a.u. in D. The limit function I = lim Ink ,
k~oo
is analytic, unless it reduces to the con~tant 00. If any sequence {[,.} of functions fn E ff contains an a.u. convergent subsequence, then ff is said to be compact. A necessary and sufficient condition for compactness of the family ff is the existence of a common, finite upper bound of the absolute values of all IE ff on each compact subset of D (Stieltjes-Osgood theorem, also called Montel's compactness condition). Clearly each compact family is normal. The real-valued !'unction p(z,f) = 2 If'(z) 1(1+I/(zW)-l is called the spherical derivative of I and has an obvious geometrical meaning (cf. Ex. 4.8.1). Now, a family ff of functions I analytic in a domain D is normal, if there exists in every compact ~ubset of D a common finite upper bound for the spherical rlerivative of all !'unctions of the family. This criterion is due to F. Marty. Another sufficient condition for normality is due to Monte! and its proof is based on the properties of the modular function: if all functions I of a family :iF are analytic in a domain D and every IE ff does not take in D two fixed, lillite values ct, {3 (ct =f. {3), then I is normal. The concept of normal family is very important in the existence questions for solutions of extremal problems. 4.8.1. Explain the geometrical meaning of the spherical derivative. 4.8.2. Verify that the family of all similarity transformations az+b is not II
normal family in the finite plane.
4.8.3. If ff is normal in a domain D and there exists a point Zo E D and a real I"llllstant Mo < + 00 such that I/(zo)1 ~ Mo for all I E ff, show that ff is a complld family. ..
4. SEQUENCES AND SERIES
76
4.8.4. If ff is a compact family of functions analytic in a domain D, show
that also ffl = {g: g = 1', IE ff} is compact. By considering the sequence {n(z2 _n 2)} verify that the derivatives of functions of a normal family not necessarily form a normal family. 4.8.5. Suppose F(w) is an entire function (i.e. a function analytic in the finite plane C) and ff is a compact family of functions analytic in a domain D. Verify that the functions F 0 I, IE ff also form a compact family in D. 4.8.6. Suppose ff is the family of all functions I(z) = az, where a is a complex constant and F(w) = eWsinw. Verify that ff is normal in C""K(O; 1), whereas the functions Fol do not form a normal family. 4.8.7. Prove Hurwitz's theorem: If {In} is an a.u. convergent sequence in a domain D and/n(z) 1= 0 for all ZED and all In, then the limiting function I is either identically 0 in D, or does not vanish in D. Hint: Verify that p(z,J) == p(z, Iff). Also prove that the terms of an a. u. sequence form a normal family. 4.8.8. Suppose {In} is an a.u. convergent sequence of univalent functions in a domain D and g = limfn. Show that either g is univalent in D, or is a constant. Hint: Consider I(z)-/(a) in D~a. 4.8.9. Show that the family of all functions I analytic in the unit disk and such that 1(0) = 0,1'(0) = 1, is not a normal family. 4.8.10. Let To be the family of all functions I analytic in the unit disk and such that 1(0) = 0,1'(0) = 1, I(z) 1= 0 for z 1= o. Show that To is compact. Hint: Consider log(f(z)/z). 4.8.11. (cont.) Show that there exists a constant ct > 0 such that any I E To takes in the unit disk any value Wo E K(O; ct). Hint: If In does not take the value ctn and ctn ~ 0, consider the sequence gn(z) = log (1-/n(z)/ct n). 4.8.12. Show that the family So of ail functions analytic and univalent in the domain D omitting one fixed value ct is normal. 4.8.13. Suppose G(M) is the family of functions analytic in a domain D and
such that ~ ~ If(z)1 2dxdy ~ M. Show that G(M) is compact in D. D
Hint: If K(zo; R) cD, verify that 2",
If(zoW
~
217':
~
o and deduce that 7':R21/(zo)12 ~ M.
If(zo+re i6WdO,
0< r
~ R,
CHAPTER 5
Meromorphic and Entire Functions 5.1. MITTAG-LEFFLER'S THEOREM
Let {a"} be an arbitrary sequence of complex numbers such that ao = 0 = 00 and let {G"(w)} be an arbitrary sequence of polynomials with vanishing constant terms. There exists a meromorphic function F which is analytic in the finite plane except for the poles ao, a1' a 2, ... and has G"(I/(z-a")) as singular parts at a".
< lall < la 2 1< ... and lima"
00
Let
L Un be an arbitrary, convergent series with positive terms and let {H"(z)}
"=1
be a sequence of polynomials such that
IG" (z 1aJ -
H"(z)
I~
U"
(n
=
1,2, ... )
for z eK(O; r"), where rn < la"1 and limr" = +00. In particular H" can be a suitable partial sum of Taylor's expansion of the singular part which is analytic In the open disk K(O; la"D. Then the function
where H is analytic in the finite plane (i.e. H is an entire function), has all the required properties. This representation due to Mittag-Leffler enables us to lind a meromorphic function with given .poles and given singular parts at these poles up to an entire function. 5.1.1. Find the most general meromorphic function F whose only singularilics are: (i) poles 1, 2, 3, ... of first order with res(n; F)
= n;
a" of first order (lal > 1, n = 1, 2, ... ) with res (a" ; F) = a"; (iii) poles l/n of first order (n = 0, ],2, ... ) with residues 1; '
77
5. MEROMORPHIC AND ENTIRE FUNCTIONS
78
(iv) poles 1, 2, 3, ... of second order with singular parts n(z-n)-2; (v) poles 1,2,3, ... of second order with singular parts n2(z-n)-2+
+(z-n)-l; (vi) poles =fI, =f2, =f3, ... with res(n; F) = Inl; (vii) poles 0, =fI, =f2, ... with residues 1; (viii) poles 0, -1, - 2, ... with residues 1. 5.1.2. Find the most general meromorphic function F whose only singularities are simple poles -1, -2, -3, ... with res(-n; F) = (_1)". 5.1.3. Find the most general meromorphic function F whose only singularities are simple poles w = m+ni (m, n = 0, 1,2, ... ) with residues 1. 1
5.1.4. Show that the function ~ e- t t Z - 1 dt is analytic for rez
> 0 and has the
o
expansion
Hint: Use the power series expansion of e- t • 5.1.5. Find the partial fractions expansion of r(z). 5.2. PARTIAL FRACTIONS EXPANSIONS OF MEROMORPHIC FUNCTIONS
5.2.1. Suppose 1 is a meromorphic function whose. only singularities are simple poles at> a2, a 3 , .. • with 0 < lall ~ la 21 ~ la 3 1 ~ ... , liman = 00 and res (an ; f) = An. Suppose there exists a sequence {Cn } of contours such that 10 Cn omits all the poles aj; 20 each Cn lies inside Cn + 1 ;
30 min Izl = Rn
--+
+00 as n
--+
+00;
ZECn
•
40 the length Ln of C~ is O(Rn); 50 max I/(z)1 = o(Rn). By considering the integral ~ C-l(C-Z)-l/(C)dC show that Cn
I(z) =/(0)+
~ An (_1_ +_1), L..... z-a an n=1
n
where the terms of the series are formed of all summands corresponding lo the poles situated between Cn and Cn + 1 •
5.2. EXPANSIONS OF MEROMORPHIC FUNCTIONS
79
5.2.2. Find the expansion off into partial fractions in case f satisfies the conditions 1°_5° and has a pole Z = 0 with a singular part G(I/z). 5.2.3. Derive the following expansions: Z
_'It_
=
2 ~ (-1) (n+ 2 ) ~
COS7tZ
(n+1.-)2-z2 '
n=O
0, =fl, =f2, ... ;
1
n
00
(ii)
=1=
Z =1=
=f-h =ft, =ft, ...
2
L [(n+t)2-z2rl, 00
(iii) 'lttan'ltz = 2z
Z =1=
=f+, =ft, =ft, ...
n=O
5.2.4. If
(X
=1=
0,
fJ /
(X
=1=
=f 1, =f 2, ... , show that
Also show that 1
1
T:2 + 4·5
1
'It
+ 7·8 + ... =
3 y3
.
5.2.5. Find the expansions of hyperbolic functions analogous to the expansions in Exercise 5.2.3. 5.2~6.
Find the expansion of (sinzsinh Z)-l into partial fractions .
. 5.2.7. Show that
for z
=1=
n'lt(=fl =fi), n
0, 1, 2, ...
=
5.2.8. If An are positive roots of the equation tanz that
=-+ 2:
=
z (cf. Ex. 3.9.16), show
00
.
zsinz
3
SlllZ-ZCOSZ
z
n=1
2z z2- A;
5.2.9. By considering the integral
r
J
clan
sinC
C(C-z)cos2C dC,
.
5. MEROMORPHIC AND ENTIRE FUNCTIONS
80
where aQn is the square with vertices n7t(=f l, =fi), show that
I
00
sinz -2cos z 5.2.10. If 0
=
r
(-l)n[z-(n+-i-) 7t
2•
"=-00
< a < l, show that
5.3. JENSEN'S FORMULA. NEVANLlNNA'S CHARACTERISTIC
5.3.1. Iffis meromorphic in K(O; R) and has m zeros a1, a2' ... , am (0 < la 1 :(; la21:(; ... :(; lam!) and n poles b1, b2, ... , bn (0 < Ib 11:(; Ib 21:(; ... :(; Ibn!) in the closed disk K(O; r), 0 < r < R, show that the following Jensen's formula holds: 21t
1 ~.
2
7t
0
loglf(re,6)ldO 2
=
rm rn 10glf(0)I+log-1- - - I -log-Ib b -b-I' a1a2 ... am 1 2'" n
-
Hint: The factor r( - az) removes a zero z = a of the function f and does not r z-a .
change If(re i6)1. 5.3.2. Show that Jensen's formula is valid also in case there are some zeros, or poles of f situated on C(O; r). 5.3.3. Write Jensen's formula for f having a zero, or pole of order A at the origin. 5.3.4. Evaluate the integral 21t
1 ('.
= ~ ~
loglf(rei6 )ldO
o
as a function of r for: (i) fez) = z3+2z+3; (ii) fez) = sinz; (iii) fez) = cotz. 5.3.5. Let f be meromorphic in K(O; R), 0
log+x
=
{
logx 0
for for
< R :(; +
00 •
x
~
1,
0
~;;
x :_~ 1 ;
We define:
5.3. JENSEN'S FORMULA. NEVANLlNNA'S CHARACTERISTIC
2°
n(r, f)
(0
<
81
= the number of poles of I in K(O; r) taken with due multiplicity
r ~ R); 2",
m(r,f) = ;7t
3°
~
log+/f(re i9)/dO;
o r
T(r,f) = m(r,f)+
4°
~
n(t;/) dt.
o
Show that Jensen's formula can be written in the form: T(r,f)-T(r,I/I) = log/I(O)/
under the assumption that I is analytic and :f:. 0 at z = O. T(r, f) is called N evanlinna's characteristic of f It plays a fundamental role in the theory of meromorphic functions. 5.3.6. Let I be meromorphic in K(O; R). Find a necessary and sufficient condition involving the distribution of zeros and poles that the integral 2",
~ log/f(re'9 )/dO o
has a constant value. 5.3.7. Determine n(r,f), nCr, l/f) for I(z) = z-ltanz. 5.3.8. Determine T(r,f) for I(z)
=
eZ •
5.3.9. If P(z) = az"+ ... +ao is a polynomial of degree show that as 5.3.10. If I(z)
=
71,
fez) = expP(z),
r-++oo.
exp{7ti/(l-z)}, z eK(O; I), show that I I+r T(r,f) = -2 log-I-' 7t -r
5.3.11. If I is analytic in K(O; R), show that 2",
tJ>(r) =
2~ ~
log/l(re i9 )/dO
o ill II
convex and incr'easing function of logr.
..
82
5. MEROMORPHIC AND ENTIRE FUNCTIONS
5.3.12. Suppose / is analytic and bounded in the unit disk, /(0) =f. 0 and fl(r) = n(r, I/f) is the number of zeros of / in K(O; r). Show that lim fl(r)logr
O.
=
r-+ 1-
5.3.:!.3. Let/be an analytic function which is bounded in the unit disk and has zeros a1 , a2, a3' ... , where 0 < la 11~ la21 ~ la 3 1~ ... and each zero is written as many times as its multiplicity. Show that (i) lim Ia1a2 ••• an I exists and is =f. 0; 00
L
(ii) the series
(I-lanD is convergent.
,,=1
5.3.14. Suppose {an} is a
of complex numbers such that lanl < 1
s~quence
00
= 1,2, ... ),0 < la 11~ la21
(n
~
... and
L (I-lanD = + 00. Iff, g are analytic ,,=1
and bounded in the unit disk and /(a n ) = g(a n ), n = 1, 2, ... , show that / = g. 5.4. INFINITE PRODUCTS
n Pn is said to be convergent if at most 00
An infinite product P1P2 ... Pn ... =
n=1
'
a finite number of the factors Pn are zero and if the sequence of products of the non-zero factors converges to a finite non-zero limit. Thus we have for a convergent infinite product limpn = 1. For this reason
n (I+an). The infinite product n=1 00
we usually write an infinite product in the form
n (I+a n), 00
I+a n =f. 0, is convergent and divergent together with the series
n=l 00
L n=l
00
is the convergence of the series
n=l
•
L
n (1 +an) 00
Log(1 +an) . A sufficient condition for convergence of the product
lanl·
n=l
If {un(z)} is a sequence of functions which are analytic in a domain G and 00
Iun(z)I
~ An for every
Z E
G with LAn < n=1
+
n [1 +u/z)] 00
C(),
then the product
n=1
is convergent in G and represents a function which is analytic in G. 00
5.4.1. If
L IU n=1
nl2
< +00, show that the necessary and sufficient condition for
convergence of the product
n (I +un) is the convergence orthe series L 00
00
"=1
"~I
U".
5.4. INFINITE PRODUCTS
83 00
5.4.2. If
Un =
L
(_I)nn-l/2, show that the series
the product
Un
is convergent, whereas
n=1
00
IT (1 +
Un)
is divergent.
n=1
5.4.3. Show that
5.4.4. If Izl
00
<
1, show that the product
IT (1 +
n
Z2 )
is convergent to (1- z) -1 •
n=O
5.4.5. Determine the domains of convergence of the infinite products (i)
(ii)
ij [1+( ~rl ij[l+(l+!(zn} n;
... ) IToo cos z (III n=1
(iv)
IT (1 + -fn) +
e- 1 / n •
11=1
5.4.6. Suppose {an} is a sequence of complex numbers such that: lanl
<
1
00
iLnd an i= 0 (n
=
1,2, ... ), am i= an for m i= n,
L
(1-lan I2) < +00. Show
n=1 00
that the product
IT z-a Z-~n1 n=1
is convergent in the unit disk and represents
n
('unction F analytic in this disk, equal to zero only at z hounded by 1 in absolute value.
II
~.4. 7.
Show that the product
il (1 +-~)
=
an (n
=
I, 2, ... ) and
e- z / n is convergent in the whole
plalle and that its limit is an entire function. /lint: Prove that 1(I-z)eZ-11 ~ Izl2 for Izl:S;; 1.
..
5. MEROMORPHIC AND ENTIRE FUNCTIONS
84
5.4.8. Show that the sequence of functions h (z) = z(z+l)(z+2) ... (z+n) n n! exp(zlogn)
is convergent in the whole plane and that its limit h(z) represents an entire function equal to zero only for z = 0, -1, -2, ... 5.4.9. (cont.) If r(z) = [h(z)rl, prove that (i) zr(z) = F(z+I); (ii) r(n) = (n-l)! for positive integers n. 5.5. FACTORIZATION OF AN ENTIRE Fl,INCTION
Every polynomial with zeros 0, a1 , represented as a product:
•.• ,
an (ak 1=
°for
k = 1, ... , n) can be
Weierstrass proved the existence of an analogous decomposition for an arbitrary entire function and he showed that it is possible to construct an entire function whose zeros form an arbitrary sequence with no limiting-point other than infinity. Let {an} be an arbitrary sequence with liman = ex). Moreover, let E(z, m) = (l-z)exp(z+ E(z, 0)
!
~ Z2+ ... + z"')
for
m ~ 1,
= l-z;
the functions E(z, m) are called the Weierstrass primary factors. There exists always a sequence of positive integers {m n} such that
f\:pr
n
1
+
n=l
is convergent in the whole plane and every entire function J with zeros at the points 0, a1 , ••• an, ... and otherwise different from zero can be written in the form of a Weierstrass product: J(z) = zmexpg(z)
IT
E(: ,mn)'
nail
n
where g is an entire function. In particular, if there eXIsts an integer q such that 00
L lanlna1
q
<
+
00
ex),
whereas
L lanl- + n=l
q 1
=
sequence {an}. Then we may choose mn
+ =
IX) ,
then q is called the genus of the
q-l for all n. Not all the terms of
85
5.6. FACTORIZATION OF ELEMENTARY FUNCTIONS
the sequence {an} must be different; if a number Zo appears k times in it, then the corresponding primary factor in the Weierstrass product appears k times. 5.5.1. Verify if the following sequences have a finite genus q and possibly, determine it: (i) -1,1, -2,2, ... ; (ii) {log(n+l)}; (iii) {[nj4W}, where [x] denotes the greatest integer n ~ x; (iv) {sn+itn}, where (sn; t n) is the sequence of all pairs of integers so ordered that max(ls.l, ItnD increases with n; (v) {(-I)· y'n}; (vi) 1; 2,2; 3,3,3; 4,4,4,4; 5, ... 5.5.2. Determine the Weierstrass product for sequences (i), (ii), (vi) in Exercise 5.5.1. 5.5.3. Derive Weierstrass theorem from Mittag-Leffler theorem. 5.5.4. Show that every meromorphic function functions.
f is a quotient of two entire
5.5.5. Let {an} be a sequence of complex numbers such that ao = 0, am =1= an for m 1= nand lima n = 00, and let {17n} be an arbitrary sequence of complex numbers. Let w be the Weierstrass product for the sequence {an} of the form: w(z) = z
fi E(:n ,mn).
Show that there exists a sequence {q.} of complex numbers such that the series (A)
is a.u. convergent in the whole plane and its sum W is an entire function such that 17k = W(ak), k = 0, 1, 2, ... This is the so-called Pringsheim's interpolation formula. 5.5.6. Find an entire function F such that F(n)
=
(n-l)! for all positive
i IItegers n. 5.6. FACTORIZATION OF ELEMENTARY FUNCTIONS
5.6.1. Derive the formula sin7':z = 7':Z
IT (1- ~:). • =1
5. MEROMORPHIC AND ENTIRE FUNCTIONS
86
5.6.2. Derive the. formulas: (i) sinh
~z = ~z
D(
1 + :: );
(ii) coshz-cosz =
(iii) eZ-1
=
Z2
n(
1+
zezt.2 n°O (I+~); 4~2n2 n=1
(iv) e""-ebz = (a-b)zexp [i-(a+ b) z]
D 00
[
] + (a-b)2z2 4n2~2 .
1
5.6.3. Derive the formula
cos~z = ll[I-(n~tr]. 5.6.4. Evaluate infinite products: (i)
(ii)
n(l+ ::); IT (1- ::).
Show that
n (1- _I)= _I co
n4
n=2
8~
5.6.5. Evaluate infinite products:
.
(i)
(ii) (iii)
fl (+ ::- + ::);
'
1
II (+ ~: ); II ( 1
1- :: ) .
Hint: Exercise 5.6.1.
5.6.6. Show that (i) n°O (1+_1 +....!.) n2 n4 n=1
= l+cosh~V3 2~2
(eTt_e-Tt).
5.6. FAC::TORIZATION OF ELEMENTARY FUNC::TIONS
87
n 1+ 1161) 1
sinh 7t ( .r ="2' ~ cosh7t-cos7t ... 3).
°o· (
(ii)
n=l
5.6.7. If An is the root of the equation tanz = z contained in the interval (n7t, (n+i-) 7t), n = 1, 2, ... , show that
sinz-zcosz=
!Z2ll(1- ~).
5.6.8. Verify the formula
n°O sin7tz 7tz(l-z) = n=l
(1 Z_Z2 ) +
n+n2
.
5.6.9. Verify the formula
sin3:: = • SInZ
~ n°O
(1-
n=-oo
4Z2 2 ) (n7t+z)
(z 1= n7t, n
= 0, =f1, =f2, '" ) .
5.6.10. Suppose F is an entire function whose zeros a1' a2 , ••• , an, ... are all simple and satisfy 0 < la 1 1~ la 2 1~ ••• , liman = 00. Suppose, moreover, there exists a sequence of contours {Gn } satisfying the conditions 10 _4 0 of Exercise 5.2.1 and such that
Prove that
where all the factors corresponding to zeros situated between Gn and C,,+l are l"Onsidered as one term. 5.6.11. Show that
s~n~(z+a) = (1+~) SIn tta
a
fr(l+-z-)e- lZ/ n (a 1= 0, =f1, =f2, ... ). a+n
n=-oo
I'he prime after a product sign indicates that the factor corresponding to n = 0 be omitted.
~hould
~.6.12.
Show that
(z)( 1- 5z) ...
.7tZ cos-7tZ 4-- SIn --4"-= (l-z) 1+ 3
5. MEROMORPHIC AND ENTIRE FUNCTIONS
88
5.6.13. Use the formula [r(z)tl
ze YZ
= ,
IT (1 + :)
e- z / n
n=l
to show that F(z)r(l-z)
7t
= -.-. sm7tZ
5.6.14. Express the following products in terms of the function (i)
IT (1 __1)e n=l
(ii) (iii)
r:
1/2n ;
2n
D(1+ 2n~1)(1- ;n); (l+-t) (I-i-) (1+i-) ... 5.1. ORDER OF AN ENTIRE FUNCTION
An entire function f is said to be of finite order if there exists a positive number A such that fez) = O(exprA) as Izl = r ~ +00. Then the number P =- 11·m 10glogM(r), r-++co logr
where
M(r)
=
M(r,!) = sup If(z)1 Izl";;'
which is finite, is called the order of the entire junction f. If fez) = CO+Cl z+ +C2Z2+ ... , then -.nlogn (S.7A) p = hm 10g(l/lcnD' 5.7.1. If k is a positive integer, show that expzk is an entire function of order k 5.7.2. Show that exp(eZ ) is not a function of finite order.
5.7.3. Show that
vz
(i) cos is an entire function of order f; (ii) an arbitrary polynomial is an entire function of order
o.
5.7.4. If j is an entire function which is not a polynomial, show that
lim 10gM(r,J) logr
=
+00.
r-++co
5.7.5. If fez)
=
Zkg(Z) is of order p
> 0, show that g is also of order
p.
5.7. ORDER OF AN ENTIRE FUNCTION
89
5.7.6. IfJis an entire function of order p, show thatJ' is also an entire function of order p, and conversely. 5.7.7. Suppose J is an entire function of order p and m(r) denotes the number of zeros of J situated inside K(O; r). Prove that m(r) = O(r p + e) , where e > 0 can be arbitrary. Hint: Cf. Exercise 5.3.1. 5.7.8. (cont.) IfJis an entire function of finite order p and ai' a2 ,
... ,
an,'"
00
are its zeros, 0 < lall :::; la 2 1:::; ... show that for every
IY.
> p the series 2..: lanl-
IX
n=l
is convergent. 5.7.9. Show that an entire function J of finite order p has a representation of the form J(z)
= t'exP(g(z))"U
E( :n ,m)
where g is an entire function and m :::; p does not depend on n. 5.7.10. Determine the order of the following entire functions:
(ii)
f eo;n r a
zn
n=l
2: e-" zn; 00
(iii)
2
n=O
(iv)
cosh lin n , z. n.
(a>
0);
CHAPTER 6
The Maximum Principle 6.1. THE MAXIMUM PRINCIPLE FOR ANALYTIC FUNCTIONS
6.1.1. If I is analytic and not a constant in a domain D, show that III cannot have in D a local maximum. Hint: Cf. Exercise 4.2.5. 6.1.2. If I is analytic and not a constant in a domain D, show that III cannot have a local minimum at Zo ED, unless I(zo) = O. 6.1.3. Suppose I is analytic and not a constant in a domain D and III has a con-
15. If I(z) i= 0 in D and m = inflf(z),l, M = sup If(z) I, < I/(z) I < M in D. 6.1.4. Suppose I is analytic in a domain D and If I has a continuous extension on J5 without being a constant. If III has a constant value on the boundary of D, show that I(zo) = 0 at
tinuous extension on z E frD, show that m
some point Zo ED. 6.1.5. If P is a polynomial of degree n, show that {z: IP(z)1 = A} cannot have more than n components for any fixed A. 6.1.6. If I is a nonconstant, analytic function in K(O; R), show that M(r,!) sup I/(z) I is a strictly increasing function of r E (0, R). \zj=r
6.1.7. Prove Hadamard's three circles theorem: .
Suppose I is analytic in B = {z: r1 < Izi < r2}, continuous in Band Mk = sup I/(z)l, k = 1,2. Izl=rk
If M(r)
=
sup I/(z)l, then 10gM(r) is a convex function of logr, i.e. Izl=r
Iog M() r
~
logr2-logr I M logr-Iogr 1 I M og 1 + - - - - - og 2' logr 2-logr1 logr 2-logr1
Hint: Consider [/(z)]pz-q, where p, q are suitably chosen integers. 90
6.2. SCHWARZ'S LEMMA
91
6.1.8. Iffis analytic in K(oo; 1), continuous in K(oo; 1) and has a finite limit at 00, show that If I attains a maximum at a point of C(O; 1). Also prove that M(r,!) = sup If(z) I strictly decreases, unless f is a constant. 6.1.9. If P is a polynomial of degree nand IP(z) I :0:;; M in K(O; 1), show that IP(z)1 :O:;;Mlzl n in K(oo; 1). 6.1.10. If P is a polynomial of degree nand IP(z) I :0:;; M for z E [-1, 1], show that for all z situated inside an ellipse with semiaxes a, b and foci -1, 1 we have: IP(z)1 :0:;; M(a+b)". Hint: Consider the mapping z = t(w+w- l ). 6.1.11. Suppose f is analytic and bounded by 1 in absolute value in K(O; 1) and tends uniformly to 0 in the angle lI. :0:;; argz :0:;; f3 as Izl -+ 1. Show thatf = O. Hint: Consider the product rp(z) =f(z)f(wz)f(w 2z) .. .f(wn-lz), where (II = exp (27ti/n) and n is a suitably chosen integer. 6.1.12. Suppose f is analytic in S = {z: Irezl < a} and there exist two real constants C, A such that If(z)1 :0:;; expAIYI for z = x+iy, Ixl < a, and lim If(z) I :0:;; C for z -+ a+iyo, resp. z -+ -a+iyo with arbitrary Yo. Show that for all z E S we have: If(z) I :0:;; C. Hint: Consider rp(z) = f(z)exp(ez2) , e > O. 6.2. SCHWARZ'S LEMMA
6.2.1. Iffis analytic and iess than 1 in absolute value in K(O; 1) and if/CO) = 0, show that either If(z) I < Izl in K(O; 1)",0, or elsef(z) = ei<x z with some real lI.. Hint: Consider z-lf(z). 6.2.2. If f is analytic in K(O; 1) and If(z) I < 1 for all z
E
K(O; 1), show that
If(Z)~f(O) I:0:;; 11-f(0)f(z)l. 6.2.3. Under the assumptions of Exercise 6.2.2 show that either 11'(0)1 = e i l7.z (0( is real).
< 1,
or fez)
6.2.4. If f maps K(O; 1) 1: 1 conformally onto a domain D containing the IInit disk and f(O) = 0, show that 11'(0)1 ~ 1 with the sign of equality for fez) l,la z only. co
6.2.5. If fez) =
2: anz"
is analytic in K(O; 1) and If(z) I :0:;; M in K(O; 1),
n=O
',how that Mlall
<: M2-la oI2.
6.2.6. (cont.) If m, n are integers, 0 ~ 2m
< n, show that Mlanl
~ M2-la m I 2.
6. THE MAXIMUM PRINCIPLE
92
6.2.7. If £0 is analytic in K(O; 1) and 1£0 (z) I < 1 in K(O; 1), £0(0) = oc > 0, show that the set of all values taken by £0 in R(O; r) is contained in K(zo; p), where
and U·
flint:
Consl·der :'4 rI() z
=
1£O(z)-oc ( ). -oc£O Z
6.2.8. (cont.) Show that any point of K(zo; p) is a value of a certain function £0 taken at a point of K(O; r). 6.2.9. (cont.) Find the set D, of all possible values taken in K(O; r) by functions £0 analytic in K(O; 1) and such that £0(0) ;;;: 0, 1£O(z) I ~ 1. 6.2.10. If Q is a polynomial of degree n with complex coefficients and IxQ(x)1 ~ M for x E [-1, 1], show that
IQ(z)1 ~ M(I+v'2)n+\
z EK(O; 1).
Hint: Cf. Exercise 6.1.10. 6.2.11. If Q is a polynomial of degree nand
1(z-'I]) Q(z) I ~ M,
z
E
('I] is a constant, 1'1]1
ceO; 1)
=
1),
show that
IQ(z)1 ~ t(n+2)2 M,
z E K(O; 1).
6.3. SUBORDINATION
Suppose J, F are analytic in K(O; R). The function I is said to be subordinate to Fin K(O; r), r ~ R, if there exists a function £0 analytic in K(O; r) such that £0(0) = 0, 1£O(z) 1 < r in K(O; r) and 1= Fo£O in K(O; r). If I is subordinate to F in K(O; r), we write 1-<, F. In. the particular case r = 1 we shall write 1-< F. If I -<,F, then obviously any value of I taken in K(O; r) is also a value of F taken in the same disk. 6.3.1. If r1
< r2 < R
and I -
6.3.2. If I -<,F, show that 11'(0)1
~
6.3.3. If 1-< R F, show that M(r,J)
IF'(O)I. ~
M(r, F), r
E
(0, R).
6.3.4. Suppose F is analytic and univalent in K(O; 1) and D = F(K(O; 1»). F(O), If I is analytic in K(O; 1), I(K(O; 1)) cD and 1(0) = wo , show that 1-< F. Wo =
6.3. SUBORDINATION
93
6.3.5. If I is analytic in K(O; 1),/(0) = 0 and Ire/(z) I < 1 in K(O; 1), show
that (i)
11'(0)1 ::(~; 7t
..) I/( ) 2 I 1+lzl (11 z I ::( --;- og l-lzl .
6.3.6. If/is meromorphic in K(O; 1),J(0) = 0 and all the values of/lie outside K(a; r), show that
11'(0)1::( rcot 2 a,
a
= arcsin(r/lal).
6.3.7. Let 9 be the class of all functions p analytic in K(O; 1) and such that
= 1, rep(z) > 0 in K(O; 1). Show that for a given, fixed z (0 < Izl = r < 1) the valuesp(z) cover the whole disk with diameter [(I-r)(1+r)-\ (1+r)(1-r)-1] as p ranges over 9.
p(O)
p
6.3.8. Given z 9.
E
K(O; 1) find precise estimates of Ip(z)1 and arg p(z) valid for all
E
6.3.9. Given z E K(O; 1) find precise estimates of rep(z) and imp(z) valid for all p E 9. 6.3.10. If I is analytic, does not vanish and I/(z) I < M in K(O; 1), show that 1-lzl
II(z) 1::( 1/(0)1 1 + 1"1 M
21z1
1-IZ I,
z EK(O; 1).
Hint: Consider log [Ml{(z)]. 6.3.11. Suppose {f,,} is a sequence of functions analytic and non-vanishing
in K(O; 1) and such that If,,(z) I < M for all n and all z E K(O; 1). If
co
l: I/n(O) I n=1
.. +00, show that l.Jfn(z)]2 is uniformly and absolutely convergent in K(O;
t).
6.3.12. Suppose I is analytic in K(O; 1) and 1(0) = O. If I does not take the vulue a (0 < IX < 1) and satisfies Iftz) I < 1 in K(O; 1), show that
11'(0)1 ::((2IXIOg
!)!(1-a
2 ).
6.3.13. Let Fbe an analytic function which maps the unit disk 1: 1 onto a condomain D so that F(O) = 0, F'(O) = 1. If I is an odd function subordinate
WI(
10
F in K(O; 1), show that
6. THE MAXIMUM PRINCIPLE
94
Hint: Consider cp(z) =
6.3.14. If fez) K(O; 1) and
f
=
-< F,
~[f( ::;z )+f( :~;z)].
a1z+a2z2+ ... , F(z)
=
0
< letl <
l.
A 1 z+A 2 z 2+ ... are analytic in
show that la 2 1:;:;:; max(lA 1 1, IA 2 1)·
Hint: Cf. Exercise 6.2.5. 6.4. THE MAXIMUM PRINCIPLE FOR HARMONIC FUNCTIONS
6.4.1. If u is harmonic in a domain D and has a local extremum at Zo ED, show that u = const in D. 6.4.2. .If u is harmonic in the finite plane and has a finite limit as z show that u = const.
--+ OCJ,
6.4.3. Suppose u is a non-constant, harmonic function in a domain D which has a continuous extension on D. If m = infu(z), M = :mpu(z), show that m < u(z) < M in D. 6.4.4. Suppose u is a non-constant, harmonic function in a domain D and there exists a constant M such that for any s > 0 and any t; on the boundary of D there exists a neighborhood K(t;; pes») of t; such that u(z) < M+s for any ZED (') K( t;; pes)). Show that u(z) < M for all ZED. 6.4.5. Supposef(z) = z-1+aO+a1z+ ... is analytic and univalent in K(O; 1)",,0 and D f is the set of all values taken by f. Suppose F(z) = Az- 1+A o+A 1z+ .. , is meromorphic in K(O; 1) and DF C D f . Show that either IAI > 1, or else F(z)
=
f(zeirT.).
Hint: Consider h(z) = loglzl-loglr 1 (F(z))I.
6.4.6. If u is harmonic in {z: R1
< Izl <
R 2} and A(r)
=
sup u(rei8 ) , show o
that A(r) is a convex function of log r. Hint: Find a function harmonic in {z: r 1 < Izl < r2} equal to A (r k ) on C(O; rk), k = 1, 2.
,
6.4.7. Letfbe analytic and univalent in K(O; R) withf(O) = O. If the image curve of ceO; r), 0 < r < R, is starshaped w.r.t. 0 (i.e. argf(re i8 ) is an increasing function of 0), show that also the image curves of all ceO; p), p < r, are starshapco
w.r.t. O. Hint:
cr.
Exercise 1.1.20 (i).
CHAPTER 7
Analytic Continuation. Elliptic Functions 7.1. ANALYTIC CONTINUATION
A function f analytic in a domain D defines an analytic element, or a function element (f, D). If (ft, D1 ), (f2, D 2) are such that Dl (') D2 is non-empty and (1 (z) = f2 (z) for all z E Dl (') D 2 , then the function elements (ft, D 1 ), (f2, D 2) are said to be direct analytic continuations of each other and (f, Dl U D 2) with fez) = jj(z) for z E DJ (j = 1, 2) sets up a new analytic element. It may, however, happen that Dl (') D2 has several components andfl = f2 in one component, whereas fl i= f2 in some other component. Then we say that (jj, D j ), ; = 1, 2, are single-valued branches of a certain global (generally multi-valued) analytic function. Two function elements (f, D), (F, LJ) are said to be analytic continuations or each other, if there exists a finite chain {Uk, D k )}, k = 0, 1, ... , n, of function dements such that (fo, Do) = (f, D), (fn, Dn) = (F, LJ) and (fb D k ) is a direct analytic continuation of Uk-I, D k - 1 ), k = 1, ... , n. All possible analytic continuations of a certain function element set up a complete analytic function. 7.1.1. Show that the sum of the power series
IOg2-~-~( 2
2
l_Z)2 2
_~(~)3 3
2
" a direct analytic continuation of the sum of the power series Z_-}Z2++Z3- .,. 7.1.2. Verify that the function elements corresponding to the sums of the following power series and their disks of convergence:
l+z+z2+ ... ,
+
I~i [I ~=.~ +(~=~r
nrc direct analytic continuations of each other. 95
+. .J
96
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
7.1.3. Verify that the sums of power series:
/l(Z)
=
Z+tz2+tz3+ ... ,
/2(Z)
=
7ti-(z-2)+t(z-2)2_t(z-2)3+ ...
have disjoint disks of convergence D1 , D2 but the corresponding function elements (fi, D1), (f2' D 2) are analytic continuations of each other.
7.1.4. Give an example of two function elements (fl, D1 ), (f2, D 2) such that Dl (') D2 has two components and /1 = /2 in one component, whereas /1 =f. /2 in the other component. 7.1.5. Show that there exists an analytic continuation of the analytic element determined by the power series /(z) = 1+-}z+tz2+ ... and its disk of convergence which is analytic in K(O; 1) "" 0 and has 0 as a simple pole. 7.1.6. If (f, D) is an analytic continuation
of the
function
element
00
(2.:
Z2 n , K(O; 1), show that Dc K(O; 1).
n=O 00
7.1.7. If the radius of convergence of
2:
anz" equals to 1 and all an are non-
11=0 00
negative, show that
(f, D) with 1 ED.
(2:
a"z", K(O; 1)) has no direct analytic continuation
"=0
2:cn
7.1.8. If is a convergent series with positive terms and {w,,} is the sequence of all rational numbers, show that
L C,,(Z-w )-l 00
/(z)
=
n
11=1
is analytic in the upper half-plane H+ and also in the lower half-plane H_. Also prove that the function elements (f, H+), (/, H_) are not analytic continuations of each other. .
7.1.9. Give an example of a function/analytic in K(O; 1) and continuous in K(O; 1) such that for any direct analytic continuation (fl, D) of we have D c K(O; 1).
(I,
K(O; 1)
7.2. THE REFLECTION PRINCIPLE
Suppose D is a domain situated in the upper half-plane whose boundary contains an open segment y of the real axis and D* is the domain obtained by reflection of D w.r.t. the real axis. If/is analytic in D and has a continuolls
7.2. THE REFLECTION PRINCIPLE
97
extension on D u y such that I assumes real values of y, then the function: F(z) = I(z) for ZED u y, F(z) = I(z) for z E D* is a direct analytic continuation of Ion D u y u D*. If Y and I(y) are circular arcs, then after suitable linear transformations we can reduce this case to the case just considered, it may, however, happen that the analytic continuation becomes a meromorphic function. If D is a Jordan domain,/is univalent in D and/(D) is again a Jordan domain, then the assumption of continuity of I on D u y can be dropped since I has necessarily a homeomorphic extension on D. 7.2.1. If I is an entire function and takes real values on the real axis and imaginary values on the imaginary axis, show that I is odd. 7.2.2. If I is meromorphic in K(O; l+h), h show that I is a rational function. 7.2.3. Supp,?se
> 0, and
I is meromorphic in a domain
axis and real on it. If a is a pole ofI and A
=
If(z) I
=
I on C(O; 1),
D symmetric w.r.t. the real
res(a; f), show that res(ii; I) = A.
7.2.4. Suppose I is meromorphic in a domain D symmetric w.r.t. an arc C of C(a; R) and real on C. If b i= a is a pole of order n with singular part n
2..:
ck(z-b)-k, show that
k=l
11* being the reflection of b w.r.t. C. 7.2.5. If I is analytic in a domain D symmetric w.r.t. the real axis, show that
r '11 + if2,
where 11 ,12 are analytic in D and real on the real axis.
7.2.6. If I is analytic in K(O; 1) except for a simple pole Zl, continuous in 1\(0; 1) and real on C(O; 1), show that I(z)
=
(Az 2 +Bz+A)[(Z-Zl)(1-Z\Z)rl
with real B. 7.2.7. Find the 1: 1 conformal mapping of the upper half of K(O; 1) onto the lower half-plane such that the points -2, 00, 2 are image points of -1, 0, 1 I'l'Sp.
7.2.8. Find the 1: 1 conformal mapping of the upper half of K(O; 1) onto Ihe' upper half-plane {w: imw > O} carrying the diameter (-1, 1) into the rlly -I- 00) and its center z = 0 into w = o.
(-!.
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
98
7.2.9. If I is analytic in K(O; 1), continuous in K(O; 1) and maps K(O; 1)
onto a n-sheeted unit disk (i.e. the equation I(z) = a has exactly n roots in K(O; 1) for any a E K(O; 1)), show that I is rational. Find its general form. 7.2.10. If I maps 1: 1 conformally the rectangle R onto a rectangle R' so that the corners of Rand R' correspond, show that I is a similarity and the ratio of sides is the same for both rectangles. 7.2.11. If
I maps 1: 1 conformally the annulus {z: 1 < Izl < r} onto
{w: 1 < Iwl < r'} so that If(z) I -'+ 1 as Izl real
IX
-+
1, show that J(z) = eia.z with
and r' = r.
7.2.12. Suppose J is analytic and univalent in the unit disk and maps it onto a domain D symmetric w.r.t. the straight line L through the origin. Show that L () D is ~the image of a certain diameter of K(O; 1), if flO) = O. 7.2.13. If J maps 1: 1 conformally the upper half-plane H+ = {z: imz > O} onto a circular polygon D so that the points Zl' ..• , Zll of the real axis correspond to the corners of D, show that I can be continued analytically along any path
omitting the points Zl' ••• , Zll. Also prove that any two branches (fl' H+), (f2' H+) of the complete analytic function so obtained satisfy:Jl = (af2+b)/(cI2+tl), where a, b, c, dare complex constants. 7.2.14. (cont.) Suppose {j, z}
=
1 (/,,)2 ' (1")' I ' -"2 7' is the Schwarzian deriva-
tive of f Verify that the Schwarzian derivative of the multi-valued function of Exercise 7.2.13 is single-valued. Hint: Show that {(al+b)/(cl+tl), z} = {j, z}.
I
7.3. THE MONODROMY THEOREM
Suppose D is a domain in the finite plane and a complete analytic function (fo, Do) of I with Do c D can be continued along any path Y contained in D. Then, according to the monodromy theorem, the continuation of the same element (10, Do) along any two paths Yl' Y2 with common end points (one of the common end points being situated in Do) leads to the same terminal element, whenever Yl' Y2 are homotopic with respect to D. Since any two paths Yl' Y2 with common end points situated in a simply connected domain D are homotopic w.r.t. D (i.e. can he continuously deformed into each other within D), we obtain for simply connected domains the following theorem: If a function el~rnent (fo, Do) of a complete
I has the property that a certain function element
7.3. THE MONODROMY THEOREM
99
analytic function I can be continued along any path y contained in a simply connected domain D, there exists in D a single-valued, analytic function coinciding ·with 10 in Do. Perhaps the most interesting applications of the monodromy theorem are connected with the so-called elliptic modular function .It = .Iter). If To is the domain {-r: 0 < re -r < 1; I-r-tl > t; im -r > O}, there exists a unique 1: 1 conformal mapping .It = .It ( -r) of To onto the upper half-plane im.lt > 0 such that the points -c = 00, 0, 1 correspond to the points .It ~ 0, 1, 00. By using the reflection principle one can continue .It(-c) onto the whole upper half-plane im -c > 0 whose image wiII be the infinitely many sheeted .It-plane punctured at 0, 1. Since .It' ( -r) #- 0, there exists locally at any point .It #- 0, 1 an inverse function clement -c = -c(.It) which can be continued along any arc in the .It-plane not passing through 0, 1. 7.3.1. If a function element (f, D),DcK(O; 1),",,0, can be continued along any path situated in the punctured unit disk K(O; 1)"-.0, show that eac;h function clement obtained by continuations within this domain has the form F(logz) where F(w) is analytic in the left half-plane {w: rew < O}. 7.3.2. If an entire function g taking all finite values has a non-vanishing derivative, show that it is a similarity transformation, i.e. g(z) = az+b (a 1= 0). 7.3.3. (cont.) Suppose I is an entire function such that f(z)/'(z) 1= 0 for all z. Prove that I(z) = exp(az+b) (a 1= 0). 7.3.4. Let I be analytic in a domain D and let K = K(zo; r) cD. Prove that
the function element ( ~
l(C)de, K) can be continued along any regular arc y
[zo.z]
~Iarting
at
Zo
and situated in D. If y joins Zo to Z in D, show that the value F(Z)
of the terminal branch equals ~ 1(1;) de . 1
7.3.5. (cont.) Prove Cauchy's theorem in homotopic form: If I is analytic in /) and the arcs Y1' Y2 with common end points zo, ZED, are homotopic w.r.t.
n.
then~/(z)dz= ~f(z)dz. 11
12
7.3.6. Let I be a function analytic and non-vanishing in a simply connected dllmain D. Verify the existence of an analytic logarithm of I in D by using the 1IIIIIlodromy theorem. II
7.3.7. Show that any entire function g omitting the values 0, 1 reduces to constant.
100
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
7.3.8. Prove that any nonconstant entire function assumes all values in the
finite plane except at most one. 7.3.9. Show that Q(w)
A( 10:iw ) is analytic in the unit disk {w: Iwl
=
< I},
00
does not depend on the choice of argw and has the form: Q(w) = with Al
> 0 and all Ak real. Also prove that
A( -r) e- im
--+
Al as im -r --+
~
.:.....J
Ak wk
k=l
+ CXl •
7.3.10. (cont.) Verify that Q has in some disk K(O; b) a univalent inverse function which can be continued along any path in the open plane omitting the points 0, 1. Also verify that Al > 1. [Remark: one can prove that Al = 16, cf. [1], or [24]]. 7.3.11. Suppose I is analytic in the unit disk,/(O) = 0, and I(z) 1= 0, 1 for Izl < 1. Show that I is subordinate to Q in the unit disk.
o<
7.3.12. Let I(z) = a l z+a 2 z 2 + ... be analytic in the unit disk and let I(z) be different from 0, 1 in 0 < Izl < 1. Show that lall :s;; 16. 7.4. THE SCHWARZ-CHRISTOFFEL FORMULAE
Let D be a simply connected domain whose boundary is a closed polygonal line L without self-intersections. There exists a function I analytic in the upper half-plane {z: imz > O}, continuous on the real axis (in spherical metric) and mapping the upper half-plane onto D. It has the foIIowing form: z
(7.4A)
I(z)
=
n
11 (C-Xk)"k-ldC+Cl , o k=l
C~
where C, Cl are complex constants, Xl < x 2 < ... < Xn are points on the real axis corresponding to consecutive vertices Wl, W2' ... , Wn and r:J.k 7t are the interior angles of D at Wk, k = 1, 2, ... , n. The integral is a line integral along any regular curve with end points 0, z situa,ted (apart from the end point 0) in the upper half-plane. On the other hand, the function F mapping {z: rez > O} on the outside of D and such that F(b) = CXJ, has the form: n
z
(7.4B)
F(z)
= C ~ (C-b)-2(C-b)-2
IT (C-Xk)Pk-1dC +C
1
k= 1 where Xl < X2 < ... < Xn are the points corresponding to the vertices ... , Wn and fh7t are exterior angles of D. o
n
We have obviously:
11'1,
W2'
n
L r:J.k = n-2, L (3k = n+2. The k=l
formula (7.4A) also
k=l
holds when D is an unbounded domain whose boundary is a polygonal line /,
7.4. THE SCHWARZ-CHRISTOFFEL FORMULAe
101
without self-intersections and two sides of L are half-lines. Sometimes a formal application of Schwarz-Christoffel formulas in a degenerate case (e.g. IXk = 2, or there exist more than two boundary rays) enables us to guess the mapping function. The obtained mapping needs verification which is occasionally possible. 7.4.1. If one of the vertices of D is the image point of the mapping function has the form
Xn =
00, show that
z n-1
fez)
C~
=
o
IT (C
-Xk)"'k- 1
dC+C1 •
k= 1
7.4.2. Find the function mapping the upper half-plane onto the inside of an equilateral triangle. Using the formula 1
~ up - 1(1-U)Q- 1du
=
reP) r(q)/r(p+q)
o
express the length a of the sides in terms of
r
(p
> 0,
q
> 0)
function.
7.4.3. Map the upper half-plane onto a rhombus with angles 7tIX, 7t(I-IX) so that its vertices correspond'to z = 0, =FI, 00. Find the side-length. 7.4.4. (cont.) Verify that the upper semicircle and the positive imaginary axis correspond to the diagonals of the rhombus. Also find the preimage of its center. 7.4.5. Map 1: 1 conformally the upper half-plane {z: imz > O} onto the part of the upper half-plane lying outside the rectangle: - k re w 0, o ~ im w « h, so that the points =Fa, =F~ correspond to the vertices. Discuss the limiting case: ~ -+ O.
«
«
7.4.6. Map I: 1 conformally the upper half-plane imz > 0 onto the domain containing the first quadrant whose boundary consists of two rays: {w: rew ~ 0, imlv = I}, {w: rew ~ 0, imw = -I} and a segment [-i, i]. The points : ·co -1, 1 should correspond to w = i, - i. 7.4.7. Map 1: 1 conformally the upper half-plane onto a part of it with the boundary consisting of two rays: {w: rew « 0, imw = 7t}, [-7tcot~, +00) II nd a segment [-7t cot ~, 7ti], where 0 < ~ < 1-7t. The points z = -1, 0 should wrrespond to w = 7ti, -7tcot~. Discuss the limiting case ~ -+ O. 7.4.8. (cont.) Find the mapping of the right half-plane reC > 0 onto the II'-plane slit along the rays: im w = =F7t, re w 0, which carries the points , '=Fi into w = =F7ti.
«
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
102
7.4.9. Find the mapping of the upper half-plane imz > 0 onto the strip domain 0< imw < 7t slit along the ray imw = Vi' rew ~ 0 (0 < Vi < 7t). 7.4.10. Find the mapping of the upper half-plane imz > 0 onto the first quadrant slit along the ray: im w = 7t, re w ;;:: h, which carries the points z = 0, a 2 into w=O, h+7ti (a>l,h>O). 7.4.11. Suppose F is a function analytic and non-vanishing in K(O; r5) which has a constant argument on (- r5, r5). Prove that the function Z
W(z) = ~ C- 1F(t;)dt; Zo
where zoEK(O; r5), imzo >0, is analytic in K(O; r5)",,-(-ir5,O] and maps the radii (- r5, 0), (0, r5) onto two paralle} rays the distance between them being equal to 7tIF(O)I. 7.4.12. Find the image domain of the upper half-plane under the mapping
-
z
~ ()-
w-Hz-
dt; t;(t;-b)}f·t;-a
(0
< a < b, imzo >0).
Zo
Find the distance between two parallel rays in each pair of boundary rays and evaluate b so as to make both distances equal. 7.4.13. Show that the mapping Z
W
= ~ t;-1(t; -a)- lj3 dt; o
carries the upper half-plane imz > 0 into a domain bounded by a polygonal line consisting of two parallel rays and a segment such that interior angles are equal -t7t, T7t. Evaluate the distance between the rays. D
7.4.14. Find the mapping of the upp~r half-plane imz = C""-(K(i; 1) u Ql) where Ql is the first quadrant. Hint: Map first D under W = w- 1 •
>
7.4.15. Find the mapping of the upper half-plane imz Dil
=
{w: 0 < imw < I} u {w: 0 < arg(w-i) < fm}
Assume that z = -1, 0 correspond to w = i,
00
0 onto the domain
> 0 onto where
0
<
It
< 1.
resp.
7.4.16. (cont.) Find the mapping of the right half-plane reZ > 0 onto Dil U (-00, +(0) U D!, where D: is a reflection of Dil w.r.t. the rea] axis. 7.4.17. Show that the function mapping the inside of the unit disk onto the
7.4. THE SCHWARZ-CHRISTOFFEL FORMULAE
103
inside of a simple, closed polygonal line has the same form as in (7.4A), replaced by the points on the unit circle.
Xk
being
7.4.18. Show that the mapping
carries the unit disk into the inside of a regular n-angle. Evaluate its perimeter. 7.4.19. Find the mapping of the unit disk It I < 1 onto the n-pointed star (i.e. 2n-angle whose all sides and alternate angles are equal). 7.4.20. Map 1: 1 conformally the unit disk It I < 1 (i) onto a pentagram (a five-pointed star obtained by extending the sides of a regular pentagon); (ii) onto a Solomon's seal (a six-pointed star formed of two congruent equilateral triangles placed one upon another). 7.4.21. Show that the mapping
A
W =
where Zk
=
expicpk,
Ck
[ l-zkC Jrrrn k=I (l-CkC)1+ k P
]
dC+B
expiVJk> 0:;:;;; CPI < VJI < CP2 < VJ2 < ... <
=
1j)n
< 27t,
n
11k> 0,
L (h =
2, carries the unit disk Izi < 1 into the w-plane slit along
k=l rays 11 , 12 , ... , In.
I he are angles between
Verify that Zk correspond to the end-points of Ik and and Ik (/n+l = 'i)·
fh
Ik+l
7.4.22. Show that the mapping z
W
r~_~_d_t===--
=
~ (1-t 4 )Jh+t 4
l"nrries the unit disk Izi < 1 into the domain
(w: Irewl < ta} u {w: limwl < ta}. I'valuate a. 7.4.23. Prove that the function J mapping 1: 1 conformally the unit disk 1.:1 ..:: 1 onto a convex domain bounded by a polygonal line and such thatJ(O) = 0 hilS the form z
J(z)
=
n
A ~rr (l-texpiOk)-a.kdt+B o k= 1
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
104
n
where 0 ~ 01 < O2 < ... < On < 21t,
OCk>O,
L
OCk
~ 2.
k=1
Verify that If'(z) I ~ 1f'(0)1(1-lzl)-2,
If(z) I ~ 1f'(O)llzl(1-lzl)-l.
7.4.24. A function f analytic in a disk K is said to be close-to-convex, if there exists in K a univalent, analytic function (/) mapping K onto a convex domain and such that re!'(z)/(/)'(z) > 0 in K. (i) Show that each close-to-convex function is univalent in K. Hint: Consider the mapping fo (/)-1 and cf. Exercise 3.1.14. n
(ii) If
L
(1Jlk-rpk)
=
1t in Exercise 7.4.21, show that the mapping function
k=1
is close-to-convex. [Remark: Exercise 7.4.24 (i), as well as its counterpart for a half-plane is a convenient tool in proving the univalence of mappings obtained by a formal application of the Schwarz-Christoffel formulas.] 7.4.25. If IC1 1= IC21 = ... = ICnl = 1 n
n
k=l
k=l
L OCk = 2, L OCkCk = 0,
and
OC k
> -1
(k = 1,2, ... , n),
show that "
IT (1-Cfk)"' dC
1
k=1
W(z) = ~ C- 2
n
1c
is continuous on C(O; 1) and maps it onto a closed polygonal line L with interior angles (1- OCk) 1t. If L has no self-intersections, show that W is meromorphic and univalent in K(O; 1) and maps it onto the outside of L. 7.4.26. Find the 1: 1 conformal mapping of the outside of the unit disk onto the outside of a triangle with exterior angles OCk 1t (k = 1, 2, 3) such that the points 'fJk on C(O; 1) correspond to the 'vertices and the points at infinity to each other. Find the relation between OCk and '17k. 7.4.27. Iff maps 1: 1 conformally the outside of the unit disk onto the outside of an equilateral triangle T so that f( (0) = 00, show that the preimages of vertices of T also form an equilateral triangle. ,
7.4.28. Find the 1: 1 conformal mapping f of the outside of the unit disk onto the outside of a regular n-angle such that f(oo) = 00 and f(r/) = 'YJk ('YJ = exp(21ti/n), k = 0, 1, ... , n-l).
7.5. FUNCTIONS sn, cn, an
105
7.5. JACOBIAN ELLIPTIC FUNCTIONS sn, en, dn
If 0 < k < 1, then the Schwarz-Christoffel integral z
U = u(z)
~ [(1-t 2)(1-k 2t 2)tl/2dt
=
o
maps 1: 1 conformally the upper half-plane imz > 0 onto the inside of ~he rectangle with vertices =fK, =fK+iK'. We have: u(O) = 0, u(=f1) = =fK, u(=fk- 1) = =fK+iK', u(oo) = iK'. The inverse function z = sn(u, k) can be continued analytically by reflections all over the open plane and becomes a meromorphic function with two periods 4K,2iK'. We have: ~2
1
K
=
K(k)
~ [(1-t 2) (1-k 2t 2)tl/2dt
=
=
o
~ [1-k 2sin2q?rl/2dq? 0
F being the complete Legendre elliptic integral. The functions en, dn are defined by the equations: sn 2u+cn 2u
=
1,
k 2sn 2u+dn 2u
=
=
.
F(k, 7t/2)
1.
All the roots of the equations sn 2u = 1, sn 2u = k- 2 are double, hence en, dn have no branch points and are merom orphic in the finite plane. z
7.5.1. If u
=
(6- 5t 2+t4)-1/2dt, show that
~
o
z z
7.5.2. If u
=
~
V 2 sn(uV3,
=
vt).
(1-t 4)-1/2dt, show that
o
z
=
cn[}i2 (u-K), 2- 1 / 2 ].
7.5.3. Let k' be the complementary modulus w.r.t. k, i.e. k' that K(k') = K'(k) and K(k) = K'(k').
=
y1-k2. Show
7.5.4. Derive. the expansions: (i) sn(u, k) = u-(1+e)u 3 /3!+(1+14k 2+e)u 5 /5!- ... ;
(ii) cn(u, k) (iii) dn(u, k)
=
=
1-u2/2!+(1+4k 2)u4/4!+ ... ; 1-k2U2/2!+k2(4+k2)U4/4!+ ... ;
lind find the radius of convergence in each case. IL
7.5.5. Express K(k) as the sum of a power series. Verify that K'(k)/K(k) is strictly decreasing function of k E (0, 1) which decreases from +00 to O.
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
106
7.5.6. Show that sn(u+iK') = (ksnu)-l. Find the initial terms of the Laurent expansion of snu in the neighborhood of u = iK'. 7.5.7. Determine the periods, zeros and poles of sn(iv, k') as depending on k. 7.5.8. Verify the identity sn- 2 (u, k)+sn- 2(iu, k') == I. 7.5.9. Find an elliptic function f(u) with periods 2, 2i which has simple zeros at points whose coordinates are either both odd, or both even and has simple poles at points whose one coordinate is even and the other one odd. Discuss the uniqueness. 7.5.10. If z moves round the rectangle with vertices iK', 0, K, K+iK', show that w = cnzj(1 +snz) moves round a quadrant of the unit disk.
•
7.6. THE FUNCTIONS a, C, f.J OF WEIERSTRASS
The function O'\:z) = a(z; £01' as the infinite ptoduct
(0
2) of Weierstrass is an entire function defined " .
a(z)=a(Z;W1,W2)=ZI)'(I-
~k)exp[~k + ~ (~kr],
where Dk = mkwl +nkw2, (mk; nk) is the sequence of all pairs of integers and the prime after the product sign indicates,that the pair (0; 0) should be omitted; £01' £02 are two complex numbers with im(wtfw2) ¥= o. The logarithmic derivative a'(z)ja(z) is the merom orphic function C(z) of Weierstrass. The function f.J(z) = -C'{z) is a meromorphic, doubly-periodic (or elliptic) function with periods £01' £02 whose only singularities are double poles mWl +nw2. We have: f.J(z) = f.J(z; £01'
(0
2) = z-~+
2:' [(Z- Dk)-2_ Dk"2]. k
7.6.1. Show that a(z)
=
z+CSZS+C7Z7+ ...
7.6.2. If £01 is real and £0 2 purely imaginary, show that a is real on the real axis and purely imaginary on the imaginary axis. 7.6.3. If imwl
~
rew 2 = 0, show that a(z) = a(z).
7.6.4. Verify that a(u+wk)ja(u)
=
-exp[(2u+Wk)C(twk)]
(k
=
),2).
go
7.6. FUNCTIONS a, "
107
7.6.5. Show that (.) cr(2u) _ _ 1
0'4(U) -
,(). fIJ u,
..
fIJ"(u) fIJ'(u)'
(n) 2C(2u)-4C(u) =
7.6.6. If C(u) has poles at 2mw+2nw' and 7:=£0'/£0, h=eiT.., 'f}=C(w), im 7: > 0, show that 2'f}w =
~~(1-24
I
k=l
7.6.7. If 1 < Izl < q, show that fIJ(logz; 2logq, 27ti)
2:
1 (q-znz+qznz-1_2)-1-12-2
0 0 ,
=
n=-oo
L: 00
(qZn+q-Zn_2)-1.
n=l
7.6.S. Suppose fIJ has periods 2a, 2bi, where a, b are real and positive. Show that (i) fIJ is real on both axes; (ii) fIJ is also real on straight lines rez = an, imz = bn (n = 0, =fl, =f2, ... ); (iii) fIJ maps 1:1 conformally any rectangle R: na
7.6.9. If fIJ(z) = fIJ(z; 2£0,2£0'), imw = rew' = 0 and fIJ(w) = e1, fIJ(w+w') e3, fIJ(w') = ez, show that ez < e3 < 'e1' and ez < 0 < e1' Prove that the converse also holds. 7.6.10. (cont.) Show that [fIJ(u+w)-e1] [fIJ(u)-etJ 7.6.11.Ifez<e3<e1,k=
V
e3-eZ
e1 -e2
=
(e1-eZ)(e1-e3)' .
2K(k)
2iK'(k)
J' e1-e 2
J' e1- e2
andflJhaspenods./,./
verify the identity: fIJ(u(e1-e2)-1/Z)
=
eZ+(e1-e2)sn-Z(u, k).
7.6.12. Derive Legendre formula: 'f}1WZ-'f}ZW1
= =f7ti
where 'f}k = C(-}Wk), k = 1,2. /lint: rntegrate C(z) round the parallelogram with center 0 and sides £01' £0 2 •
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
108
7.6.13. Suppose a(z) is the Weierstrass a-function with zeros m+n-r (im T > 0; m, n = 0, =f1, =f2, ... ) and 1}(z)
=
exp{-1]lZ2} a(z) ,
where
1]1 =
C(+)
a'(+)/a(+).
=
Show that (i) 1} is an entire function with the same zeros as a; (ii) 1}(z+l) = -1}(z), hence 1} has the period 2; (iii) 1}(Z+T) = -1}(z)exp[-7ti(2z+T)]. 7.7. CONFORMAL MAPPINGS ASSOCIATED WITH ELLIPTIC FUNCTIONS
7.7.1. Suppose a, b are real and positive, {z: -a < rez < 0, 0< imz < b} = Rand w = fez) is the univalent mapping of R onto {w: im w > O}"", K(O; 1) such that the points z = 0, ib, ib-a, -a and w = 00, -h, -1, 1 (h > 1 depends on a, b) correspond to each other. Use the reflection principle to show thatfhas a double pole at the origin and is doubly-periodic with periods 4a, 2ib. If ~(z) = ~(z; 4a, 2ib) and e1 = ~(2a), e2 = ~(ib),e3 = ~(2a+ib), show that
7.7.2. Find the function mapping 1: 1 conformally the annulus 1 < lei < Q onto K(oo; 1)"",(-00, -h] where h> 1. Also find the relation between Q and h. 7.7.3. Show that the mapping w = yksn(u, k) carries the open segment (-K++iK', K++iK') into the upper semicircle of C(O; 1). 7.7.4. (cont.) Show that the image domain of the rectangle with corners =fK=f~ iK' is the unit disk Iwl < 1 slit" along the radial segments (-1, -Vf], [Vk, I). . K'(k) 1 a+b 7.7.5. If 0 < b < a and k IS such that 2K(k) = --;-log a-b show that the function
w = w(z)
=
V k sn [2K(k) 7t arcsin 1-
Va z_b 2
2
2
is analytic in H, where H is the ellipse with boundarY;i-
,
k] 2
+ -t-2-
=
1.
Also prove that w = w(z) maps HI: 1 conformally onto the unit disk.
7.7. CONFORMAL MAPPINGS
7.7.6. Given k, 0
109
< k < I, show that the function w = w(z) =
yIC sn ( 2:i
is analytic in the annulus A = {z: 1 < Izl 1:1 onto K(O; I) jlkj.
logz,
< R}, R
k)
= exp(7tK'/4K), and maps it
"'[-Yk,
7.7.7. If A = {z: 1 < Izl
<
7tK'(k) Q} and k is such that 2K(k) = 10gQ, show
that W= T1 [ sn
('K ~logz, k )]-2
maps AI: 1 conformally onto K( 00; 1) '" [k- 1 ,
+ 00 ).
7.7.8. Find the mapping of the right half-plane rew > 0 slit along [1, k- 1] onto {z: 1 < Izl < Q}. Find the relation between k and Q (0 < k < 1). 7.7.9. Find the conformal mapping of the extended plane slit along two circular arcs with end points a1, a 2 and b1 , b2 situated on the same circle onto a concentric I:ircular annulus. 7.7.10. Suppose w = f(z) is a 1: 1 conformal mapping of the triangle with vertices 0,2K, (1+i)K (K = K(1/V2» onto the lower semi-disk of K(O; 1) 'iucb that the points z = 0, K, 2K and w = 1, 0, -1 correspond to each other. Show that w = cn(z, l/y'2). 7.7.11. Given the rectangle R: 0 < x < a, 0 < y < b and a point ~+ifJ E R. Hxpress in terms of a(z; 2a, 2ib) the function mapping 1: 1 conformally R onto "'(0; I) so that the origin is the image point of ~+ifJ. Hint: Use the reflection principle and evaluate zeros and poles of the mapping function.
CHAPTER 8
The Dirichlet Problem 8.1. THE RIEMANN MAPPING THEOREM
If D is a simply connected domain whose boundary contains at least two points, there exists a function I analytic in D which maps D 1: 1 conformally onto the unit disk. This theorem, due to Riemann, has the following completion, first conjectured by Osgood and then proved by Osgood, Taylor and Caratheodory: If D is a Jordan domain (Le. a domain bounded by a closed Jordan curve) then the mapping function I admits a homeomorphic extension to D.
8.1.1. Show that the finite plane C cannot be mapped 1: 1 conformally onto the unit disk. 8.1.2. Give an example of a locally univalent mapping I of the unit disk such that 1(0) = 0, /,(0) = 1 and I[K(O; 1)] = C. Hint: Verify that q(w) = we w takes every finite value in G = C"'-( - 00, -1]. 8.1.3. Suppose D is a Jordan domain and g is a real-valued function harmonic zo, continuous vanishing on fr p and tending to + ex) as Z --+ Zo' Find the in univalent function mapping D onto K(O; 1) in terms of g.
D"'-
8.1.4. Suppose D is a Jordan domain and it ,12 are two mapping functions of Riemann's theorem such that II (Zk) = 12(Zk) at three boundary points Zk of D. Show that II = 12' 8.1.5. Suppose Dk are Jordan domains symmetric w.r.t. circular arcs Yk (k = 1, 2). Suppose, moreover, I is a univalent mapping of Dl onto D2 which carries two boundary points ZI' Z2 symmetric w.r.t. Yl into WI. W2 symmetric w.r.t. Y2 and one end point of YI into an end point of Y2' Show that I(Yl) = Y2' 110
B.2. POISSON'S FORMULA
111
8.1.6. Find the images of arcs of C(1; 1), C(O; 1), re r = t- contained in To under the mapping l = l( r), where l is the modular function. Also evaluate lct+ it y3), lei): l(l+i). 8.1.7. Construct by means of Riemann's mapping theorem a univalent function having C(O; 1) as its natural boundary. 8.1.8. Suppose D is a simply connected domain whose boundary contains at least two points and a ED. Show that D can. be mapped onto K(O; r) under a univalent function rp so that rp(a) = 0, Irp'(a)1 = 1. Verify that r = rea; D) is uniquely determined by D and a. The number rea; D) is called the inner radius of D at a. 8.1.9. If I is univalent in K(O; 1), D = I[K(O; 1)] and a = I(zo) , show that rea; D)
=
(l-lzoI2)1f'(zo)l.
.
8.1.10. Find the inner radius of: (i) the disk K(O; R) at a, lal
<
R;
(ii) the upper half-plane at ih (h > 0); (iii) the plane slit along a ray at a point on its prolongation whose distance from the end of the ray is equal to d. 8.1.11. If rp is a univalent mapping of a simply connected domain Do onto D and a = rp(a o), show that rea; D) = Irp'(ao)lr(ao ; Do). Otherwise, the metric Idwl/r(w; D) remains unchanged under 1: 1 conformal mappings. It is called the hyperbolic metric. 8.1.12. Consider all univalent functions in a simply connected domain Do which satisfy: rp(a o) = 0, rp'(a o) = 1, where ao is a fixed point of Do. Show that the area of the image domain has a minimum, if rp(Do) = K(O; r), r = r(ao; Do). Hint: Exercise 4.2.12. 8.1.13. Suppose that G is a domain in the extended plane n
union
U
r
k
Cand C",G is a finite
of n disjoint continua. PrOve that G can be mapped 1: 1 conformally
k=l
onto a bounded domain whose boundary consists of n closed, analytic Jordan I:urves. 8.1. POISSON'S FORMULA
Let U be a real, bounded and piece-wise continuous function of 0, 0 (1(0) = U(271:). The function 2"
(X.2A)
u(z) ==
1 \ (R2_r2) U(O) u(r, rp) = 271: J R 2-2Rrcos(O-rp)+r 2 dO o
~
0 ~ 271:,
8. THE DIRICHLET PROBLEM
112
is harmonic in the disk K(O; R) w.r.t. X = rcosrp, Y = rsinrp and satisfies lim u(r, 0) = U(O) at any continuity point of U. In case U is continuous, the r-+R-
Poisson formula (8.2A) gives the solution of the Dirichlet problem for a disk. 8.2.1. Show that the Poisson kernel J
=
R2_r2 --=--<---=:-=---:-;;----,:--;;-
R2-2Rrcos(0-rp)+r2
can be written as re[(z+C)/(z-C)] with z = Rei9, C = reirp • 8.2.2. Given a continuous, real-valued and periodic function U (U(O) = U(27t)), find a function f analytic in K(O; R) and continuous in K(O; R) such that ref(Rei9 )
=
U(O).
•
8.2.3. If u is harmonic in K(O; R), continuous in K(O; R) and nonnegative on C(O; R), show that R-izi R+lzi u(O) R+lzi ~ u(z) ~ u(O) R-izi .
8.2.4. Find the function u harmonic in K(O; 1), continuous in K(O; 1) and such that u(ei9 ) = cos 2 0. 8.2.5. Given two complementary arcs C(, (J of C(O; 1) and a function U vanishing on (J and equal to 1 on C(. Find a corresponding function u according to (8.2A). Show that 27tu(z) is equal to the length of an arc 'Y of C(O; 1) cut off by the straight lines through z and the end points of C(. 8.2.6. Show that the limit function u of an a.u. convergent sequence {un} of harmonic functions is harmonic. 8.2.7. Derive the formula (8.2A) under the assumption that U has in [0, 27t] a Fourier series representation:
L (ancosnO+bnsinnO). 00
U(O)
=
t
a o+
n=l
Hint:
an
(~r cosnrp,
bn (~r sinnrp are harmonic functions of z = re 'rp in
K(O; R).
8.2.8. Derive in a similar way a formula analogous to (8.2A) for a function u harmonic in C"'-K(O; R) with given bowldary values U on C(O; R).
B.l. THE DIRICHLET PROBLEM
113
8.2.9. Give a Fourier series representation of a bounded, harmonic function = 1 for 101 < oc, u(Rei~ = 0 for oc < (J
u(rei'P) with boundary values: u(Rei~ < 27t-OC (0 < oc < 7t, r < R).
8.3. THE DIRICHLET PROBLEM
A domain G is said to be regular with respect to the Dirichlet problem, if any real-valued function U defined and continuous on the boundary fr G of G has a continuous extension u to d which is harmonic in G. The function u is called the solution of Dirichlet's problem for G with given boundary values U. If we can associate with any point' E frG a continuum He (i.e. a closed, connected set containing at least two points) such that, E He C C,,", G, then G is regular w.r.t. Dirichlet's problem, cf. [19], p. 93. 8.3.1. Show that the domain K(O; 1),,",0 is not regular w.r.t. the Dirichlet problem.
8.3.2. Find the solution of Dirichlet's problem for the annulus Rl < Izi < R z with boundary values (i) constant on each boundary component: U(Rle i6 ) = A, U(R 2 e i6 ) = B; (ii) equal r:Jk(O) on C(O; R k), k = 1, 2, where Uk(O) are represented by unirormly convergent Fourier series. Hint: Put
8.3.3. Show that any Jordan domain is regular with respect to the Dirichlet pl'Oblem. If z = $(w) maps G 1: 1 conformally onto the unit disk Izl < 1, express Ihe solution by means of $. 8.3.4. Find the solution h of the Dirichlet problem for the domain (,' K(-1; vi) (") K(1; )i2) with boundary values H(w), W EfrG. Hint: Exercise 2.9.6. 8.3.5. Prove that the visual angle w(z) of the segment [a, b] of the real axis ill a point z = x+iy of the upper half-plane is a harmonic function. Find lim W (z)
Ii II'
,
situated on the real axis.
.-+C
H.3.6. Given a system of n+ 1 points on the real axis Xo < Xl < X2 < .. , < x" n real numbers Ul, U2, ••• , Un' Find the function u bounded and harmonic
IIl1d
III Ihe upper half-plane whose boundary values on the real axis form a step
B. THE DIRICHLET PROBLEM
114
function equal to 0 on (- 00, XO), (Xn , k=I,2, ... ,n.
+ 00)
and equal to Uk on (Xk-l, Xk),
8.3.7. Find the solution of Dirichlet's problem for the upper half-plane and continuous boundary values U, where U vanishes outside a finite interval [a, b]. 8.3.8. Suppose U is a real-valued function of t +00
\'
J
!U(t) 1 dt
1+ltl
<
+ 00
E ( - 00,
+ 00)
such that
.
-00
Show that +00
.
u(z) = -
y
~
7t
(
U(t)
)2 2 dt x-t +y
-00
is harmonic in the upper half-plane and lim u(z)
=
U(to) for any continuity
z-+to
point to of U. 8.3.9. Suppose H is the interior of the ellipse 3X2+4y2 = 12 and L is the segment with end points at its foci. Find the function u harmonic in H~L, continuous in ii, equal to 1 on fr H and equal to 0 on L. 8.3.10. Find the function u harmonic and bounded in C~ {( - 00, -1] u u [1, +oo)} which is equal to -h on (-00, -1] and equal to h,on [1, +00). 8.3.11. Find the function u harmonic outside the ellipse 'x2 y2 £l2 +[;2 = 1 (a> b > 0) such that u(z) -+ A as z approaches the ellipse and u(z)-loglzl = 0(1) as Z -+ 00. Discuss the uniqueness. Find the value A for which u(z)-loglzl = 0(1) as z -+ 00. Hint: Consider first an analogous problem for the unit disk. 8.3.12. Suppose u is harmonic and bounded in the upper half-plane slit along [i, + ioo) and has the boundary values 0 on the real axis and boundary values 1 on the ray [i,
+ioo).
Find the derivative
~:
on the real axis.
8.4. HARMONIC MEASURE
r
Let G be a domain such that the spherical image of = fr G is a finite systcm of simple, closed curves. If y is a finite system of open arcs on r, then the function w(z; y, G) harmonic and bounded in G which tends to 1 as z -+ CE Y and lcnds to 0 as z -+ CE r~y is said to be harmonic measure of y at z.
8.4. HARMONIC MEASURE,
115
8.4.1. Find a function harmonic in K(O; 1) and tending to 0 as z -+ !;, 1)",1 which is not identically O. Verify that the boundedness condition in the definition of harmonic measure is essential for its uniqueness. ~ E C(O;
8.4.2. Determine the lines of constant harmonic measure w(z; y, G), if: (i) Y = (a, b), G = {z: imz > O}; (ii) Y = {z: Izl = 1, fJ 1 < argz < fJ 2 }, G = K(O; 1). 8.4.3. Find the harmonic measure of each of the boundary rays 11 , 12 of the angular domain G= {z: O<argz
Go
8.4.7. Find the harmonic measure w of the arc: Izl = 1, 0 < argz < 27tA., w.r.t. the annulus {z: 1 < Izl < R}, 0 < A. < 1. Hint: Exercise 8.3.2 (ii).
8.4.8. The boundary of G consists of two circular arcs Y1' Y2 with common end points a, b (a, b real, a < b) both Yk situated in the upper half-plane. Determine w(z; Yk, G) (k = 1,2). 8.4.9. Prove the monotoneity of harmonic measure: (i) Y1 c Y2 implies w(z; Y1' G) :;:;:; w(z; Y2' G);
r r r
(ii) if the boundaries 1 , G1 c G2 , then for any Y c w(z; y, (;2)'
2
1
.
of G1 , G2 have a non-empty intersection and n 2 and any z E G1 we have: w(z; G1)
r
y,
8.4.10. Prove the following "two constants theorem": Let f be analytic and hounded by M in absolute value in a domain G whose boundary consists of It finite system of Jordan curves. Suppose that for a system of open arcs a c we have: lim If(z) I :;:;:; m < M
r
r
for any CE a. If (3 =
z--..c
r",Ci., then
log If(z) I :;:;:; w(z; a, G)logm+w(z; (3, G)logM.
8.4.11. Use Exercises 8.4.5 and 8.4.10 to prove the following PhragmenI.indelof theorem:
116
8. THE DIRICHLET PROBLEM
Suppose f is analytic in the upper half-plane and lim If(z) I ~ 1 for any real Z-->-X
and finite x. If M(r)
=
sup If(re i8 )1, then either lim 10gM(r) > 0, or else 0<0<"
If(z) I ~ I in the whole upper half-plane.
'-->-+00
r
8.5. GREEN'S FUNCTION
Given a domain G in the extended plane and a point Zo E G. A function g harmonic in G"'-..zo, continuous in G'"zo, vanishing on frG and such that g(z)+loglz-zol = 0(1), as z ~ Zo 1= 00 (resp., g(z)-loglzl = 0(1) as z ~ Zo = 00) is called (classical) Green's function g(z, Zo; G) of G with pole zoo If a domain G has a Green's function, the function is unique. If the boundary of G is a finite system of analytic Jordan curves, then Green's function of G exists and has at each point Cof fr G a normal derivative ogf one and the solution u of the Dirichlet problem for G with boundary values U(C) has the following form: (8.5A)
8.5.1. Show that any bounded domain regular w.r.t. the Dirichlet problem has Green's function. Hint: Consider Dirichlet's problem with boundary values 10gIC-zol. 8.5.2. Show that any simply connected domain G whose boundary contains at least two points, possesses Green's function. Express it by means of the univalent function mapping G onto K(O; 1). 8.5.3. If the domain Gz has the classical Green's function and z = z(w) is a I: 1 conformal mapping of Gw onto Gz show that also Gw has the classical Green's function and g(w, Wo; Gw ) = g(z(w) , z(wo); Gz). 8.5.4. Suppose G is a simply connected domain whose boundary is an analytic Jordan curve rand f is a univalent mapping of G onto K(O; 1). If Zo E G and c!>(z) = [1-f(zo)f(z)]f[f(z)-f(zo)], show that the interior normal derivative of g(z, Zo; G) is equal to Ic!>' (C) I, C E frG. 8.5.5. Determine g(z, Zo; K(O; 1» and derive Poisson's formula from (8.5A). 8.5.6. Determine Green's function for the upper half-plane and derive the formula of Exercise 8.3.8 from (8.5A).
B.S. GREEN'S FUNCTION
117
8.5.7. Find the Green's function g(z, CXl; G) for (i) G
=
K( CXl; 1);
(ii) G
=
C", [-1, 1];
(iii) G =
{z = x+iy:
::
+ ~:
> I}.
8.5.8. Suppose 0 < h < c < 1 and take the constant T determining the function {} of Exercise 7.6.13 equal to 10gh/7ti. Verify that the function F(w) = {}
(2~i log ;
) : {}
(2~i 10gCW)
has the following properties: (i) F is single-valued and analytic in A = {w: h (ii) F has exactly one simple zero in A; (iii)
IFI
< Iwl < I};
has constant values on each of two components of fr A .
8.5.9. (cont.) Find the Green's function g(w, c; A). 8.5.10. (cont.) Show that the interior normal derivative
J3g an Express
~~ in
= \'
([J'(w) \ ([J(w) ,
loge
where
([J(w) = w logh /F(w).
terms of Weierstrass C-function.
8 5.11. Suppose G is a simply connec,ted domain in the finite plane whose complement contains at least 2 different points and p( w, Wo; G) i~ the hyperbolic distance of two points w, Wo E G (which is defined as hyperbdlic distance of their image points after a univalent mapping onto the unit disk). Find the relation between pew, Wo; G) and the Green's function of G. 8.5.12. Let G be a domain whose boundary r is a finite system of analytic Jordan curves. Show that g(z, Zl; G) is a harmonic function of Zl in G"'z. Also verify the symmetry of the Green's function: g(z, Zl; G) == g(Zl' z; G). Hint: Cf. formula (8.5A) and Exercise 8.5.1.
8.5.13. If Go c G, show that g(z, Zl; Go) :S;; g(z, Zl; G) for any z, Zl
E
Go.
n
8.5.14. Let G be a domain whose boundary G =
U rb where r k are
analytic
k= 1
Jordan curves with positive orientation w.r.t. G. If h(z, zo) is the (multi-valued) conjugate of g(z, zo; G), show that the increment i1h (C, zo) as C describes r k
118
B. THE DIRICHLET PROBLEM
is equal to - 27tWk(ZO) , where Wk(ZO) = w(zo; Fk> G) is the harmonic measure of Fk w.r.t. G. Hint: Express Wk by means of (8.5A). 8.5.15. Find the increment of 10gF(w), where F is the function of Exercise 8.5.8 and w moves over vA. 8.5.16. Suppose a, b, c are complex numbers different from each other and such that c ¢:. [a, b]. Determine the domain yielding the maximal value of g(a, b; D), among all convex domains D such that a, bED, C E C"'-D. 8.5.17. Let G be a domain such that H = C~ G is a finite system of bounded continua. Given two real constants A, B, show that there exists a unique function cp harmonic in G, continuous in C, which takes the value B on H and has the form Aloglzl+O(1) as Z -+ 00.
8.6. BERGM,AN KERNEL FUNCTION
In this chapter G denotes a domain in the finite plane whose complementary set is a finite system of disjoint continua and L 2 (G) is the family of all functions
I analytic in
G and such that the integral
HIf(x+iy)1
2 dxdy
(taken in the sense
G
of Lebesgue) is finite. 8.6.1. Verify that L2(G) is a complex, linear space with the usual definitions of addition and multiplication by numbers. 8.6.2. If f, g EL 2 (G), show that the integral
Hfgdxdy =
(f, g)
is finite
G
and has usual properties of an inner product. 8.6.3. Show that for ~ny 1; E G there ~xists f E L2(G) such that 1m = 1. Hint: If is the unbounded component of C"'-G consider the univalent function cp mapping c",-r onto a disk.
r
8.6.4. If A(M) is the family of all (f,f) = IIfll2 :;:;:; M2
f
E
L2(G) such that
and
K(a; p)
c:
show that there exists a uniform bound of If(z) I for f Hint: Cf. Exercise 4.8.13.
G, E
A(M),
Z E
K(a; p).
8.6.5. Evaluate a common bound for If(z) I in terms of M, a, p (Ial+p < I), when G = K(O; 1).
8.6. BERGMAN KERNEL FUNCTION
119
8.6.6. show that for any ~ E G there exists fo E L 2 (G) which has the minimal norm (f,f)1/2 among all f E L 2(G) taking the value 1 at the point 1;. 8.6.7. (cont.) Prove that for any g E L2(G) with g(1;) = 0 we have (g '/0) = O. Hint: Consider f* =fo+ee i8g (e > 0, 0 ~ () ~ 27t). 8.6.8. An analytic fun~tion k of G and defined by the formula
Z E
G depending on a complex parameter
~ E
k(z, 1;) = Ilfoll- 2fo(z, 1;),
where fo is the solution of the extremal problem considered in Exercise 8.6.6 is called Bergman kernel function. Express k(z, 1;) in case G is simply connected, by means of the univalent function w mapping G onto K(O; 1) and such that w(1;) = O. Also verify the uniqueness. Hint: Exercise 8.1.12. ~)
8.6.9. Determine k(z,
for: (i) G = K(O; R); (ii) G
=
{z: imz
> O}.
8.6.10. CAven k(z, 1;) for a simply connected domain G, find the univalent function w mapping G oItto K(O; 1) so that w(1;) = 0, w'(1;) > O. 8.6.11. Verify the reproducing property of k: fm =
~~ f(z)k(z, C)dxdy,
1; E G,/E L2(G).
G
Hint: Consider f(z)/f(1;)-fo(z, 1;); (cf. Ex. 8.6.7).
8.6.12. Show that k(1;,
~)
=
~~ Ik(1;,
z))i 2 dxdy
and
Ik(1;, '])1 2
~ k(1;,
1;)k('], ']).
G
8.6.13. Verify that the reproducing property of Exercise 8.6.11 implies the uniqueness of k. 8.6.14. Suppose {IP.} is a complete, orthonormal set of functions in L 2 (G) (i.e. (lPj, IPk) = ~jk with ~jk = 0 for'j i= k, ~kk = 1 and the linear combinati~ns
• \ ' CklPk
("":"'0
rr f,
g
form a dense set in L 2 (G)). E
L2 (G) and
ab
bk are Fourier coefficients off and g, resp., show that
,,,
00
~ a.b. is convergent and the Parseval identity (f, g) ,.
0
=
L n=O
/lint: Exercise 8.6.11.
L anb,. holds.
n=O 00
8.6.15. (cont.) Prove that k(z, ~)
=
IPn(z) IPn (1;).
8. THE DIRICHLET PROBLEM
120
8.6.16. Show that the functions fPn(z)
=]1:
fPo (z) = (27t log
(l-h 2n
r zn1
1
(n
=
=fl, =f2, ... ),
~) -1/2 Z-l
form a complete, orthonormal system for the annulus A
=
{z: h < Izl < I}.
8.6.17. Find Bergman kernel function for the annulus A
=
{z: h < Izl < I}.
8.6.18. If A
= {z:
h < Izl < I} and J
has the Laurent expansion J(z)
00
=
L
bnzn in A, find the conditions for coefficients bn in order that
J
n=-co
belong to the class L2 (A). 8.6.19. Examine the behavior of k(z, C) under conformal mapping. Show that
yk(" C)ldCl
is a conformal invariant.
CHAPTER 9
Two-Dimensional Vector Fields Potential theory in two dimensions is usually concerned with fields of force in which the vectors of the field are always parallel to a distinguished plane. Moreover, the vectors associated with the points of any straight line perpendicular to the distinguished plane are equal; hence the whole field is characterized by the field in the distinguished plane. If the field does not depend on time, it is called stationary. 9.1. STATIONARY TWO·DIMENSIONAL FLOW OF INCOMPRESSIBLE FLUID
Stationary two-dimensional flow of incompressible fluid is characterized by two harmonic functions: the velocity potential rp and the flow function 1p. If w = u+iv is the vector expressing the velocity of a particle of the fluid past a given point z = x+iy, then u = rpx, V = rpy. On the other hand, the difference 1p(Xl' Yl)-1p(X O ' Yo) yields the volume of fluid passing in one second through a face of height 1 whose basis is any arc joining xo+iyo to Xl +iYl' In absence of singularities and in case the velocity field is a simply connected domain, I - rp+i1p is an analytic function which is called complex potential of flow. The lines rp = const are called equipotential lines and their orthogonal trajectories 'y'(x, y) = const are called the lines of flow. In a stationary velocity field the lines of flow are the paths of the particles of the fluid. 9.1.1. Express the velocity wand its absolute value JwJ in terms of the complex potential of flow. 9.1.2. Prove the following uniqueness theorem for the velocity potential: if the domain D is swept out by regular arcs y and on each y the functions u, U have both a constant value (depending on y), then U = Au+B, where A, B II rc real constants. /lInt: Consider ux-iuy. 121
9. TWO·DIMENSIONAL VECTOR FIELDS
122
9.1.3. The lines of flow are circles x 2+y2-2ax = O. Find the complex potential of flow. Evaluate the ratio of velocity of two particles of fluid passing the points 2a, a(1+i). 9.1.4. The lemniscates r2 = a 2cos2(J (r, (J are polar coordinates) are the lines of flow. Evaluate the ratio of velocity of particles passing the points z = a, z = aC}irt + iJl't) (a is real). 9.1.5. Find the loci of constant velocity for the flow pattern as considered in Exercise 9.1.3. 9.1.6. Discuss the equipotential lines and the lines of flow and evaluate the velocity, if the complex potential of flow is equal: (i) az; (ii) aiz; (iii) Z-l; (iv)
z-b
log--; (v) Z2; (vi) alogz; (vii) ailogz (a z-c
> 0).
9.1.7. If the volume of fluid flowing outside across a cylindrical face of height 1 and basis C(a; r) is equal Q for all r sufficiently small, the point a is called a source of intensity Q. If the flux Q is negative, the point a is called a sink of intensity IQI.
The line integral ~ wsds, where Ws is the tangent component of velocity and y y
is a contour, is called the circulation along y. If the circulation along all contours containing the point a inside and sufficiently close to a is equal 1= 0, then a is called a vortex. If y is a contour containing inside one singular point and f is the complex potential of flow, show that
r
r+iQ
=
~f'(z)dz. y
9.1.8. Given a flow with complex potential log(z2 +Z-2). Evaluate sources and sinks and the corresponding intensities. 9.1.9. Discuss the flow pattern with the complex potential 2ilog(Z2_ a2). Evaluate the circulation along the cir~les Iz=Fal = a. 9.1.10. The complex potential of flow is equal log sinh 1tZ. Evaluate the flux across the circle C(O; t) and the circulation along it. 9.1.11. Evaluate the complex potential of flow of fluid from the lower halfplane into the upper half-plane through an orifice in the real axis between 1 and -1 supposed the velocity at 00 is equal 0 and the flux across the orifice is equal Q. Find the pressure at various points of the orifice. 9.1.12. Evaluate the complex potential of flow past a circular cylinder with cross-section CeO; I) immersed in a parallel stream, assuming that the velocity
9.1. FLOW OF INCOMPRESSIBLE FLUID
123
Woo at infinity is parallel to the real axis. Evaluate the velocity at z = i and z = 2, as well as the difference of pressure at these points.
9.1.13. Evaluate the -complex potential of flow past an elliptical cylinder x2 y2 7 lJ2 = 1 (a > b) immersed in a parallel stream, if the velocity at in-
+
finity is equal
Woo e i «
(0( real, woo> 0).
9.1.14. Consider analogous problems for cylinders with cross-sections:
(i) K(-i; y'2) n K(i; (ii) [-ih, ih].
]/':2);
9.1.15. Evaluate the complex potential of flow with a source of intensity Q at z = a, assuming Woo = O. 9.1.16. Evaluate the complex potential of flow with a source of intensity Q at -1 and a sink of the same intensity at z = 0, assuming Woo = O. 9.1.17. Evaluate the complex potential of flow in the first quadrant with a source of intensity Q at l+i and a sink of the same intensity at z = 0 assuming Woo = O. 9.1.18. Discuss the limiting case as h --+ 0 of flow with a source of intensity p/h at z = h and a sink of the same intensity at z = -h (a dipole of the momentum p). Find the loci of constant velocity. 9.1.19. Evaluate the complex potential of flow with a vortex at z = a involving the circulation Assume Woo = O. .
r.
9.1.20. Evaluate the complex potential of flow in the upper half-plane under the assumption there exists a source at z = ai of intensity Q and the velocity at 00 is parallel to the real axis. Find the velocity at z = O. 9.1.21. Consider analogous problems as in Exercise 9.1.20 assuming that:
(i) z = ai is a dipole of the momentum p and Woo > 0; (ii) z = ai is a source-vortex of intensity Q involving the circulation
r.
9.1.22. Determine the complex potential of flow past a cylinder with the cross-section K(O; 1) with a vortex z = ai (a > 1) involving a circulation Assume Woo = O.
r.
9.1.23. Evaluate the complex potential of flow past a cylinder with the crossNcction K(O; 1) assuming that z = a (a > 1) is a source of intensity Q and II'", . O. 9.1.24. Consider an analogous problem, the source being replaced by a vortex with circulation
r.
124
9. TWO-DIMENSIONAL VECTOR FIELDS
9.1.25. Show that the complex potential function
r logz ( R2) + 27ti
J(Z) = a z+--z
corresponds to an asymmetric flow past a circular cylinder with the cross-section K(O; R). Determine the real constant a so that z = iR is the only point where the stream lines enter (and exit) the cylinder cross-section. Find the velocity at 00 in this case and evaluate the force acting on the cylinder according to the Joukovski lift formula. 9.2. TWO-DIMENSIONAL ELECTROSTATIC FIELD
Two-dimensional electrostatic field is produced by a system of charged, long, parallel wires and cylindrical conductors. The potential is constant throughout each conductor and there are no charges in the interiors of conductors. The electrostatic field can be characterized by an analytic function J = qJ+i'IjJ or the complex potential oj the electrostatic field. The real, single-valued harmonic function 'IjJ is called the (real) electrostatic potential. The field vector w is equal -grad'IjJ = -if'(z). The complex z-plane playing the role of a system of coordinates may be any plane perpendicular to all conductors involved. One single wire with a charge of q units per 1 cm of length which intersects the coordinate plane at z = a gives rise to the electrostatic field with complex potential -2iqlog(z-a) (cf. Ex. 9.2.2). Given the complex potential f= qJ+ i'IjJ , the lines 'IjJ = const correspond to equipotential lines, whereas the lines qJ = const are the lines of force.
9.2.1. An infinite, straight-line shaped wire I is charged with a positive charge of q electrostatic units per 1 cm of length. Show that the Coulomb force w produced by the wire and acting on a unit 'charge at a distance rcm from the wire is perpendicular to the wire and Iwl = 2qr- 1 • 9.2.2. (cont.) Show that w can be derived from the complex potential J(z) = - 2qilogz, where the z-plane is perpendicular to I and cuts it at the origin. 9.2.3. Evaluate the complex potential of the electrostatic field due to two long parallel wires each bearing a positive charge q per unit of length, the distance of wires being 2h. Discuss the equipotential lines. 9.2.4. Evaluate the complex potential of an electrostatic dipole, i.e. the limiting case of electrostatic field due to two long parallel wires each bearing charges of"
9.Z. TWO-DIMENSIONAL ELECTROSTATIC FIELD
125
opposite sign =Fq per unit of length, the distance of wires being 2h, 2qh as h -+ O. Find the field strength and the loci of a constant field strength. 9.2.5. If
(1
-+
M
is the density of charge on the surface of a cylindrical conductor
and "P is the real electrostatic potential, show that
(1
= - :7t .
~~ .
Hint: The flux of the electrostatic vector field across a closed surface S is equal to 47t times the total charge inside S (Gauss theorem). 9.2.6. Given the potential "P outside a charged cylindrical conductor with the cross-section evaluate the total charge on the conductor.
r,
9.2.7. A n-tuply connected domain G whose boundary consists of two disjoint systems r o , r 1 , of closed analytic curves determines an electrostatic condenser. The systems ro, r 1 correspond to the cross-sections with the plane of reference of two systems of conductors (outer and inner coatings), the potential of conductors of either system being kept on the same level. Usually the inner coating is charged and the outer one is grounded so that its potential is equal to zero. Show that the total charges qo, ql on both coatings are equal but have opposite signs.
Hint: Consider the
int~gral ~
ro+r,
~ ~~
hds = -
W
:"P ds and apply the Green's formula: n
~~ (hx"Px+hY"Py}dxdy - ~~ hLJ"Pdxdy G
with
h = 1.
G
9.2.8. (cont.) If "Po, "P1 are potentials of ro, r 1 resp., and q is the charge on 1'1' the ratio c = q /("PI -"Po) does not depend on "Po, "P1 and is called the capacity of a condenser. If w(z} = w(z; r 1 , G) is the harmonic measure of r 1 w.r.t. the domain G, show that
47tC =
~~ (w~+w;}dxdy. G
Hint: Consider
~
iJG
w
~w
un
ds.
9.2.9. (cont.) Show that the capacity of a two-dimensional electrostatic l"Ol1denser is invariant under conformal mapping. 9.2.10. Find the complex potential of the electrostatic· field between two l'o:txial circular conducting cylinders of radii r and R. Show that the capacity or sLich a condenser is equal to t(log(R/r)t 1 cm per unit oflength of gen-
n:ttrix.
9. TWO-DIMENSIONAL VECTOR FIELDS
126
9.2.11. Show that the capacity c of a pair of long conducting parallel wires of radius a separated by a distance b is equal c=
~ (10 2
b+
vbC4a )-1 T
g b-vb 2 -4a 2
9.2.12. Find the complex potential of the electrostatic field between two con-
ducting circular cylinders with cross-sections K(O; 1), K(I; 4) assuming that the potential on the first cylinder is equal 1, whereas the second one is grounded. Evaluate the capacity of the cylinder. 9.2.13. The potential on the cylinder with cross-section 1(5; 4) is equal 1,
whereas the cylinder corresponding to K( - 5; 4) is grounded. Find the extremal densities of charge on both cylinders. 9.2.14. A horizontal wire of negligible radius bearing a charge A per unit length is suspended at a distance h above the surface of a conducting plane. Find the complex potential and the charge per unit area in the plane. Hint: The field will not change, if we replace the plane by a symmetric wIre bearing a charge - A per unit length. 9.2.15. Show that the electrostatic potential outside a system of long, conducting cylinders with cross-sections k (k = 1, 2, ... , n) bearing jointly a positive charge A per unit length and kept on the same potential B, has the form VJ(z) = 2Aloglzl- 1 +o(1) as z --+ 00, if B is suitably chosen. Prove that the ratio BI2A = Y does not depend on A. Express y by means of the Green's function
r
n
g(z) of the exterior of
U
r
k•
The constant
y is
called Robin's constant.
k= 1
9.2.16. Prove that Robin's constant for the exterior of the ellipse x2 y2
(i2+V= 1 is equal to -Iogi-(a+b).
CHAPTER 10
Univalent Functions 10.1. FUNCTIONS OF POSITIVE REAL PART
10.1.1. Iffis a complex-valued function of a real variable t
E
[a, b],J = u+iv, b
and g is a real-valued function of bounded variation, the Stieltjes integral ~ f dg b
a
b
is defined as ~ udg+q vdg. By using the well-known estimate for real-valued/: a
a
h
I~ fdgl ~
V~(g)max If I ,where V~ (g) denotes the total variation of g on the
a
interval [a, b] show that an analogous estimate also holds for complex-valued f. b
I
b
I
Hint: Take real oc such that ~ fdg = e ia ) f dg and transform the last expression. a
a
10.1.2. If the sequence of complex-valued functions {fn} of a real variable t E [a, b] is uniformly convergent in [a, b], and the sequence {gn} of real-valued functions with uniformly bounded total variation on [a, b] converges to g, b
b
show that the limit lim ~ f"dg" exists and is equal to ~ fdg. a
a
10.1.3. Using the Schwarz formula (Ex. 8.2.2) prove the following theorem due to Herglotz: Let f be analytic in K(O; R) with f(O) = 1 and ref(z) > 0 for all z E K(O; R). There exists an increasing function {t of t E [0,2,,], {teO) = 0, {t(2,,) = 1 such that 2" . Ret+z ~ fez) = -R-'t- d{t(t). e -z o
t
flint: Consider the sequence {t,,(t) = (27tr 1 ~ u(R" ei8 ) d() , where u(z) = ref(z),
/(" (I -- .!)
o
R, and write Schwarz's formula in a Stieltjes integral form. 127
to.
128
UNIVALENT FUNCTIONS
10.1.4. Let f!J' be the class of all functions analytic in K(O; 1) and such that = 1, ref(z) > 0 in K(O; 1). Show that the function (1+z)/(1-z) belongs to !?J and cannot be represented by the Schwarz formula in the unit disk. Write a corresponding Herglotz representation.
f(0)
10.1.5 If H is a fixed, continuous, complex-valued function of t E [a, b] and I-' is a variable, increasing function of t E [a, b] subject to the conditions 1-'(0) = 0, 1-'(1) = 1, show that the set of all possible values of the Stieltjes inb
tegral ~ H(t)dl-'(t) is identical with the convex hull of the curve F: w = H(t),
a~ t
•
~
b.
10.1.6. If Cn is the 11th Taylor coefficient at z = 0 off E t!I', express Cn in terms of the function I-' of Exercise 10.1.3. Show that en E K(O; 2). Also show that for any Wo E R(O; 2) and any positive integer n there exists hE!?J such that h(n) (O)/n! = wo. 10.1.7. Using Exercises 10.1.3 and 10.1.5, find the region of variability of the point fez) for fixed z E K(O; 1) and f ranging over !?J. 10.1.8. If f
E
f!J' and argf(zo)
= oc, show that
I
f'(zo) 2cos oc \ f(zo) ~ I-lzol2 . Hint: Consider qJ(z)
= [f( ttz:oz) -f(zo)] [f( t:z:oz) +f(zo)T
1
10.1.9. Let f be analytic in K(O; R) and :1= 0 in K(O; R)",O. Suppose "P is a positive, differentiable function of r E (0, R) such that
I d If'(z) fez) ~ dr log"P(r~ =
"P'(r) "P(r) ,
Izl = r.
Show that for any real, fixed 0 the function If(re i6 )""P(r) decreases in (0, R) and the finite limit lim If(rei 6)""P(r) exists. ' ....R-
10.1.10. If f
E
f!J', show that for any fixed () the finite limit lim (1- r )f(reiO ) ' .... 1-
exists. 10.1.11. If the function I-' of ExerciselO.1.3is continuousatt show that
lim (l-r)f(re i6 ) ' .... 1-
= O.
= ()
(0:1= 0, 21t).
10.2. STARSHAPED AND CONVEX FUNCTIONS
129
10.1.12. If f-t has a jump h at t = () «() #- 0, 27t), I.e. f-t«()+ )-f-t«()-) show that lim (l-r)/(re'6) = 2h.
= h,
Y---=;l-
10.1.13. If () = 0, 27t, show that lim (1-r)/(r) = 2[1-f-t(27t-)+f-t(0+)]. r-+-l-
10.1.14. If IE :!J, show that for any fixed, real () the finite limit lim (l-r )/(rei6 ) = rx«()) exists and is real, nonnegative. Also show that the set r-+I00
{(): rx«())
> O} is at most countable; if ()k are. its elements, show that
L rx«()J :;:;:; 2. k=l
10.2. STARSHAPED AND CONVEX FUNCTIONS
10.2.1. If I is analytic and does not vanish in the annulus {z: r- r5 < Izi < r+ r5}, show that
a
16 zl'(z) aoarg/(re ) = re I(z) .
10.2.2. If I is analytic in K(O; R), does not vanish in K(O; R)""O, 1(0) = 0, f'(O) #- 0, and re{zl'(z)/I(z)} > 0 for any z E K(O; R), show that I is univalent in K(O; R). Hint: The argument principle.
10.2.3. Show that (i) fl(Z) = z(l-z)-3 is univalent in K(O; f); (ii) I(z) = z(1-Z)-1X is univalent in K(O; 1) for 0:;:;:; rx :;:;:; 2. 10.2.4. Let S* be the family of all normalized starlike univalent functions, i.e. IE S* means that I is analytic in K(O; 1),/(0) = 0,1'(0) = 1 and zl'(z)/J(z) has positive real part in K(O; 1). Show that the domain D f =/[K(O; 1)] is starlike (or starshaped) with respect to the origin, i.e. W E Df implies [0, w] c D f . Also prove the converse. 10.2.5. Using Herglotz's formula derive structural formula for IE S*. 10.2.6. Find precise estimates of ill and 11'1 on C(O; r) for IE S*. Hint: Exercise 10.2.5, or Exercise 10.1.8. 10.2.7. Find the region of variability Gz of [z/l(zWJ2 for a fixed z (0 I ranging over S*. /lint: Exercise ]0.2.5 and Exercise 10.1.5.
II nd
< Izl < 1)
10. UNIVALENT FUNCTIONS
130
10.2.8. If IE S*, show that rerJ(z)/z]1/2
t.
>
10.2.9. If I maps the unit disk 1: 1 conformally onto a convex domain B, show that any smaller disk K(O; r), 0 < r < 1, is also mapped onto a convex domain B,. Hint: Consider the function "p = 110 cp, where
cp(z) =
if(;:
assume 1(0)
z)+(1-t)/(Z),
zEK(O; 1) and 0
1, IZ11'::;:; IZ21;
= O.
10.2.10. Let SC be the family of all normalized convex univalent functions, i.e. lEse means that I is univalent in K(O; 1) ,f(0) = 0, f' (0) = 1 and I[K(O; 1)] is a convex domain. Show that lEse, iff zi' E S*. 10.2.11. If/ESc , show that tet'f'(z»+. 10.2.12. If Zl' Z2 EK(O; 1), show that
re[(Z2- Z 1)-l L Og
!-;:] > ~.
Hint: Integrate F(z) = (l+z)/(l-z) over the segment [Zl' Z2]. 10.2.13. If lEse, show that re [f(z)/z] > t. Hint: Using Exercises 10.2.11,10.1.3 find a formula expressingf' as a double Stieltjes integral. Integrate under the sign of integral and use Exercise 10.2.12. 10.2.14. If lEse, show that 11-zf(z)I'::;:; Izl, zEK(O; 1). 10.2.15. If IE SC, show that re[zi'(z)/I(z)]
Hint: Consider g(z) = [f'(C)(l-1C1 2)]-1
> t.
[/(C)-/( lC~;Z )l
10.2.16. If IE SC, show that: (i) (l+lzl)-l,::;:; Izf'(z)[f(z) I ,::;:; (l-IZ!)-l;
(ii) larg[zf'(z)/I(z)] I ,::;:; arcsin Izl. 10.2.17. Find sharp estimates of If(z) I and argrJ(z)/z] for IE Sc. 10.3. UNIVALENT FUNCTIONS
10.3.1. Suppose that r is a closed Jordan curve with parametric representation = R«()), (/J = (/J«()), 0'::;:; () ,::;:; 21t, where R, (/J are polar coordinates; R(O). tP«() are supposed to be continuous; moreover, tP«() is piecewise monotonic R
10.3. UNIVALENT FUNCTIONS
131
and the index n(r, 0) = 1. If h is a nonnegative continuous and increasing function of R E (0, +(0), show that 2"
~ h (R«())) dtP«())
> O.
o
10.3.2. Let f be analytic and univalent in the annulus {z: a < Izl < b} and let rr be the image curve of C(O; r) under J, a < r < b. If g is continuously differentiable and If(re i9 ) I = R(r, ()), argf(re i9 ) = tP(r, ()), show that 2"
~
2"
R(r, ())g'(R(r, ())) d 9 tP(r, ())
= r :,
o
~ g(R(r, ()))d(). 0
10.3.3. (cont.) If n(r" 0) = 1, a
2"
b, show that ~ If(re i9Wd()
IS
o
an
increasing function of r in (a, b). 10.3.4. Let 2: be the family of all functions F(z) = z+bo+b1Z-1+b2Z-2+ ... analytic and univalent in K( 00; 1). If r > 1, show that 00
r- .2.:nlbnI2r-2n-l
> O.
n=1 00
10.3.5. (cont.) Show that
2: nlb l2 ~ 1. n
This inequality is sometimes called
n= 1
the area theorem. 10.3r6. Let .Eo be the subfamily of .E consisting of all FE.E which do not vanish in K(oo; 1) and let S be the family of functions f(z) = z+a2z2+ ... analytic and univalent in K(O; 1). Find the relation between the functions of S and .Eo. Verify that z(1 +Z-3)2/3 E.Eo. 10.3.7. If F(z) = z+b o+b1z- 1 + ... and FEIo , show that Ibol ~ 2. Discuss the case of equality. Hint: Consider G(z) = YF(Z2) . 10.3.8. IfF E .E and does not assume the values W1' W2' show that Hint: Consider F(Z)-Wk' k = 1,2.
IW1-W21 ~
4.
10.3.9. If f(z) = z+a2z2+ ... , fE S, show that la 2 1~ 2. Also show that eq uality holds only for the Koebe function fr.(z) = z(1 +ei "z)-2. 10.3.10. If fE S does not assume the value h, show that Ihl equality holds only for f". /lint: Consider the development of rp(z) = hf(z)/(h-f(z)).
I hltt
> t. Also
show
10. UNIVALENT FUNCTIONS
132
10.3.11. Suppose that a, bE K(O; 1) and z = z(t) is the function mapping the unit disk {t: It I < I} onto the domain D(a, b) of Exercise 2.9.20 so that z(O) = a. If f E S, verify that also
cp(t)
=
[f(z(t»-f(a)] :
[~ f(z(t» 1=0
belongs to S. Deduce the inequality
, [ a-b [II-libl II-abl-Ia-bl If (a) I f(a)-f(b) I-lbI2' II-abl+la-bl
< 1.
10.3.12. (cont.) Find the following estimates: (i) the upper bound of logarithmic derivative 1f'(a)!f(a)l; (ii) the upper and lower bound of If(a)l; (iii) the lower bound of logarithmic derivative. Hint: Consider in (iii) the mapping
1p(t)
=
J
[f( ::;t )-f(Z) [f'(z) (I-lzI2)]-1.
Verify that equality can be attained in each case. 10.3.13. If f
S, show that (I-Izl) (l+lzl)-3
E
< If'(z) I < (1 + Izl) (I-lzl)-3.
Verify that in either case equality can be attained by a Koebe function. 10.3.14. Let a, b (a 1= b) be two fixed points of K(O; 1). Find the lower and the upper bound of the ratio If'(a)jf' (b) I for f ranging over S.
Hint: Consider cp(C)
=
f( C+b ) I+bC .
10.3.15. Find for FE Eo an inequality analogous to the inequality of Exercise 10.3.11.
10.3.16. Find for FE Eo sharp estimates of: (i) !F(z)l; (ii) IzF'(z)jF(z)l. 10.3.17. If FEE, F(z) = z+bo+b1z- 1+ ... , show that (i) lim IF(z)-bol < 2; Izl-->l+
(ii) 1.F(z)-bo-zl
< 3Izl-1.
10.3.18. If FEE, show that: (i) IF'(z)-II < GzI 2-1)-1; Hint: Use the area theorem and Schwarz inequality; (ii) IF'(z)1 < IzI2jGzI 2-I).
10.4. CIRCULAR AND STEINER SYMMETRIZATION
133
10.3.19. If F(z) = z-(h-h- 1 )(hz-l), h> 1, show that FEE and verify that for z = h the equality in Exercise 10.3.18 (ii) actually holds. Find the image domain of K(oo; 1). 10.3.20. Suppose that f for all z E K(O; 1). Show that (i) la21
=
-}If''(O)1
E
S and there exists a constant M such that If(z) I <: M
<: 2(l-M-1);
(ii) -fM(-lzl) <:If(Z)1 ~fM(lZI), wherefM is the Pick function which maps K(O; 1) onto K(O; M) slit along a radius from -M to -M[2M-I-2J!M(M-l)]. Hint: Derive first the inequality (ii) by considering g(z) = f(z)[1+e i <XM-1f(z)]-2; w = fM(Z) satisfies the equation z(l-z)-2 = w(l-w/M)-2. 10.3.21. If f
E
Sand 1; E K(O; 1), show that the function g(z)
f'(1;)(I-ICI 2) [f(
=
~:~C )-fmr
1
belongs to the family E. Find the initial coefficients in the Laurent development of g and verify that l{f, !;}I ~ 6(1-ICI 2)-2, where {f, !;} is the Schwarzian derivative of f (cf. Ex. 7.2.14). Verify that the same inequality also holds for an arbitrary function meromorphic and univalent in K(O; 1). 10.4. THE INNER RADIUS. CIRCULAR AND STEINER SYMMETRIZATION
The definition of the inner radius r(zo; G) of a simply connected domain G in the finite plane (cf. Exercise 8.1.8) can be restated in terms of Green's function g(z, zo; G) as follows: (lO.4A)
logr(zo; G) = lim[g(z, Zo; G)+loglz-zol]. Z-+Zo
The limit on the right-hand side exists for any bounded domain regular with respect to the Dirichlet problem. If G is an arbitrary domain in the finite plane, there exists an increasing sequence {G n } of bounded domains, each regular with respect to the Dirichlet 00
problem, such that G
=
U
G n • Each point Zo of G belongs to Gn for all n suffi-
n=1
dcntIy large and one can prove that the increasing sequence {r(zo; Gn )} has always a limit, finite or infinite, which does not depend on a particular choice of {ON}. This limit may be assumed as the inner radius of G. Moreover, if r(zo; G)
10.
134
U~IVALENT
FUNCTIONS
is finite at a certain point Zo E G, it is also finite at any other point of G (cf. Exercise 10.4.4). Circular (or P6lya) symmetrization of an open set G in the extended plane is determined by a fixed ray A with the origin at a and associates with G a symmetrized set G* which is defined as follows. 1° If G contains a, or 00, or both, so does G* and conversely. 2° If C(a; r) c G, then C(a; r) c G*, and conversely; if C(a; r) n G = 0, then C(a; r) n G* = 0, and conversely. 3° If the intersection C(a; r) n G is neither empty, nor coincides with C(a; r) and has the total angular Lebesgue measure I(r) , then C(a; r) n G* consists of a single open arc bisected by A and subtending at a an angle equal to I(r). Steiner symmetrization is determined by a straight line which we take to be the real axis in the complex plane. Let a(x) be the straight line {z: rez = x}. If G is an open set in the plane, then the symmetrized set G* is defined as follows. 1° If a(x) n G = 0, then a(x) n G* = 0. 2° If a(x) n G has the linear Lebesgue measure l(x) , then a(x) n G* is a single, open segment of length lex) bisected by the real axis. If G is a domain, so is G* for both Steiner and circular symmetrization. Also the area of G and G* is the same. As shown by P6lya and Szeg6, for any point Zo situated on the line (resp. on the half-line) of symmetrization and any domain G containing zo, we have: r(zo, G) ~ r(zo, G*). 10.4.1. Show that both Steiner and P6lya symmetrization carries a domain G into a domain G*; if G is simply connected, so is G*.
10:4.2. If G is starshaped with respect to the origin and the ray A of circular symmetrization emanates from the origin, show that also G* is starshaped with respect to the origin. 10.4.3. Show that Steiner symmetrization preserves the convexity of domains. Give an example showing that circular symmetrization does not have an analogous property. 10.4.4. Iffis univalent in the unit disk K(O; 1) and the area of the image domain f[K(O; 1)] is finite, show that
lim (1-lzOIf'(z)1 Izl..... '
Hint: Cf. Exercise 4.2.12.
=
O.
10.4. CIRCULAR AND STEINER SYMMETRIZATION
135
10.4.5. If G is a bounded, simply connected domain, show that the inner radius r(w; G) is a continuous function of w which tends to zero as dist(w; frG) -+ Hint: Cf. Exercise 8.1.9.
o.
10.4.6. Let G be a bounded domain, regular with respect to the Dirichlet problem, and let h(z, zo) be the solution of the Dirichlet problem with boundary values log Ie -zol. If ZO, ZI E G, Zo =F ZI and dist(zk; frG) ~ tJ for k = 0, 1, show that (i) Ih(z, zo)-h(z, zl)l < IZI-Zol/tJ for any z E G; Hint: Use the maximum principle and the inequality logll+zl ~ Izl; (ii) Ilogr(zo; G)-logr(zl; G)I < 2Iz1 - Zol/tJ . 10.4.7. If the inner radius r(z'; G) is finite for some z' E G, show that it is also finite for any z" E G. Show that r(z; G) is a continuous function of z. 10.4.8. If H is an ellipse, show that r(z; H) has a maximum at the center of H. 10.4.9. Let C§ = C§(r, R) be the class of domains G, K(O; r) c G c K(O; R), regular with respect to the Dirichlet problem and such that each circle ceO; p), r":;; p ,,:;; R, meets C"",,G. Find supr(O; G). G
10.4.10. Let C§ = C§(R), R > 1, be the class of domains G, 0 E G c K(O; R), such that r(O; G) = 1. Find the least possible distance tJ of the point w = 0 from the component of E"",, G containing the point at infinity. 0,
10.4.11. Let C§(wo) be the class of simply connected domains containing Wo and satisfying r(O; G) = 1. Show that sup{r: r = r(w o; G), G E C§(wo)} ,= 1+4Iwol.
10.4.12. Suppose that G is a simply connected domain of hyperbolic type (i.e. G can be mapped 1: 1 conformally onto a disk) and G* is the domain obtained hy circular symmetrization. Show that G* is also of hyperbolic type. Show that Steiner symmetrization does not have this property. 10.4.13. If G is of hyperbolic type, WI, W2 E G (0":;; WI < W2) and G* arises rrom G by circular symmetrization with respect to the real axis, show that 1/( 1111 , W2; G*) ,,:;; P(WI' W2; G) where p denotes the hyperbolic distance (cf. lix. 8.5.11). Hint: Show that p(wI ,
W2;
G)
'=
~ r~~), where r is the image of the h-seg-
r r w,
lIIent under the 1: 1 conformal mapping of K(O; 1) onto G.
10.4.14. (cont.) If g denotes the Green function, show that g(Wl. w2; G*) ~g(WI' w2 ; G).
10. UNIVALENT FUNCTIONS
136
10.4.15. If J maps the unit disk 1: 1 conformally onto G and G is invariant under circular symmetrization with respect to the real axis, moreover, J(O) = 0, ['(0) > 0, show that M(r,f) =J(r). Hint: If Wo =J(zo), then g(O, Wo; G) = -loglzol. 10.4.16. Suppose thatJ,f* are analytic and univalent in the unit disk and the image domain G* = J*[K(O; 1)] contains the origin and arises from G = J[K(O; 1)] by circular symmetrization with respect to the positive real axis. If J(O) =1*(0) = 0, show that M(r,f) ~ M(r,f*) for all r E (0, 1). 10.4.17. Suppose that JE S and Df = J[K(O; 1)]. Verify Koebe one quarter theorem (Ex. 10.3.10) by considering r(O; Df ). 10.5. THE METHOD OF INNER RADIUS MAJORIZATION
10.5.1. Let D be a domain possessing the classical Green function g( w, ao ; D) = g(w) and let ff(a o ; D) be a family of functions analytic in the unit disk and such that J(O) = ao , J[K(O; 1)] = Df c D. Prove that 1[,(0)1 ~ r(a o ; D) for any JEff(a o ; D). Hint: Consider h(z) = g(f(z»)+loglzl, z EK(O; 1). 10.5.2. (cont.) Prove an analogous theorem for arbitrary D. Hint: Consider D(P) = J[ K(O; p)] . 10.5.3. Prove following symmetrization principle. Let J be analytic in the unit disk and let J(O) = ao . If the symmetrized domain D* of Df = f[K(O; 1)] with respect to a line (Steiner symmetrization), or a ray through ao (circular symmetrization) is situated in Do, show that 1[,(0)1 ~ r(ao ; Do). 10.5.4. Suppose that 0 < oc < 2 and that the function J(z) = aO+alz+ ... is analytic in K(O; 1) and Df n C(O; r)is for any r > 0 a system of arcs of total length 7tocr at most. Show that lall~~ 20c laol, with equality holding for
( I+Z)" .
f(z) = ao l-z
10.5.5. If J(z) = aO+alz+ ... is analytic in K(O; 1) and R = R f is the least upper bound of all r > 0 such that C(O; r) c Df , show that lall ~ 4(la ol+R) with equality when ao ? 0 and J(z) = ao+4(a o+R)z(l-z)-2. 10.5.6. If J(z) = z+a 2z 2+ ... is analytic (not necessarily univalent) in the unit disk, show that for any 8 > 0 there exists r > {--8 such that C(O; r) c Df . 10.5.7. Deduce Koebe one quarter theorem from Exercise 10.5.6.
10.5. THE INNER RADIUS MAjORIZATION
137
10.5.8. Suppose that/(z) = aO+alz+ ... is analytic in the unit disk and that there exists h > 0 and a real-valued function v(u) , - 00 < u < + 00 such that on any straight line re w = u each point u+i(v(u)+nh) (n = 0, =fl, =F2, ... ) belongs to C,,-Df . Show that lall :;:;:; 2h/1t. Hint: Consider g(z) = exp {27tf(z)/h}. 10.5.9. Suppose that any circle C(O; r), r > 0, contains a point of and I is analytic in K(O; I). Show that (1-lzI2) If'(z) I :;:;:; 4 I/(z) I
C',Pf
for any z E K(O; 1). Hint: Exercise 10.5.5. 10.5.10. If I is analytic, univalent and does not vanish in the unit disk, show that l- lzl )2 ( 1+IZI)2 1/(0)1 ( 1+lzl :;:;:; I/(z) I :;:;:; 1/(0)1 l-lzl . 10.5.11. If cp is analytic in the unit disk and never assumes both values w, -w, show that (1-lzI 2)lcp'(z)1 :;:;:; 2Icp(z)l· 10.5.12. Suppose that I is a Bieberbach-Eilenberg function, i.e. I is analytic in the unit disk, 1(0) = 0 and I(Zl)f(Z2) '" 1 for any Zl' Z2 E K(O; 1). Show that (i) I is bounded in the unit disk; (ii) (1+/)/(1-/) never assumes both values w, -w. 10.5.13. Show that for any Bieberbach-EiIenberg function I we have: (i) (1-lzI2) If'(z) I :;:;:; II-P(z)l; (ii) 11'(0)1 :;:;:; I. 10.5.14. If IE Sand L(r, f) is the angular measure of values omitted by I and situated on C(O; r), then L(r, f) is equal to zero for r :;:;:; -} and any IE S and can be equal to 21t for r;;:;:: 1 (e.g. for I(z) == z). If r E (!f, 1), show that L(r)
=
sup L(r,J)
=
4 arc sin
(2vr -1).
/eS
Show that the extremal function maps K(O; 1) onto
C",({w: Iwl
=
r, largwl :;:;:; 2arcsin(2vr
-In u
[r,
+(0»).
"jnd the extremal function (cf. Ex. 2.9.22). 10.S.15. Solve an analogous problem for starlike univalent functions.
10. UNIVALENT FUNCTIONS
138
Hint: Verify first that the mapping
W=F(t)=(I+Vt)1J I-OV t + t , I-Vt I+OV t + t
0<0<2,
carries 1: 1 conformally the upper half-plane im t > 0 onto the domain H(O) = Iw: Iwl < I} u Iw: 0 < argw < On}. Next find the mapping J: K(O; 1) -+ H(O). Using Exercise 10.4.2 show that J(z)/f'(O) is the extremal mapping.
SOLUTIONS
CHAPTER 1
Complex Numbers. Linear Transformations 1.1.1. t+ti; 64; i; 2+i2
Y3.
1.1.2.0, 1, -t=t=itYI 1.1.4. (i) z is real; (ii) z 1.1.5.
=
cosO+isinO, 0 is real.
C= y~ [(X+yx2+y2)1/Z+iy(x+ yx2+y2tl/2].
1.1.6. Use the identity la+bl 2 = (a+b)(a+b); the sum of squares of all sides in a parallelogram equals the sum of squares of both diagonals. 1.1.7. Square both sides and use Exercise 1.1.6. 1.1.R. Use the identity la+bl 2 = (a+b)(a+b). 1.1.9. The left-hand side is equal to
{exp [i(n+I)O]-l}: (eiB_I)= {sint(n+I)Oexp[t(n+I)Oi]}: (sintOexp-}Oi). 1.1.10. ZlZ2+Z2Z3+Z3Z1 = ZlZ2Z3(Z1+Z2+Z3) = 0, X (z-Z3) = z3_ a with lal = 1.
L Xl X'X m =
0, hence
n (Z-Zk) = (z2-a 2)(zZ_b2),
where
1.1.11. We have as before
..
I
hence (Z-Zl) (Z-Z2) X
lal = Ibl = 1.
i=/
1.1.12. ZlZZ = x1xz+YtYz+i(x1Yz-Y1X2)'
1.1.13. The area is equal to t[Zl' zz]x [Zl. Z3] = t im (zz-zl) (Z3- Z1) = t(Z2Z3+Z3Z1+Z1Z2)
Mince ima
= im(-a).
1.1.14. Induction and Exercise 1.1.13. 1<11
SOLUTIONS
142
1.1.15. cosO( = IZ2-Z11-1im(ZlZ2+Z2Z3+Z3Z1), cf. Exercises 1.1.12, 1.1.13. 1.1.16. Take C(O; 1) as the circum-circle and use Exercise 1.1.10. 1.1.17. (i) Logarithmic spiral; . (ii) the circle C( with the point
t; -}) Z = 0 removed; (iii) ellipse with semi-axes la=Fa- 11; (iv) cycloid; (v) involute of C(O; a); (vi) cardioid; (vii) the right-hand half of parabola y = x 2 ; (viii) a branch of hyperbola xy = 1.
1.1.lS. If Ix'I+ly'l
> 0, then
z'(t) is the tangent vector.
= arg [z'(O)/z(O)] = arc cot [r'(O)/r(O)]. 1.1.20. (i) If x =F 0, then argz = arctan(y/x); (ii) Izl = 1.1.19.
0(
YX2+y2.
1.1.21. (i) The line of symmetry of [a, b];
(ii) closed ellipse with foci =Fe; (iii) a part of the right half-plane outside C(I; y'2); (iv) a strip bounded by the real axis and the line y = -1 with the, real ~xis included; (v) the domain to the right of the right-hand branch of equilateral hyperbola
x 2 _y2 =
0(;
(vi) the left half-plane; (vii) the closure of the domain to the left of parabola y2_1 = -2x; (viii) the interior domains of both loops of a lemniscate; (ix) the union of it part of the exterior of C(i; y2) contained in the right half-plane and of a part of K(i; V2) contained in the left half-plane. 1.1.22. If z = kZ+l, a = k+l, b = -k+l, then the given set becomes {Z: arg(Z2-I) = const} which is an equilateral hyperbola center at the origin. 1.1.23. R
>
Ib-ia21.
1.1.24. Circle C(BA-l; IAI-1VIBI2-AC). 1.1.25. The radius: kla-bIII-k 21- 1; the center: (a-k 2b)(I-k2)-1. 1.1.26.
Izl2 IZl12 IZ212 IZ312
z Z Zl Zl Z2 Z2 Z3 Z3
1 1 =0. 1 11
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
143
1.1.27. Cf. Exercises 1.1.26, 1.1.24. 1.1.28. (i) If m 1 = 0, then Zo E [Z2' Z3]; if m 1 = 1, then Zo = Zl; if m1 =f. 0, 1, then Zo = m1z1+(1-m1H'1' where C1 = (l-m1)-1(m2z2+m3z3) E [Z2, Z3] and therefore Zo E [Zl' Cd E T. (ii) If the change of mj by hj gives the same point zo, then h1Z1+h2Z2+h3Z3 = 0, or h1 (zl-z3)+h2(z2-z3) = since h1 +h2+h3 = 0. This means that z} are collinear.
°
1.1.29, 30. Induction with respect to n and ExerCise 1.1.28. 1.1.31. If ft =
L" IC- Zkl- 2,
mk = ft- 1!C-Zkl- 2, then mk
k=l
and
L" mk(C-zk) = 0,
k=l
or C =
L"
k=l
>
0,
L" mk =
1
k=l
mkzk'
1.1.32. If P(z) = A(Z-Zl) ... (z-z,,), then P'(z) =_1_+_1_+ + __ 1_. P(z) Z-Zl Z-Z2 ... z-z" '
(cf. now Exercise 1.1.31). 1.1.33. logz" = ;
[2; +0 (:2)] ~
x, or Iz,,1
~ eX;
argz" = narctan(y/(n+x)) ~ y. 1.1.34. The sequence is divergent; tz,,1 2
~ R2 =
Ii (1 + -\-), n
,,=1
however,
. a term 0 f a d'Ivergent series; . . 0 f C(O ; R) arg -z"- = arctan -1 IS every pomt Z,,_l n is an accumulation point for {z,,}.
1.1.35.
I
b"C" is absolutely convergent, since {e,,} is bounded.
1.1.36. The convergence of I Z" implies the convergence of I x,,; since .\'" ~ 0, is also convergent. The convergence of ~>; follows now from the l'OIlVergence of both I
Ix;
Ix;, z; .
1.1.37. Consider first the case
lime"
=
°and show using (i), (ii) that limz"
1.1.38. Put a"k = Pk/(P1+P2+, ... , +p,,) for k,;;;; n, and a"k apply Exercise 1.1.37. 1.1.39. Express
W"
by means of
Z"
and use Exercise 1.1.37.
=
0 for k
= O.
>
n;
144
SOLUTIONS
n
1.1.40. Express lVn in terms of
L
Sn =
Zk
and use Exercise 1.1.37.
k=l
1.1.41. Put p. 1.1.42. If v
=
#;;1 and cf. Exercise 1.1.40.
1 0, consider the sequence lVnlv; note that nv
=1=
1 VI' 0, 0, '" are lines of a matrix of Exercise 1.1.37. If v nv by Vn = g+v n with g =1= O.
•.. , Vn
1.2.1. Use the parametric equation of the straight line NA: X
Z
=
1 nv
V n,
V n_ l , •..
0, replace
=
= Xl t,
Y
=
x 2 t.
I+(x3-I)t.
1.2.2. (coscx; sin'cx; 0), (--j-; -j-; i-), (,;; - 1~;
!;-).
1.2.3 .•C'"K(O; 1); K(O; 1). 1.2.4. Note that all projecting rays are situated in one plane through N and the given straight line. 3
1.2.5. The points of S situated on a circle satisfy:
L
AkXk+B =
0; use now
k=l
Exercises 1.2.1, 1.1.24.
1.2.6. Use the formulas of Exercise 1.2.1: 1.2.7. (i) opposite points on the same parallel; (ii) points on the same parallel symmetric w.r.t. the plane' OXIX3; (iii) opposite poi~ts on the circle Xl = C, x~ +x~ = 1- C 2 • 1.2.8. Put
Xl =
cos()cosrp etc. into the formulas of Exercise 1.2.1.
1.2.9. Cf. Exercise 1.2.1. 1.2.10. Note that the equation of the stereographic projection of the great circle is identically satisfied by -z-I·(Ex. 1.2.9). 1.2.11. The equations Iz-zol2 = I+lzoI2, zz-zoz-zoz-I = 0, are equi. valent (cf. Ex. 1.2.10). 1.2.12. C(zo; R), Zo
=
-14-
2;
i, R
=
-}
VI5I7.
1.2.13. Apply the formula for the distance of two points and express their coordinates by means of formulas of Exercise 1.2.1. da 1.2.14. -d S
=
. a(zl z) lim '2 (cf. Ex. 1.2.13); hence da 2 Zl--+Z
IZ1-zl
this implies preservation· of angles.
=
2
2
A(z)(dx +dy ) alld
f. COMPLEX
~UMBERS.
LINEAR TRANSFORMATIONS
1.2.15. O'(z, a) = 0'([;, a), O'(z, _a-I)
=
145
0'([;, -a-I), hence
+;[; 1= I ;+~z I;
11C
moreover, z and [; are situated on circles througli a, _71- 1 intersecting at an angle cp, hence
C-a C+a- l
arg
arg=---
1.2.16. IZ-alllz-a21-1 with limit points ai' a2' 1.2.17.
t
O'(lal-r,
z-a z+a 1
=
cpo
(1 + laI1 2)1/2(1 + la 212)-1/2 which is Apollonius circle
=
lal+r).
1.2.18. Evaluate e.g. the length of inscribed polygonal line using Exercise 1.2.13 and consider its limit for a normal sequence of partitions. =
1.2.19. The stereo graphic projection of a rhumb line intersects all rays argz const at a constant angle hence it must be a logarithmic spiral (cf. Ex. 1.1.19). 1.2.20. The equation of stereographic projection is r
r I ek6 , k
=
= ~ log ~ , rJ.
.
rl
hence l(r)
=
Jr
22 yl+k- 2 l+r2 dr
--
=
2 j/l+k- 2 (arctanr 2-arctanrl )·
"
The numerical example corresponds to arc tan r1 = 'Ttj6, arctanr2 l(r)
1.2.21. d0'2
=
=
;
[I + ( 61~3
4(dx 2+ dy2) (I + Iz12)2 ' hence IDI
=
rr
,
Tt/3, hence
/2 •
~~ YEG-Pdxdy = ~~ (l+~ZI2)2 LI
=
dxdy.
LI
1.2.22. Without loss of generality we may take the south pole as one of the I'ertices of T. The stereographic projection To of T is bounded by two straight line segments emanating from the origin and an arc of a circle: z = zo+ I ~/ 1-=f-lzoI2e''P, CPI::( cP ::( CP2 (cf. Ex. 1.2.11). Using the formula of Exercise
1.2.21 in polar coordinates we obtain:
~ ~~ (l-~;'iF drdfJ = ~2 1~:i dfJ;
ITI ..
Tn
6,
146
SOLUTIONS
we now introduce a new variable g; putting rei8
,2 =
=
zo+Yl+lzoI2ei'l'. We have:
1+2IzoI2+2 yl+lzoI2(xocosg;+yosing;),
dO d (./ .. dg; = dg; arg zo+ V 1+l z ol2 el'l') =
where xll+iyo
=
r- 2 [1
+ Izol2+yl + IZoI2(xocosg;+Yosing;),]
zo, (cf. Ex. 1.1.20). This gives: '1'2
ITI = ~ dg; = g;2-g;1 = (oc+p+y)-7t, '1'1
which follows from elementary geometric considerations. 1.2.23. The stereographic projection of T is bounded by arcs of C(O; 1), C( -1 + i; y3); hence ITI =
7t
2
[i, i(l+v'Z)] and
.r 1 +2arctan V 2-7t = arctan 2 {i.
1.3.1. The angle of rotation: arga; the ratio of homothety center at the origin: lal· 1.3.3. W = 1ti(z-t). 1.3.4. (i) w = az+b, a> 0, b real; (ii) w = az+b, a < 0, b real; (iii) w = -i(az+b), a> 0, b real.
1.3.5. w =
(b-a)-l[(B-~)z+Ab-aB].
1.3.6. w = (l+i)(l-z). 1.3.7. w =
b:~~l
1.3.8. Zo
b/(I-a); the angle of rotation: arga, the ratio of homothety:
=
(z-ib2)·
lal·
1.3.9. (i) Zo = t(3+i), arga = --t 7t , lal = l/i; (ii) Zo = b2(l-ik)/(b 2-b1-k-i), a = (k+i)/(b 2-b1). 1.3.10. It is sufficient to consider the case W = aZ (cf. Ex. 1.3.8); if Z(O) AlaI 8/"e i8 , -CXJ < 0 < +CXJ, A > 0, arbitrary, and oc = arga ¥: 0, then evidently aZ(O) = Z(O+ oc). Hence the spiral Z = Z(O) remain unchanged after the transformation W = aZ. =
I. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
1.4.1. If e ¥= 0, then w =
be-ad e
147
a -1 Zl +-, ZI = Zz , Z2 = ez+d. e
1.4.2. If z(O) = r(0)ei6 is the equation of y, then the image curve has the equation w(O) = (r(0)t1e-16. The angle oc. between y and the radius vector satisfies cot oc = r'(O)/r(O); a corresponding value for the image curve after taking the opposite orientation is the same. 1.4.3. Use Exercises 1.4.1 and 1.4.2. 1.4.4. The equality is equivalent with the equality [w-a(laI 2-r 2)-1] [w-aClaI 2-r 2)-I] = r2(laI 2_r2)-2; the image curve of CCa; laD is a straight line. 1.4.5. Use Exercise 1.4.4 with a = bi, r2 = b 2+1. 1.4.6. Use Exercises 1.4.1 and 1.4.4. 1.4.7. Verify that w = az+b, w = Z-l do not change the cross-ratio and cr. Exercise 1.4.1. b-e z-a 1.4.8. w = - - ' - - . b-a z-e 1.4.9. (i) Straight lines rew = (2a)-1; (iii) circles b(U 2+V2)+U+v = 0; (iii) straight lines v = - ku; (iv) circles through w = 0 and w = ZOl; (v) cissoid u2(v+l)+v 3 = O. 1.4.10. Linear transformation defined by the equation (ZI' Z2' Z3' z) c (a, b, e, w), where a, b, c are real and different from each other carries the l'ircle (or possibly the straight line) C determined by Zl' Z2' 'Z3 into the real axis. Since the r.h.s. is real iff w , is peal, so is the l.h.s. iff z E, C. 1.4.11, 12. Verify that identity, inverse transformations and superposition are also transformations of the same type. 1.5.1. The equation satisfied by z* is equivalent to the conditions: Iz-al .: Iz*-al = r2, arg(z-a) = arg(z*-a). 1.5.2. If Z = a+re i6 , z = a+pe''l', then IZ-z*IIZ-zl-l
= /re l6 -
~
el'l'!lrei6_pei'l'l-l
= ;.
X
SOLUTIONS
148
1.5.3.
-t+i.
1.5.4. (i) The straight line re z = t; (ii) lemniscate (.~2+y2)2 = x 2_y2. 1.5.5. If Zo is the center of the circle orthogonal to C(O; 1), its radius is JI Izo12_1. Its reflection satisfies (C- 1-zo)(C- I -zO) = IZoI2-1, which is equivalent to (C-zo)([-zo) = IZo12-1. 1.5.6. This is an immediate consequence of the fact that symmetry is defined in terms of circles and their orthogonality and both properties are preserved under linear transformations.
1.5.7. z*
=
1.5.S. If Lk
(02-o1)[(a2-al)z+a1 a 2-a1a 2]. C(ak; rk), k = 1,2, then
=
Z
=
a2+d(z-al) [(a l -ti 2) (z-al)+rtr1.
Hence (z = a1 ) +-+ (Z = a 2 ). If the same holds after interchanging the order of reflections, then rr+d = lal-a212 which shows that L 1, L2 are orthogonal. The sufficiency is easily verified after an additional linear transformation carrying L 1, L2 into perpendicular straight lines. 1.5.9. An immediate consequence of Exercise 1.5.1. 1.5.10. The roots of the equation. 2 .
a+ _r_ z-a
=
(a2-al)-1[(a2-al)Z+ala2-aI02]'
1.5.11. Circles through -7=f4V3 which are symmetric points with respect to both circles. 1.6.1. (i) The lower half-disk of K(O; 1); (ii) K(O; 1) n (C",K(-i-i; i-)); (iii) {w: imw < O} n {C"-K(t(1-i); +V2)}; (iv) K(t; i-),,-K(i-; -t); (v) {w: rew > i-} n (C,,-K(-t; t)). 1.6.2.
IV
= (z-4)(z-I)-1.
1.6.3. If A (w, b, c, d)
=
(b-a)(d-c)(c-b)-l(d-a)-t, then k
=
2A+I-2 JlX(A-I-1);
= (z, -1, 1, k- 1).
1.6.4. If a, b are real and (z = a) w
+-+ (w = b),
= (a+l)bz[(a-b)z+a(b+l)tl,
then ab(a+l) (b-H)
> O.
f. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
1.6.5.
C""K(-f; 1); {w: imw < O}
n
149
[C",K(-}; 1)].
1.6.6. z = (1-2w) (1+W)-1. 1.6.7. Z= i(w+1)(w-1)-1; Apollonius circles Iw+11/lw-11
1.6.S. If (z = a)
~
(w = 0), lal
=
r.
< 1, then
w = ei
-(1+i) (w-i)/(w+1-2i).
=
1.6.10. The points symmetric w.r.t. either circle are =f3, (cf. Ex. 1.5.9). Hence w = 2(z+3)/(z-3), R = 4. 1.6.11. The points symmetric w.r.t. either circle are:
Zl
=
f(3+i), Z2
=
w = 2- 1/2(2z-3+i)/(z-2+i),
2+i;
R
=
Y3/2.
1.6.12. w = (z-yh 2-R2)/(z+l/h 2 -R 2 ); P is found by putting z 1.6.13. w 1.7.1. w
=
=
=
h-R.
a(z-2)-1+b, a is real. (az-fY.(J)/(z+a-fY.-(J), a#- fY., (J.
1.7.3. ewz+wd-az-b
== czw+dz-aw-b == 0, iff a+d = 0 since w #- z.
1.7.4. If e = 0, then w = -z+A and fA, 00 are invariant points, if e #- 0, then the discriminant of the quadratic equation for invariant points: ez2-2az-b = 0, i.e. a2+be = be-ad #- O. 1.7.5. W
0,
00
=
W(Z), where W
W-fY. Z w-(J'
= --
Z-fY. has two invariant points z-(J
=
---
(cf. Ex. 1.7.2).
1.7.6. A
=
1.7.7. A =
W-fY. z-(J w- (J Z-fY.
=
w'(fY.)
f( -1+iy3), fY.
=
=
a+d+yLf a+d- Y LI .
f(1-{3)(1+i), f3
=
f(1+Y3)(1+i).
1.7.S. The equation ez 2+(d-a)z-b = 0 for invariant points has no finite roots, iff e = 0, d = a, b #- 0, i.e. w = z+h; if W = (w-fY.)-t, Z = (z-fY.)-t, I hell necessarily W = Z h, since 00 is the only invariant point.
+
1.7.9. (R-W)-l
=
(R-Z)-l+ih, h is real.
(.7.10. The angle subtended by [fY., f3] at z II' E
C% we have:
7t
E
Cz is equal to =fT' hence for
SOLUTIONS
150
ffJ
note that
2R
=
W-IX
=
arg
11X- f3llcosec
w-f3
'It
argA =fT
=
71;
=
0 =f2";
(0 =F ; )1.
1.7.11. Consider the sequence Zn = (zn-IX)-l; if the trarisformation is parabolic and Zn = (zn-IX)(zn-f3)-t, it has two finite, invariant points IX, f3. In the former case Zn -+ 00, or Zn -+ IX (cf. Ex. 1.7.8); in the latter case the behavior of {zn} depends on the constant A in the canonical representation of Exercise 1.7.5. If IAI 1= 1, {zn} is convergent to one of the invariant points, otherwise {zn} is divergent. . d'Ivergent; A 1.7.12. T h e sequence {Zn } IS
=
--l+ill3 -2--
. 0, - 1, t h e pomts
- i are points of accumulation of {zn}.
1.7.13. If A is real, the circle through IX, f3 is mapped onto itself; if A = ei'P the circles of Apollonius with limit points IX, f3 are mapped onto themselves. 1.7.14. Put W = (W-IX)-t, Z = (Z-IX)-l and consider the family of straight lines parallel to the vector h (cf. Ex. 1.7.8), which remain invariant under the transformation W = Z+h. 1.7.15. If 0( ¢ C and IX* is its reflection w.r.t. C, then IX* is invariant, too. However, there are at most two invariant points, hence f3 = Ii*. 1.7.16. If C is mapped onto itself, then either IX, f3 are symmetric, or both lie I W-IX on C. In the former case !, w- f3
means that
1
=
1
z-p I for
Z-IX
I
w,
Z E
W-IX
.
C (cf. Ex. 1.5.2), whtch Z-IX
IAI = 1. In the latter case arg w-f3 = arg z-f3
(w,
Z
E
C)
which means that A is real. 1.7.17. The discriminant of the quadratic equation for invariant points equal to (a+d)2-4 (cf. Ex. 1.7.6, 1.7.5).
IS
1.7.18. Cf. Exercise 1.2.15. 1.7.19. We find h such that lX+h = a; f3+h = _12- 1 which is possible, when IP-O(I = la+a- 1 1 = lal+lal- 1 ;? 2. The elliptic transformations with invariant points a, -12- 1 correspond to rotations of the sphere (cf. Ex. 1.2.15). 1.7.20. If
+2ihz+b =
2ih, invariant points 0(, P are roots of the equation bz 2 I 0, and consequently O(P = -1; A = (a+a+2i v'jbJ2~~/1-2)/(a-Hl
a-a
=
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
151
\..
-2tVlb\2+h2) = ellf', where cp is the angle of rotation of the sphere. Hence the transformation has the form as in Exercise 1.2.15 with
cp
=
2arg(rea+iVl b2 1+(ima)2).
1.7.21. (w-I-i)/(2w+I+i) 1.7.22. w
=
=
exp(i7t/3)(z-I-i)/(2z+l+i).
(ihz+b)/( -liz-ih) , h is real, a is purely imaginary (cf. Ex. 1.7.20).
1.7.23. Since w, z may be interchanged, A2
From W = (az+b) (cz-a)-l we have +(ct+P)
=
1 and consequently A = -I.
=!!-.-
which is the pre image of c W = 00. Now, ct, fl, +(ct+P) , 00 are situated on one straight line, their images being ct, fl, 00, ct+ p), resp., which means that the straight line considered is mapped onto itself.
+(
1.7.24. The straight line through the invariant points +C1=y3-i), as well as the symmetry line of the segment whose end points are the invariant points (cf. Ex. 1.7.15). 1.8.1. If Zl 1= 0, the circle through Zl' Z2 which intersects C(O; I) at a right angle, also contains Z1 1 , hence it is uniquely determined. 1.8.2. It is determined by three different points: Zl' Z1t, e ilZ . 1.8.3. L meets C(O; I) in two points e ilZ , e iP ; the pairs {Zl' eilX }, {Zl' e'P} determine two h-lines L 1 , L2 parallel to L. Each h-line through Zl situated between Ll and L2 does not meet L.
1.8.4. C(O; I) remains unchanged, hence zo, Zo 1 are invariant points and the corresponding linear transformation must be elliptic (cf. Ex. 1.5.2, 1.7.13, 1.7.15). Hence (w-zo) (w-zo1)-1 = e 8 (z-zo) (Z-zol)-t, () is real. 1.8.5. If Zl' Z2 are invariant points, IZll = IZ21 = I, the linear transformation must be hyperbolic because C(O; I) is mapped onto itself (cf. Ex. 1.7.13, 1.7.15). Hence (w-z 1)/(W-Z 2) = k(Z-Zl)/(Z-Z2) with k real. If z = 00, then k-~ (WO-Zl)/(WO-Z2) is real and Wo is outside [Zl' Z2], hence k > O. This mndition is also sufficient. 1.8.6. zt!(w-z 1 ) = Zl/(Z-Zl)+ih,
with real hand Zl = 1 (consider W
zd(z-zd). e'P(z-a)/(I-az)-l, with real fl and lal < 1. Hence z = e-'P X -- (II' I-ael/i)/(I+awe-//l), thus the inverse transformation is again an h-motion
I.S.7.
IV
=
SOLUTIONS
152
If w = e'Y(Z-b)/(I-bZ), Z = eifJ(z-a)/(I-az), the superposition has the form: w = exp {i({J+ y)} (1 +abe-' fJ) (z-e) [(I+abe ifJ ) (l-cz)]-1,
where e = (a+be-ifJ)/(I+abe- 1fJ ). Note that lei < I and II+abe-ifJl
=
II+abe'fJl.
°
=
[t(Z2-Z1)+Zl(I-Z1Z2)] [l-ZlZ2+tZ1(Z2-Z1W1, ~ t ~ I; the Z)-1 substitution C = (Z-Zl)(1-Z1 gives the Euclidean segment [0, C2] with parametric representation C = tC 2. 1.8.8. z
1.8.9. If w(z) is the given h-motion, (w(z)-b)/(I-bw(z) is a linear transformation with a zero at a and a pole at a-1, hence it has the form A(z-a)/(I-az). Moreover, for Iwl = Izl = I absolute values of both expressions are equal I, hence IAI = 1. 1.8.10.
~8 =
(l-l zI2)/(1-ICI 2) (cf. Ex. 1.8.9); if
u(1.
•
~=
max(tk+1-t,J, then k
with 1.8.11. If the arc-length 8 on y is the parameter and C(o; r) intersects y at two points corresponding to 81' 82, then after removing the open arc 81 < 8 <'82 and rotating suitably the remaining parts so as to join these, we obtain a new, shorter regular curve. Hence we may take (J = (J(r) as the equation of y in polar coordinates; R
R
~ (1-lzI2r 1d8 = ~ (l-r 2)-1[I+r 2((J'(r)2r,2dr o
0
which is a minimum for (J(r) = const = 0. 1.8.12. After a suitable h-motion e'1p(z-zl)(1-z1Z)-1 the end points Zl' Z2 0, R = IZ2-Z11/II-Z1Z21. The h-Iength of Cremains uncha,nged after the h-motion (Ex. 1.8.10), hence (Ex. 1:8.11) the geodesic lines are h-lines and be~ome
P(Zl' Z2)
=
p(O, R)
= flog((1+R)/(I-R) = artanhR.
1.8.13. The points z on the h-circle satisfy
..
Iz-zol/lz-z-11 = IZoitanhR
which is an Apollonius circle with limit points Zo and its reflection w.r.t. C(O; I). After a suitable h-motion h-circle becomes C(O; tanhR), thus lh = 7tsinh2R. 1.8.14. Only the triangle inequality is non-trivial. It follows, however easily from Exercise 1.8.12.
I. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
153
1.8.15. The linear transformation considered in Exercise 1.8.12 gives: (Zl' Z2' e P , eiIX) = (0, R, 1, -1) = (I+R)/(l-R);
(cf. Ex. 1.8.12, 1.4.7). 1.8.16. Note that the Jacobian
o(~, 1')) o(x, y)
= (da)2 = ( l_I~12)2
l-lzl2
ds
and use the formula for the integral transformed to new variables. 1.8.17. IQlh
=
t
7tR2(1- R 2 1.
1.8.18. Suppose the angle at z = 0 is equal y. Using the polar coordinates we obtain:
where r = reO) is the equation of the side AB in polar coordinates. Let C(s; R) be the circle containing the side AB and N a variable point on AB. We now
A FIG. 1
introduce the new variable qJ d
-=
=
~ OsN;
. obvIOusly tanO
=
RsinqJ d R ' where - cosqJ
151, thus
Ilowever, hence
dO = (d 2+R2_2dRcosqJ)-1 R(dcosqJ-R)dqJ.
Now. r2 = d 2+R 2-2dRcosqJ, (orthogonality condition).
l-r2
= 2R(dcosqJ-R) because I+R 2 = d 2
SOLUTIONS
154
Finally (l-rZ)-lr ZdO
=
i: dtp.
Considering the quadrangle OAsB we easily verify that
y+(P+t 7t)+(oc+i: 7t)+1tp =
27t
and this gives ITlh = -l-~dtp = fL1tp = H7t-(oc+,8+y)]. 1.8.19. The angle oc .at Zl is equal to
arg{(z3-Z1) (l-ZZ Zl) [(ZZ-Zl) (l-z3 Z1)r 1}
because after an h-motion considered in Exercise 1.8.12 two sides meeting at Zo become segmet;lts. The remaining angles are obtained by cyclic permutations. From Exercise 1.8.18 it follows that ITlh =
i- arg(l-zl zZ) (l-ZZZ3) (l-Z3 Z1)
if the orientation is positive.
•
CHAPTER 2
Regularity Conditions. Elementary Functions 2.1.1. (i), (iv) are continuous at z = 0; (ii), (iii) are discontinuous at z = 0: if z = x(1 +mi) and x -+ 0, the limit depends on m. 2.1.2. Suppose that B > 0 is arbitrary and ~ > 0 is such that If(z')-f(z") I < B for any z', z" eK(O;l) with Iz'-z"l <~. If IZn-CI < -~-~ for all n > N, then IZm-znl < ~ and consequently If(zm)-f(zit)1 <.B which means that {flzJ} is convergent for any {zn} -+ C e C(O; 1). 2.1.3. ux(O) = uy(O) = vx(O) = v,(O), however 1'(0) does not exist for f is discontinuous at z = O. In fact, the limit of u(x, y) along the line y = mx depends on m. 2.1.4. Cauchy-Riemann equations are satisfied at 2.1.5. Consider L1z =
~x
Z
= 0 only.
and L1z = iL1y.
2.1.6. If L1z = h+ ik, then
L1u = hux+kuy+o(Yh2+k2) ,
L1v = hVx+hvy+o(Yh 2+k2);
thus
Assume that k = 0 and h -+ 0, and vice versa which gives lim lL1wl2 = L1z
(A)
""z->o
luke now k (Ii)
lim
11.-... 0
U 2+V 2 x
x
2. = Uy2+Vy'
= mh which gives
I~wz 12 = u!+v!+2m(I +m2)-1(uXVX+UyVy) , ...
155
or
uxvx+uyVy = O.
156
SOLUTIONS
It follows from (A) and (B) that (u x+iUy)2 = (vx_iVy)2 which means that either J, or I satisfies Cauchy-Riemann equations. 2.1.9. Ux = Vy = 2uuy, Vx = 2uux = -un thus (1+4u 2)ux Hence Vy = Vx = uy = 0.
=
0, i. e. Ux = 0.
III = }/U2+V 2 ; I' = ux+ivx = vy-iuv • Hence Ltl/l = 111-1(u~+v~+u; +v; )-1/1- 3[U2(U~+U;)+V2(V~+V;)- 2uv(ux v x+uyv y)] = 1/1-12If'12-1/1-3(u2+v2)11'12 = 11'1 21/1-1, 2.1.10. If 1= u+iv, then
because uxvx+uyVy = 0. 0 22 I12 2.1.11. ox 1 --
2 2). 0 2 I12 C 2(uuxx+vvxx+ux+vx, oy2 I has an analogous lorm;
note that Ltu = Ltv = 0. 2.1.12. Ur = r- 1 V9 , Vr = -r- 1 U9 ; I'(z) = e- 19 (U,+iVr). 2.1.13. U
r"cosnO, V
=
=
r"sinnO;
I' = e-i 9nr"-1(cosnO+isinnO)
=
nz"-l
(cf. Exercise 2.1.12). 2.2.1. In the ri~t half-plane u has the form rp(y/x) and the condition Ltu =
°
gives ddt [rp'(t)(1+t2)] = 0, where t = y/x. Hence rp(t) = Aarctant+B, or . u(z) = Aargz+B, where A, B are real constants. 2.2.2. u has the form rp(X 2+y2) and Ltu =
d ° gives Yt[trp'(t)] = 0, where
t = X2+y2. Hence rp(t) = Alogt+B, or u(z) = 2Alogz+B, where A, Bare
real constants.
2.2.4. Ux =
~cosy
= v y, uy =
exp(loglzl +iargz)
=
-~siny
= -v x ;
exp(loglzl)[cos(argz)+isin(argz)]
= Izlexp(iargz) Logexpz = 2.2.5. I'(z)
=
=
z;
log~+iArg(cosy+isiny)
= x+iy = z.
ux-iuy, hence
x-iy (Logz)' = - - - = X2+y2
Z-1
and
(expz)'
=
eXcosy+iexsiny =, expz.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
157
2.2.6. If f = u+ iv and Ux = v y, uy = -vx, then
v u
V= Arg-.
Hence Ux
= (uux+oo x) (U 2+V2)-1 = (uvy-VUy) (U 2+V 2)-1 = Vy
and similarly Uy = -Vx ; F' = Ux-iUy =f'1lfr 2 =f'/f. 2.2.7. Uxx = -uy;because L1u = 0; uxy = uyx because the partials of second order are continuous.
2.2.8. Cauchy-Riemann equations imply
2.2.9. From Euler's formula for homogeneous functions: u(x, y) = m- 1(xux+Yuy)
it follows by differentiation; Ux = m-l(ux+xuxx+YUyx) = Vy = m-l(ux+YuXY-XUyy),
and similarly, uy = -vx • 2.2.10. (i) u is homogeneous of degree 2, hence by Exercise 2.2.9 fez) = (1-t·i)z 2.
vex, y) = -Hy(2x+y)-x(x-2y)].= t(y2_X2)+2xy;
(ii) v(x,y) = -2X3+3x2y+6xy2_y3, fez) = (1-2i)z3;
(iii) vex, y) = _Y(X2+y2)-1, fez) = Z-l; " 2 4 2 (iv) u = rez z- = rez- , v = i~z-2, f(i) ~ Z-2. 2.2.11. The denominator has the form "(1+Z2)(1+Z2), whereas the numerator is equal t(z+z)(1+zZ) = rez(1+:z2). Hence u v
=
rez(1+z2)-1,
'/= z(1+Z2)-1,
= imz(1+z2)-1 = y(1_x2_y2)[1+2(X2_y2)+(X2+y2)2]-1.
2.2.12. If v exists and F = u+iv J then F' = tix+iv x tive F = P+iQ does exist, then F'
= Px+iQx =
=
ux-iuy = f. If a primi-
Qy+iQ.~ =f= ux-iu,.
lienee Ux = Qy, uy = -Qx which means that Q is a conjugate harmonic function for u. 2.2.13. ux·-iuy = (z+l)e= has a primitive ze", hence v
=
imze" = eX(ycosy+xsiny).
158
SOLUTIONS
[} (
[}F)
2.2.14. rTr rTr 2.2.15. (i) u (ii) u = A
+ [}2F [}e 2
Axy+B, v
=
O.
= =
{-A(y2_ X2)+C;
-Vx+y'X2+y2 +B, v = Ay(x+ y'X2+y2tl/2+C,
i.e. u+iv = AZ 1/2+B+iC (cf. Ex. 1.1.5); (iii) do not exist. 2.2.16. Put cp(t) = ~exp(-bJ(t)dt)dt which means that CP"+CP'''P = 0 in (a, b). Hence L1(cp 0 F) = (F;+F;)cp"
0
F+L1F' cp' 0 F = (F;+F;)(cp" +CP'''P) 0 F = O.
r [} ( [}F) + [}eF]
2.2.17. ( r 2 Fr2 +F92 ) -1 r Tr r ar
[}2
2
=
"Po F.
2.2.18. The equation of the family of parabolas is x+ y' x 2+y2 If F(x,y) = X+J/X2+y2, then (F;+F;)-lL1F= (2F)-1;hence u (cf. Ex. 2.2.15 (ii)) .
= =
P = const. Areyz+B
2.2.19. fez) = exp(Az-1+B) with A real. 2.2.20. fez)
=
exp(AiLogz+B), A is teal.
2.2.21. u(r, e)
=
2.2.22. fez)
exp(-Aiz- 2+c), A is real.
=
A(loglzl-hrgz)+B; A, B are real.
2.3.1.1'(-1) = --t, hence
(X
= 7t, A =-t.
2.3.2. (i) The circle Iz+d/el = lel- 1Iad-ber 1 / 2 ; (ii) the straight line z = -d/e+e- 1y'ad-be t, -
00
< t<
+ 00.
2.3.3. The tangent vector ~f r = f(y) , i.e. f'(z(t))z'(t) has a constant direction. If s is the arc length on rand w = u+iv = fo z, then du/ds = at> dv/ds = a2 and hence w = as+b, a = al+ia2' .
2.3.4. argf'(z) = 0 on the straight line of Exercise 2.3.2 (ii) hence on its image line: argf'(z)z'(t) = argz'(t) = const. 2.3.5. Infinitesimal segments are expanded (i) outside C(O; -t), (ii) outside C(-I;-t), (iii) inside C(O; 1).
2.36. 1f'(zo)1 = 75, argf'(zo) = arg(7+24i)+7t.
2 2 = 1u'" +.IV", 12 2•3•• 7 [}(u, [}(x, v) y) = uxvy -uyvx = u",+v",
=
If'I2
h' h IS . equu I
W lC
to the ratio of area of infinitesimal squares corresponding to each other under f
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
159
2.3.8. The mapping is locally I: I because the Jacobian is =F 0 at Zo; note that the curves u(x, y) = const, vex, y) = const are mapped onto perpendicular I segments. 2.3.9. (i) equilateral hyperbolas x 2_y2 = const, xy = const; (ii) circles ceO; r) and rays Argz = const. b
HIf'(z)1 2dxdy (cf. Ex. 2.3.7).
2.3.10. ~ 1f'(z(t»)llz'(t)ldt;
D
a
2.3.11. e(l; I)
= {z: z = 2cosllei 6}, hence w = 4cos 2 l1e 2 i 6 = 2(I+coslP)ei'l',
IP = 2l1;
n/2
~ 2Iz(lI)llz'(lI)1 dll = 16,
~~ 41z1 2dxdy = 67t.
IQI =
-n/2
K(l; 1)
2.3.12. Under w = Z2 D is mapped opto the rectangle L1: I hence
<
u
< 4,0 <
v
< 2,
d 12 I 3 ~~ Izl 2dxdy = ~~ Iwl 1d~ dudv = ""4 ~~ dudv = 2 . D
D
d
2.3.13. -} 7t .
2.4.1. w = 1+2iz; IZ2_(1+2iz)1 = Iz-il 2
<
1~O'
2.4.2. The boundary of the image domain consists of the segment [-I,ll and two arcs of parabolas symmetric w.r.t. Ov: =fu = 1_-}V2, u+iv = w;
1= 2+V"2+log(I+V2). 2.4.3. w = (z2-2i)(Z2+2i)-1. 2.4.4. w = (I+Z)2. 2.4.5. w
=
1-2(a/z)2 (consider first Z
= Z2).
2.4.6. Superposition of two transformations: w
2.4.7.
W
2.4.8. If
= (vi-C)(C-y'P(y"2-I))-l,
z
= tp-C 2 •
= i(2z 2-1). W1, W 2
have the same image point, then
2a2w~(1+wn-1
=
2a2wHI+w~)-1,
I.e.
w~ = w~.
Sincef(-w) = -f(w), we have W1 = W2 which means that z =f(w) is univalent. rr Iw2 1 < I, then re{2a 2w2(1 + W2)-1} < a 2 [consider the linear transformation 2 21/ W(l + W)-J in K(O; I)l.
160
SOLUTIONS
Note that the domain between both branches of hyperbola is characterized by the inequality 'rez 2 <:: a2 • 2.4.9. z = p-2(w 2_a 2). 2.4.10. The mapping W = w2 carries the inside of lemniscate into the double sheeted disk K(a 2 ;p2) with the branch point W = O. A subsequent linear transformation Z = p2W[p4+a 2(W-a 2)rl yields the double sheeted unit disk with branch point Z = O. Ultimately z = wp-l[l+a 2p-4(w 2_a 2)rl/2. 2.4.11. The transformation W
=
vz
(im W
>
0) yields a strip parallel to the
real axis; w = (V2-1tl[2vz-i(I+V2)]. 2.5.1. The linear transformation z = (1]w+ 1)(1]w-l)-1 with suitably chosen 1] (11]1 = 1) carries C into the imaginary axis rew = O. If Wk are image points of Zk> k = 1,2, then Z1Z2 = 1 implies WI +W2 = 0, i.e. the imaginary axis separates Wk, and consequently C separates Zk'
2.5.2. The roots ZI' Z2 of the equation w = t(Z+Z-l) satisfy Z1Z2 = 1 and this means (Ex. 2.5.1) that the mapping is univalent in either component of C. If C intersects the real axis at an angle 0( its image under the mapping C= (z-I)2(z+ 1)-2 is a ray subtending with the real axis an angle 20( which is mapped under C= (w-l)(w+ 1)-1 onto a circular arc y with end points -1, 1 inclined to the real axis at an angle 20(. Hence both components of Care mapped 1: 1 onto C"-,y.
C"-"
C""
2.5.3, 4. Particular cases of Exercise 2.5.2, 0( = 0 and 0( =
i-7t,
resp.
2.5.5. A simple, closed Jordan curve containing the image arc C1 of C inside and having a cusp at w = 1 with a common tangent with C1 . 2.5.6. (i) Ellipses u 2 [i-(R+K 1]-2+ V 2[t(R-R- 1)r2 = 1 with foci at =fl; (ii) half-branches of hyperbolas u2 coS- 2 0-V 2 sin- 2 0 = 1 with foci at =r-= I. 2.5.7. w = z+z-t, R = 7 (cf. Ex. 2.5.6 (i)). 2:5.8. Take first R so that the circle C(O; R) is mapped under w = onto the given ellipse and put zR = Z;
t c(Z + Z
1)
w = t[(a+b)z+(a-b)z-l]. 2.5.9. The mapping Z = Z+Z-l carries the given domain onto the outside 01 some segment and this is mapped by a suitable similarity onto C"" [_. I , II. Finally t(w+w- 1) = [2(z+z-1)_(b+b- 1 -a-a- 1 )] (a+a- 1-b--b- 1)-I.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.5.10. (i) w = z+z-I-2; (ii) W =
Vw =
161
(1'-Z)Z-I/2.
2.5.11. The mapping ZI = t(Z+Z-I) yields the upper half-plane slit along [ii, + i 00) and this is mapped under Z2 = Zf onto C,,{(- 00, - 1~] U U [0, + oo)}. Again, the linear transformation Z3 = 1+ ~ Z2 1 yields C,,[O, + 00). Ultimately, w
2.5.12. w
=
=
kyZ3
Z+Z-I; Rl
=
=
(4z 4 +17z2+4)1/2(l+Z2)-1.
-Hy5+1), R2 = t(y5+3), R2/Rl = -HY5+1).
2.5.13. The mapping w = t(Z+Z-I) , imz > 0, carries the rays argz rt/2 onto the right-hand branches of hyperbolas u 2cos- 2()-v 2sin- 2() hence it is just the desired mapping.
°< () < W
2.5.14. w is the image of z E ceO; r) under the mapping w = Z+Z-1, where Z = Z3.
=
(),
=
1,
Z+Z-I; note that
=
2.5.15. The quadrant considered is swept out by arcs of confocal ellipses with foci =f=2 and each arc is mapped 1: 1 onto an arc of confocal ellipse situated in the complementary domain of the first quadrant (cf. Ex. 2.5.14). 2.5.16. The transformation Z = 2(z2-1) carries the upper half of the domain considered into the domain of Exercise 2.5.15 which is mapped under Z = w3 -3w onto the quadrant (+; -) in the w-plane. One proves in an analogous manner that the same transformation carries the lower half of the domain considered into (+;+). Ultimately Z2= l+t(w 3 -3w) maps {w: rew>O} onto the given domain. 2.6.1. (i) lezi = exr/sin2y+cos 2 y = c; (ii) exp(z+2rti) = eX[cos(y+2rt)+isin(y+2rt)] = e (iii) (expzl) (expz2) = eXl+~2[coSYI cosyz - sinYI sinY2 + i(sinYI COSY2 + +COSYI sinYz)] = C , +X 2 [cos(Yt +Y2)+isin(Y1 +Y2)] = exp(zl +Z2)' Z ;
2.6.2. Putting w = R(cosW+isinW), w = e and comparing absolute values, we obtain x = 10gR; also Y = W+2krt is determined uniquely, W being given. Z,
2.6.3. C,,(-oo, 0]; circles ceO;
eXo ) with deleted points _exo ; rays argw =
2.6.4. Logarithmic spiral R
exp[(W-rt)m- 1]; R, Ware polar coordinates.
=
2.6.5. w-plane slit along the logarithmic spiral R
=
Yo.
exp[(W-rt)m- 1].
2.6.6. (i) On straight lines y =.c krt; (ii) on straight lines Y I I, r2, ... ; exp(2+i) 3.992 ... +i 6.218 ...
=
(k++)rt, k =, 0,
SOLUTIONS
162
2.6.7. The image domain of the square is the domain largwl
<
S,
e a- €
< 1)1'1
< ea+€, with area s(e2a+2€_e2a-2€), hence
2.6.8. The chord [rle i6 , r 2e i6 ], -arcsinp ~ () ~ arcsinp, of K(1; p) is mapped 1 : 1 under w = Log z onto the segment [log r1 + i(}, log r2+ iO] whose center is -Fog r 1 r 2+ i(} = {- 10g(1- p2) + i(} which shows the symmetry of the image domain w.r.t. the straight line rew = tlog(1-p2). Symmetry w.r.t. imw = 0 is obvious. The parametric equation of the boundary of the image domain is: w«(}) = Log(1+pe 6 ); hence argw'«(}) = -t'lt+(}-arg(1+pei6 ) , d
d(}
argw'«(})
=
(l+pcos(})/(1+2PCOS(}+p2)
>0
(cf. Ex. 1.1.20) which shows the convexity. 2.6.9. The image domain of C"..jcx,,8] under Z = (z-cx)(z-,8r 1 is C',,(-oo,O] which is mapped under LogZ onto {w: limwl < 'It}; (i) straight lines im w = const; (ii) straight lines re w = const; (iii) w = O. 2.6.10. The mapping is univalent as a superposition of linear transformation and the univalent mapping of Exercise 2.6.9. The mapping W = (,8-cx)w- 1 carries the strip lim wi < 'It onto the outside of two circles with diameters 1,8-cxl'lt- 1 tangent to each other externally at the origin with the common tangent parallel to [cx,,8].
2.6.11. Logf is analytic in D, hence 10g
d 'd dx 2 log
=
2a
(a is a real constant). Hence
log
= ax 2+b1 x+C1 ,
log'!j!(y)
log If(z) I = a(x2_y2)+blX+b2Y+Cl +c2 ,
=
-
ay2+b 2Y+C2;
argf(z) = 2axy+b 1 y-b 2x+d 1
and finally fez) = exp(az 2+bz+c)
with real a and complex b, c. 2.6.12. The rays argz = () are image lines of meridians, the circles Izi
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
163
are image lines of parallels. Hence w = u(O)+iv(r); vCr) as a harmonic function constant on circles Izl = r has the form kloglzl (cf. Ex. 2.2.2), thus
w = ikLogz = ik(loglzl+iargz) = -kO+iklog (tan+7t+-!-q.», (cf. Ex. 1.2.8). The conformal mapping of the sphere so obtained which arises by a superposition of stereographic projection and logarithm is called Mercator projection. =
2.7.1. It follows from the definition of eZ that eix cosx-isinx (x is real).
=
cosx+isinx; e- ix
2.7.2. cosz = -}(e-y+ix+ey-ix) = coshycosx-isinhysinx, sinz = sinxcoshy+icosxsinhy; tanz = (sin2x+isinh2y)(cos2x+cosh2y)-1. 2.7.3. Icoszl 2 = cosh 2ycos 2x+sinh 2ysin 2x = sinh2y(l-cos 2x)+cosh 2ycos 2x
= cos 2x+sinh 2y; similarly Isinzl2 = sin2x+sinh2y; cosz = 0 means Icoszl 2 = 0, hence y = 0, x = (k+-!-7t), similarly' sinz = 0 for y = 0, x = k7t (k is an integer). ~
2.7.4. On vertical sides Isinzl 2 = 1+sinh2y ~ 1; on horizontal sides Isinzl2 sinh 2y ~ sinh2-!-7t > (2.29)2 > 1. 2.7.5. Cf. Exercises 2.7.4, 2.7.3.
<
2.7.6. E.g. Isinzl = -}Ieiz-e-izi -}(JeiZI + le-izl) = coshy +iy). sinz 12 1+sinh 2y '_2 2.7.7. E.g. - h :(: 1+(smh~) . cosz sm 2y 2.7.8. sin 2z+cos 2z = +(eiz+e-iZ)2_+(eiz_e-iZ)2 = 1.
I
<
< coshR
(z = x+
.
2.7.9. E.g. sin(x-iy) = coshysinx+isinhycosx = sin(x+iy). 2.7.10. (i) sin, cos, tan are real on the real axis; moreover, sine is real on lines x = (k+-!-)7t and cosine on lines x = k7t. (ii) Sine is purely imaginary on lines x = k7t and cosine is purely imaginary on lines x = (k+-})7t, whereas tangent is purely imaginary on lines x = -!-kn (k 0, =fl, =f2, ... ). 2.7.11. cos(5-i) = cos5coshl-isinhlsin5 ~ 0.438-1.127 i; sin(1-5i) ~ 62.45-40.99i; sinzo = sin-} 7tcosh [log(4+ V 15)] = -!-[4+ y15+(4+ .115)_1] = 4. 2.7.12. Use the formulas of Exercise 2.7.1 and the property proved in Exercise 2.6.1 (iii).
SOLUTIONS
164
coshxcosy+isinhxsiny, sinhz = sinhxcosy+icoshxsiny. 2.7.14. Isinhzl2 = sinh 2x+sin 2y, Icoshzl2 = sinh 2 x+cos 2 y. 2.7.13. coshz
=
2.7.15. cosz = coshiz, isinz = sinhiz, itanz = tanhiz; hence cos arises from cosh by rotation of z-plane by an angle i-7t; similarly sine and tangent arise from corresponding hyperbolic functions by rotations of z-plane and w-plane by i-7t and -f7t, resp.
2sin2IX . . sin[(IX+i,8)+(IX-i,8)] 2.7.16. COt(IX+I,8)+COt(IX-I(3) = [sm . (IX + I',8)'sm (IX-I'(3) = cos2i(3 -cos21X 2sin2IX cosh2,8-cos21X ' .
.
.
I [cot (IX+I,8)-COt (IX-I,8)]
=
isin[(IX-i,8)-(IX+i,8)] . (IX+I',8)'sm (IX-I',8) sm
=
2sinh2(3 cosh 2,8 -cos 2 IX
2.7.17. The image line of the segment (-T7t+Yo, f7t+Yo) is a half-ellipse u = sin x cosh Yo , v = cos x sinh Yo with foci =r= 1 for Yo i= 0, or the segment (-1, 1) for Yo = 0, resp. The correspondence between half-ellipses and segments is 1: 1, because u is a strictly increasing function of x. Moreover, the segments with Yo varying sweep out the whole strip Irezl < +7t, whereas their images sweep out the domain C",{(-oo, -1]u[l, +oo)}. This implies univalence of sine in {z: Irezl < t7t}. The image lines of lines x = Xo are branches of hyperbolas u = sinxocoshy, v = cosxosinhy for Xo i= 0, and the imaginary axis for Xo = O. 2.7.18. A quarter of the ellipse with semi-axes sinha, cosha and foci =F1; the right angle at z = +7t corresponds to the angle 7t at w = 1 (note that the derivative is = at this point).
°
2.7.19. Note that cosz
=
sin(t7t-z) and cf. Exercise 2.7.17.
2.7.20. W= (1+w)J(I-w) maps 1:1 the right half-plane onto K(O; 1) in the W-plane and [1, + (0) (-jo (-1, 0]. Hence W = (1-cosz)J(1 +cosz) = tan 2 ; z. <
2.7.21. The mapping Z = +7tl/i carries D"---,(-oo,O] onto {Z: 0 < reZ < i-7t} which is mapped under w = tan2i-Z = tan 217t onto K(O; 1)",( - I, OJ (cf. Ex. 2.7.20). However, tan2-t7t Vi is analytic on (-00,0] and maps it I: I onto (-1, 0] so that both slits can be removed.
yi
2.8.1. w = =filog(z+ Vz2-1).
2.8.2. Cf. Exercise 2.7.19.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
165
2.8.3. -iLog(I+i) = -t7t-tilog2. 2.8.4. Evaluate w from the equation iz = (e 2iw -1)/(e2iw + I); the mapping Z= (1+iz)/(I-iz) carriesC'-,-.,{(-ioo, -i]u[i, +ioo)} into C""(-oo,O] I and the latter domain is mapped under w = 2T Logz onto Ire wi < -}7t. 2.8.5. The image lines of Apollonius circles I(z-i)/(z+i)1 = const are segments im w = const; their orthogonal trajectories, i.e. circles through i, - i are straight lines re w = const. 2.8.6. (i) ~-}Arg(-2+i)+-!-ilog5;
(ii) -}7t++ilogltan(-!-7t+-}O)I. 2.8.7. The strip Irewl
< -!-7t (cf. Ex. 2.8.6 (ii».
2.8.8. w = tLog((1+z)/(I-z»; the strip lim wi 2.8.9. The half-strip: u
< -!-7t, (cf. Ex. 2.8.7).
> 0, 0< v < tv.
°
2.8.10. The mapping Z = e% carries the strip < y < 7t into the upper halfplane which is mapped by w = ArcsinZ onto the half-strip Irewl < t7t, imw > 0. The vertical boundary rays correspond to the rays: x» 0, y = and x» 0, y = 7t (x+iy = z). Since the mapping function has symmetry property w.r.t. the boundary rays, we can propagate the mapping onto the whole z-plane with removed rays y = k7t, X < (k = 0, =FI, =F2, ... ) and the values will cover the upper half-plane im w > 0.
°
°
2.9.1. The linear transformation Z = (z-a)/(z-b) carries the wedge into an angular domain with vertex Z = and a subsequent mapping w = zrt/lZ gives a half-plane.
°
2.9.2. Mter the transformation Z = zexp[ - ti(IX+P)] we obtain an angle symmetric w.r.t. the real axis; hence
w=
zrt/(P-IZ)
=
z!I/(P-rr.)
exp[ - t7ti(IX+P)/(IX-P)].
2.9.3. The linear transformation Z = (z+ l)/(z-l) carries the semi-disk into lhe quadrant (-; -) and a subsequent mapping w = Z2 gives the upper halfplnne. Hence w = [(z+I)/(z-IW. 2.9.4. The mapping W = Z3 carries the given sector into the upper semi-disk which is mapped under Z = [(W-J-I)/(W-I)]2 onto the upper half-plane. On
SOLUTIONS
166
the other hand, Z imZ > O. Hence
=
i(l+w)j(l-w) maps K(O; 1) onto the upper half-plane
, i(l+w)j(l-w)
=
[(z3+1)j(z3-1)]2.
2.9.5. w = i(z+I)3(z-1)_3 (cf. Ex. 2.9.1). 2.9.6. The mapping Z = (z+i)j(z-i) carries 'the wedge into the angle {-1t < argZ < t1t which is mapped under W = Z2 onto the right half-plane. Hence
w
=
i(W-l)j(W+l)
=
2z(I-Z2)-1.
2.9.7. w = tanht1tz (cf. Ex. 2.8.7); Apollonius circles with limit points -1, 1; circular arcs joining 1 to -1. 2.9.8. The linear transformation Z = (l-Z)-l maps the given domain onto the strip t < reZ < 1 which is carried under similarity W = 2i(Z-1-) into the strip lim WI < we now apply Exercise 2.9.7; finally
t; w
= tanh1ti((l-z)-l_1-) = itan1t(I-z)-l_1-).
2.9.9. The mapping Z = Z-l carries the given domain into the strip limZI
< t which is mapped under w = cotht1tZ onto
C",K(O; 1) (cf. Ex. 2.9.7).
Hence w = coth(1tj2z). 2.9.10. Two cjrcles intersect at an angle -t1t at 0, 2 and the third circle being an Apollonius circle with limit points 0, 2 is orthogonal to former ones. The mapping Z = zj(z-2) yields the circular sector: 0 < IZI < 2, t1t < argZ < f1t which can be mapped onto im w > 0 similarly as the sector of Exercise 2.9.4. Hence . w = [16(z-2)4-z4]2[16(z-2)4+z4t2.
v'
2.9.11. The mapping Z = w+t carries the slit w-plane into the right halfplane reZ> 0 so that (w = 0) (-jo (Z = t); the mapping Z = +(I+z)j(l-z) carries K(O; 1) into the right half-plane, hence w = z(l-z)-2 is the desired mapping. This is the Koebe function which plays an important role in various extremal problems in conformal mapping. 2.9.12. The linear mapping Z = (1+2w)j(I-2w) carries the given domain into ~(-oo, 0] and a subsequent mapping W = liz gives {W: re W > O}. Hence w = z(1 + Z2)-1 .
2.9.13. w = (l+Z)2(l-Z)-2. 2.9.14. The linear transformation Z = (z-ih)j(z+ih) carries the slit halt'plane into K(O;I)",(-I,O] which is mapped under W=Z(l-Zr 2 on(o C",(-oo, 0]; finally w = 2ihVW = 1/z2+h 2.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.9.15. After the inversion Z thus w = (hZ)-1(Z2+h2)1/2.
=
167
Z-1 we obtain the domain of Exercise 2.9.13,
2.9.16. By Exercise 2.9.11 the given mapping carries K(O; 1) into C",,(-oo, _+p-1(I_p)2]. Note that (-1,0] corresponds to (-tp-1(I-p)2,0].
2.9.17. w = (I+1]z)2(1-1]z)-2, \1]\ I+t 2.9.18. ( l=t
)2
(I_p)2
P
=
1.
Z
(1-Z)2 (cf. Ex. 2.9.17, 2.9.16).
. (1+p)2 2.9.19. The mappmgs Z = 4p
z (I-z)2 ' Z = (I-t)2 carry the slit
disk and the full disk, resp. into the Z-plane slit along (- 00, sired mapping has the following implicit form:
-t] so that
the de-
t(l+p)2p-1 z (I-z)-2 = t(1_t)-2.
2.9.20. The linear transformation
z = -e-ill(Z:-!X)/(I-aC),
where
(b-a)/(I-ab) = eill\b-a\/II-abl,
maps D(a, b) onto K(O; 1)",,(-1, -p] with p = \b-al/ll-abl; moreover (C = a) ~ (z = 0). The mapping function is obtained by substituting the above given values of z, p into the formula obtained in Exercise 2.9.19. 2.9.21. (dC) = (dZ) : (dZ) = _4eill Ib-al\I-ab\(lI-abl-\b-a\) . dt 1=0 dt 1=0 dC C=a (I-\b\2) (ll-abl+\b-al) 2.9.22. If Z1' Z2 are roots of the equation Z2+(p-W)Z-Wp-l.= 0, Z1Z2 = -Wp-1; moreover,
then
Z1(Z1 +p)(I+pz1)-1 = z2(z2+p)(I+pz2)-1 = Wp-1.
Hence Z2 = -(z1+p)(I+pz1)-1 and this means that IZ11 < 1 implies \z21 > 1, i.e. the mapping is univalent in K(O; 1), as well as in K( 00; 1). If Iz\ = 1, then 111'\ = p. The end points of circular slit on' C(O; p) are found by solving the equation dw/dz = O. 2.9.23. Put P = y2 in Exercise 2.9.22; w = z(z+ y2)(1+z }lilt1.
2.9.24. w = Rz(z+p) (1+pZ)-1, p = cosect!X.
CHAPTER 3
Complex Integration In this chapter I denotes the integral to be evaluated
3.1.1. (i) z(t)
=
(l+i)t, rez(t)
t, t
=
[0, 1];
E
1
I
~ t(1+i)dt
=
=
-}(1+i);
u
(ii) z(O) = reiD, rez(t) = rcosO, 0 E [-1t', 1t']; 1<
I = ~ rcosO· ire iO dO = i1t'r2. -1<
3.1.2. For example b
~ u(x, y)ds = ~ u(x(t), y(t))
VX'2(t)+ y'2(t) dt =
a
y
The same holds for v
3.1.3. z(O)
a
=
im f and the result follows.
cosO+isinO, \z-l\
=
b
~ u(x(t), y(t))\z'(t)\dt.
=
2sin-}O, 0,;;;; 0,;;;; 21t';
21<
I = ~ 2sintOdO = 8. o
3.1.4. (i) z(t)
= it,
-1';;;; t,;;;; 1; 1
= ~
I
i\t\ dt = i;
-1
(ii) z(O)
=
-sinO-icosO,
0';;;; 0';;;; 1t'; 1<
I = ~ (-cosO+isinO)dO = 2i; o
(iii) z(O)
=
sinO - i cosO, 0';;;;
(J :(.
1t';
7r
I
=~
(cosO+isinO)dO = 2i.
o 168
3. COMPLEX INTEGRATION
169
11
3.1.5. (i) If ~f(t)dt = 0, the inequality is obvious; it is easy to verify that b
..
b
~ cf(t)dt
= c ~f(t)dt for any complex c, hence for real () we have:
a
a b
b
b
re[e- i8 ~f(t)dt] = ~ re[e- i8f(t)]dt::;;; ~.lf(t)ldt. a
a
a
b
b
Choose now () such that e- i8 V(t)dt
=
IV(t)dt I;
a
I
(ii) ~f(z)dzl
..
b
=
b
Ia~f(z(t)z'(t)dtl::::;; ..~ If(z(t)llz'(t)ldt = ~ If(z)lldzl "I
7
(cf. Ex. 3.1.2). 3.1.6. By Exercise 3.1.5 (i): b
b
" I "
a
IV(z)dzl ~ ~ If(z(t)llz'(t)ldt::;;; M~ Iz'(t)1 dt = ML. 3.1.7. On the lower side of aQ we have: z(t) = zo-a(1+i)+2at (0::;;; t::;;; 1), whereas on the upper side z(t) = zo+a(I+i)-2at (0::;;; t::;;; 1), thus the sum 1
of both integrals is equal to 4i ~ 1+
(;:-1)2 =
7ti. The sum ot; integrals over
two remaining sides is also equal 7ti. 3.1.8. The existence of a primitive in D would imply ~ (z-zo)-ldz = 0 fQf c
any closed curve C c D which contradicts Exercise 3.1.7. 3.1.9. z has in E a primitive tz2, hence ~ zdz = 0, and consequently ~ xdz
r
=
-i) ydz, ~ xdx+ydy ~ 0;
r
)xdz =
r
r
t
~ zdz =
r
t
r
~ xdx+ydy+
r
ti ~ xdy-ydx = r
t r~ (z+z)dz = 1- r~zdz. .
3.1.10.
t.
3.1.11. 7ti. 3.1.12. From the existence of the total differential at f(zo+re I8)
for small r
>
\Zo it follows
= f(zo)+f~(zo)rcos()+if;(zo)rsin()+ro(1)
O. Hence
~ f(z)dz = [f~(zo)+if;(zo)] i7tr2+21tT 2o(I).
C( .o,r)
that
iA;
SOLUTIONS
170
Note that Cauchy-Riemann equations at Zo are equivalent to the equation
o.
I~(zo)+if;(zo) =
1
~ I(z)dz
3.1.13.
= F(b)-F(a) = (b-a) ~/[a+(b-a)t]dt; consider now a
[a,b]
partition to
=
0
0
<
t1
< ... < tm
=
1 of the interval [0, 1] and put Ik
= I[a+
m
+L1tk(b-a)], Lltk = tk-tk_1' The point t; =
J'lkLltk, as the center of k7J. mass of a system of particles 11, ... ,1m with masses Llt1, ... , Lltm belongs to conv[f1, ... ,fm] e convr (cf. Ex. 1.1.29). Consider now a normal sequence of b
partitions and a corresponding sequence {en}. Then t;n ~ t;
=
~ I(z)dz, and on a
the other hand t;n
E
conv r which is a closed set. Hence t; E conv r.
3.1.14. Suppose that Z1 i= Z2 and F(Z1) = F(Z2)' It follows from convexity of Ll that [Z1,Z2] eLl. If r is the curve: W=F'[Z1+(Z2-Z1)t] (O~t~I), then F(Z2)-F(Z1) = (Z2-Z1)t;, where t; = 0 E convr. However, reF'(z) > 0 implies re t; > 0 for any t; E convr, and this contradicts t; = O.
3.1.15. K(O; 1) is a convex domain; reF'(z) 3.1.16. The half-plane rez for x < 0 (z = x+iy).
=
nre(I+z"-l)
> 0 for z
< 0 is a convex domain; reF'(z)
=
E
K(O; I).
I+excosy
>0
3.1.17. A similar proof as in Exercise 3.1.16.
If dist(zo; y) = 15
3.1.1S. Con tin u it y. we have: IF(z)-F(zo) 1= Iz-zo I
> 0, then for any
z
E
K(zo; -}b)
II ~ (t;-zr1(t;- ZO)-1ep(t;)dt; I< Iz-zo 12MLb-2, 1
where M = sup Iep(t;) I on y and L is th~ length of y. This implies continuity. Differentiability. We have: F(z)-F(zo) Z-Zo
where ep1(t;)
=
= \'
ep(t;)
J (t; -z)(t;-zo) 1
dt; = \' ep1(!;) dt; = F1(z), J t;-z . 1
(t;-zo)-1ep(t;). Since F1 is continuous at any point not on y,
the limit lim F1(z) exists and is equal to ~ (t;-zo)-1ep1(t;)dt;. Z~Zo
y
3.1.19. We have: h'(t) = [z(t)-ar 1z'(t) at continuity points of z'(t). Consequently, u'(t) = 0 except for a finite set of values t, and from the continuity of u(t) it follows that u(t) = const = U(IX) = z(IX)-a = exp( -h({J) [z(fJ)-a] , hence, by Z(IX) = z({J), it follows that h({J) = 2k-ri.
3. COMPLEX INTEGRATION
171
3.1.20. n(y; a) = const as a continuous function of a ED (Ex. 3.1.18) whose all values are integers (Ex. 3.1.19). in
3.1.21. C"'-K is a connected set disjoint with y hence n(y; a) = const = C",-K, since 00 E C"'-K. 3.1.22. n(y; a)
=
n(y; (0)
=
°
0.
3.1.23. If zo = re i8 and r < r(O) , the segment [0, zo] does not contain any points of y, hence 21<
n(y; z)
=
const
=
n(y; 0)
= (27t;)-1 ~ [i+r'(O)jr(O)]dO
1
=
o
for any z E [0, zo]. On the other hand, for Z1 = re i8 with r > reO) the ray: argz = 0, Izl ~ r, does not meet y and n(y; z) = const = n(y; (0) = 0 for any z on this ray. The corresponding sets Eo, Eoo are obviously domains being open and arc-wise connected. Eo n Eoo = 0, since n(y; z) takes different values in either set. 3.1.24. If Izl < b < bk for all k > ko, then the points 0, z are situated in K(O; b) which does not contain any points of Yk. Hence n(Yk; z) = n(Yk; 0) = 1 (Ex. 3.1.20). 3.1.25. (i) 2,1,0; (ii) 0, I, O. 3.1.26. If a#- b, then (z-ar 1(z-b)-1 consequently,
=
(a-b)-1[(z-a)-1-(z-b)-1] and,
1= 27ti(a-b)-1[n(y; a)-n(y; b)] = O.
If a = b, again 1=
° by continuity of I as a function of a (Ex. 3.1.18).
3.1.27. Differentiate both sides of the formula of Exercise 3.1.26 (m-I) times w.r.t. a and then (n-I) times w.r.t. b (cf. Ex. 3.1.18). n
3.1.28. I
has the form
~
L k=O
~
ak
z-n-l+k(z-a)-1dz. Since n(C(O;r);O)
C{O;,)
n(C(a; r); 0) (cf. Ex. 3.1.20), we have 1=0 by Exercise 3.1.27.
3.1.29. Use the equality 1= 27ti(a-b)-1[n(y; a)-n(y; b)]. 3.1.30. Differentiate both sides of the formula of Exercise w.r.t. a and (n-I) t~mes w.r.t. b. This gives T
3.1.31. n(C(O; 2);
i)··
=
(-I)n27ti (m;;;~]2) (b_a)-m-n+l. n(C(O; 2);
-i) = 1 (cf. Ex. 3.1.26).
3.1.~9
(m-I) times
172
SOLUTIONS
3.1.32; Any polynomial has a primitive in the open plane, hence ~ W(z)dz = 0 y
for any closed curve y.
3.1.33. (z-a)-n has a primitive in C"'-a for any integer n
~
2.
3.1.34. After developing R in partial fractions we obtain a polynomial plus terms of the form Ajn(z-aJrn and the integration cancels the polynomial, as well as alI terms Ajn(z-aj)-n with n ~ 2 (cf. Ex. 3.1.32, 3.1.33). 3.1.35. If z
E
C, Z = r 2 z-1, hence
Iz-al 2 = zz-az-az+laI 2 = r 2-ar 2z- 1-az+laI 2,
or I
!!.- 'i a~
=
---,~d-:-z_--:-::::::--
(z-a)(z-r 2/Q) .
Now, one of the points a, r 2 /a is situated inside C, another one is outside C, hence 1= 2r\laI2-r2\-1 (cf. Ex. 3.1.29).
3.1.36. (21t)-1
~
/z-W2Idzl
=
r/(I--;r2) (cf. Ex. 3.1.35).
C(O;r)
3.1.37. If y is a curve consisting of [1, r] and of a circular arc: z = re i9 (0 ~ () ~ qJ), then ~ z- l dz = logr+iqJ; note that r-y is a cycle, hence y
(21tW1
~
.
,
d; = n(r-y; 0) = k,
r-y
3.2.1. 1= 0 since
~ z- l dz =
l.e.
21tik
+ ~. z- l dz .
r
r
y
is contained in the rectangle Ixl
< 1-,
Iyl
< i-
where
(1 +Z2)-1 is analytic.
3.2.2. 1= 0 since n(C; a) = 0 for all a being the zeros of the denominator. 3.2.3. If 0 < r1 < r 2 < R, D = {z: 0 < Izl < R}; hence
~ z-lf(z)dz
=
0,
r
or
then. ~
r
=
C(O; r1)-C(0; r 2 )
z-lf(z)dz
=
~
'"
0 (mod D),
z-lf(z)dz.
C(O;r2)
C(O;rl) 2",
3.2.4. f(O) = (21ti)-1
~ C(O; r)
z- l f(z)dz = (21t)-1 ~ f(re i9 )dO , hence 1= 21tf(O). 0
3.2.5. Given U, find its complex conjugate v in K(O; R), apply the formula of Exercise 2.3.4 for u+iv and compare real parts of both sides.
'9
3.2.6. loglre -al = rel~g(z-a) is harmonic in C(O; 101), hence I=- 21tloglal. \
3. COMPLEX INTEGRATION
173
3.2.7. Put J(z) = (z-a)-1(A+s(z») and z = a+re iO (s(z) hence
-+
0 for r
-+
0);
~ J(z)dz = 21tiA+o(1) = 21tiA C(a;r)
by Exercise 3.2.3. 3.2.8. Let C = C(O; R) be a circle containing y inside and leaving all an with n ;?: N outside. If Ck is a circle center at ak such that all an> n 1= k, are situated outside Ck , then N
r = y- ) nkCk '" 0 (mod D) ~ k= 1
for
nk = n(y; ak).
Hence V(z)dz r
=
0,
(cf. Ex. 3.2.7). 3.2.9. 21ti[fn(C; l)+fn(C; -l)-n(C; 0)]. 3.2.10. f1ti. 3.2.11. 21tia-1sina. 3.2.12. f(zeZ)~:'a
ea(l +fa).
=
3.2.13. (i) 1; (ii) -f(z-1ez);~ 1 = -fe. 3.2.14. 1= 21ti(a-b)-1 [f(a)-J(b)]; on the other hand III:S;; 21tRM(R-lal)-1(R-lbl)-1
-+
0
as
R
-+ 00
since J is bounded: IJ(z) I :s;; M. Hence 1= 0, or J(a) =J(b). 21ti d m- 1 -1 3.2.15. 1= (m-l)! dZ"'-l [(z-b) ]z=a
.-In
=
-2m(b-a)
.
3.2.16. By Cauchy integral formula for the cycle r = C-C(O; R) which is '" 0 mod (C"'O), and for the function z- 1J(z) we have: n(r; z)z-Y(z)
=
(21ti)-1 ~ [C(C -z)r 1J(C)dC
r since
~
-+
0 as R
=
(21ti)-1 ~ [C(C -z)r1J(C)dC C
-+ +00.
C(O; R)
Now, n(.l'; z) ated outside C.
=
0 for z situated inside C and 1= 0, while n(r; z)
=
1 for z situ-
SOLUTIONS
174
Z
-+zk
+ ...
-zn)rl, then 1= 27ti(EIAI +EnAn) where Ek = n(C; Zk) = 0, 1. Now, Al +Az+ ... +An = 0, which is verified by taking C = C(O; R) and R --+ +00, hence I can take at most 2"-1 different values (all Ek = 0 and all Ek = I give
the same value 0). 3.3.1.
r
y-n(y, a)C(a; r) '" 0 (mod K(a; R)"-.a) for any 0
=
~J(z)dz
=
0
~
Note that
C(a; r)
1
J(z)dz
0
=
R, hence
~ J(z)dz.
V(z)dz-n(y; a)
=
r
(Ex. 3.2.7).
C(a; r)
3.3.2. Suppose that 0 (mod K(a; R)"'--.a) , hence J(C)
=
<
r1
< IC-al; then r
(27ti)-1 ~ (z-C)-IJ(z)dz
=
(27ti)-1
r
because
\
=
r
=
C(a; r)-C(a; rl) '" 0
(z-O-Y(z)dz,
C(o;r)
0 by Exercise 3.3.1.
qo; r,)
~
3.3.3. The integral (27ti)-1
(z-C)-Y(z)dz = q;(O is an analytic function
C(a;r)
of CE K(a; r) (cf. Ex. 3.1.18). 3.3.4. lim z-IJ(z) z--+O
=
lim [f(z)-J(O)]/(z-O)
=
1'(0).
z-".O
3.3.5. For any positive integer p and real x hence lim z Pexp(1/z) does not exist.
--+
+ 00
we have x-Pe X --+
+ 00,
z--..o
3.3.6. Put z = I/iy; if y --+ +00, then zPsin(l/z) = (2iP+l)-ly-p(e-Y-eY) has no finite limit for any positive integer. p. 3.3.7. (sinz), = co~z, hence lim sinz/(z-k7t) = cosk7t = (-It and conz-'Jok",
sequently lim (z-k7t)Z(sinz)-2
=
I.
z--+k~
3.3.8. (i) 0, =f I are s~mple poles, z = 00 is a removable singularity; (ii) z = I is a pole of second order, z = 00 is a pole of order 3; (iii) z = =fi are simple poles, z = 00 is an essential singularity; (iv) z = (2k+I)7ti, k = 0, =fl, =f2, ... are simple poles, z = 00 is a point of accumulation of poles; (v) z = I is an essential singularity, z = 00 is a removable singularity;
3. COMPLEX INTEGRATION
175
(vi) z = I is an essential singularity, z = 2k7r;i are simple poles, z = ex:> is an accumulation point of poles; (vii) z = 2[(2k+I)7'Cr 1 are ess~ntial singularities, z = 0 is their accumulation point; (viii) z = 2[(2k+I)7'C]-1 are essential singularities, z = 0 is their accumulation point, z = 00 is a removable singularity. 3.3.9. I is analytic in K(O; 2R), hence If(z) I ,::;:; M1 for all z E K(O; R); 1(r;-1) is analytic in K(O; 2R- 1), hence If(C- 1) I ,::;:; M2 for ICI ,::;:; R- 1, and consequently If(z) I ,::;:; max(M1, M 2) (cf. Ex. 3.2.14).
3.3.10. If ak are poles of order nk, k = 1,2, ... , N, then
n N
cp(z) = I(z)
(Z-ak)n k
k=l
is analytic except for a pole at
00
and consequently cp is a polynomial.
3.3.11. We have: I(z) = /(a)+ (z-a)f1 (z), 11 having a removable singularity at z = a; similarly 11(Z) =!t(a)+(z-a)/2(z), ... ,J,._1(Z) =J,._1(a)+(z-a)fn(z) and this implies: I(z) =/(a)+(z-a)/1(a)+(z-a)2/2(a)+ ... +(z-a)nJ,.(z).
Differentiating and putting z
=
a we obtain j'C.k>(a) = k!fk(a).
3.3.12. lim(z-a)-m/(z) = (ml)-1j'Cm)(a) (cf. Ex. 3.3.11). 3.3.13. lim(z-a)"'/(zr 1 = m![J<m)(a)r1 #: O. Z-+Q
3.3.14. By Exercise 3.3.10 I(z) = Pm(z)/Pn(z), where Pm, P n are polynomials of order m and n, resp.; if e.g. m > n, then I has q = m zeros, n finite poles and z = 00 is a pole of order m-n. 3.3.15. If 0
< r < R and
M = suplf(rel 9l) I then rp
J,.(z)
=
(27ti)-1
~
(C-a)-n(C-z)-1/(C) dC,
Iz-al
<
r
C(a;r)
therefore IJ,. (z) I ,::;:; r-n(r-lz-aD- 1Mr. By Exercise 3.3.11 and J
E
I
\
=
0 we
K(a; r).
3.3.16. In case no such m exists, we should have 1= const by 3.3.15.
Exercise
176
SOLUTIONS
3.3.17. Iff(a) "# 0, then Ilfis analytic in some neighborhood of a; iff(a) = 0, then f has a zero of order m at z = a which means that 1If has a pole of order m at a (cf. Ex. 3.3.16, 3.3.l3). 3.3.18. We have If(z)-wol ~~, hence lim[f(z)-wo]-l(z-a)
=
0 which
means that q;(z) = [!(z)-w o]-l is analytic in K(a; R); by Exercise 3.3.17, z is either a pole, or a removable singularity for both l/q; and f
=
a
• 3.3.19. By Exercise 3.3.18 the values taken by f in any annular neighborhood of a (Le. in {z: 0 < Iz-al < ~}) form a set dense in C which is mapped under any non-constant polynomial P onto a set dense in C. This implies that Po f cannot have a as a pole, or regularity point. 3.3.20. The assumption If(z)-wol ~ tradiction, similarly as in Ex. 3.3.18.
> 0 for all z E K(a; r)""a gives a con-
~
3.4.1. If n(y; a) = m and 0 < ~ < r, then the cycle r = y-mC(a; ~) "" o(mod K(a;r)'",a) and by Cauchy's theorem V(z)dz = O. Now, w = ~ f(z)dz r
~
does not depend on
V(z)dz-mw
q~~
(Ex. 3.2.3), hence
0,
=
~ [!(z)-w(27ti)-1(z-a)-1]dz
or ~
Hence A = (27ti)-l w = (27ti)-1
=
o.
f(z)dz.
qa:~)
3.4.2. Note that f-A 1(z-ar 1 has a primitive in K(a; R) and co~sequent1y
~[!(Z)-Al(z-arl]dz
0
=
y
for any closed curve y c K(a; 3.4.3. res(O; f)
=
~)""a.
2, res(l; f) =
-i-,
res(-l; f)
=
-to
3.4.4. Cf. Exercise 3.2.7. 3.4.5. Obviously res(O; F) = 1; if for some nonnegative integer k we have hr( -k+h) -+ (k!rl( -l)k as h -+ 0, then hr(-k-l+h)
=
(-k-l+h)-lhr(-k+h)
-+
[(k+l)!]-l(-lt+l.
3.4.6. (i) res(O; f) = -1, res(!; f) = e; (ii) res(i; f) = ti, res(-i; f) = -ti. ~
3.4.7. res( -a; f) = (27ti)-1
C( -a: r)
odd f Similarly res(-a; f)
=
f(z)dz = (27ti)-1
"~ qa: r)
-res(-a; f) for even f
fendc = res(a; f) fOl
177
3. COMPLEX INTEGRATION
3.4.8. Use the representation: fez)
(z-a)m[A+(z-a)cp(z)],
=
g(z)
=
(z-a)"[B+(z-a)V'(z)],
where A i=. 0, B i= 0, and cp, V' are analytic in some neighborhood of a; m, n are positive for a being a zero and negative for a being a pole. 3.4.9. lim (Z-Zy)(Z4 +a4)-1 3.4.10. res(zv;f)
=
lim 1J4z 3 = -i-zva-4 (cf. Ex. 3.4.4, 3.4.8).
r
lim z"-1(z-zv)(z"+an 1 = lim [nz"-1-(n-1)zyzn-2]jnz"-1
=
Z--+Zy
=
Z-70-Zy
lin. 3.4.11. By Exercise 3.3.11 : (z-a)-Y(z)
=
(z-a)-kf(a)+(z-a)-k+1!'(a)+ ... +(z-a)
-1 j
where fk is analytic in some neighborhood of a (cf. Ex. 3.4.2). 3.4.12. The point z = Zk is a removable singularity for (Z-Zk)2f(z)(cOSZ)-2, hence by Exercise 3.4.11, d res (zk;f(z) (COSZ)-2) = dz [fez) (Z-Zk)2(COSZ)-2]z=Zk = !'(Zk) , the corresponding limit can be evaluated by using Exercise 3.4.8. 3.4.13. res(O;f) = lim 21 z-->O
3.4.15.
~
P(z)dz
=
0,
(s~nfJrt.z)" = SIn z
rt.fJ-1(p2-rt. 2)J6.
cp(z)z- 2dz
~
=
0(r- 1) and having a constant
C(o: r)
C(O; r)
value, the latter integral must vanish. 3.4.16. (i) Log[(l-aC)/(l-bC)] = (b-a)C+C 2cp(O, where cp is analytic in some neighborhood of C = 0 (cf. Ex. 3.3.11); hence res(oo; f) = a-b; (ii) (z-a)(z-b) = =r= [z-+(a+b)-{-(a-b)2 z-1+ Z-2cp(Z)] , where cp(z) =0(1) as z --+ 00. Hence res(oo; f) = =r={-(a-b)2. The above representation is obtained by putting z = and applying Exercise 3.3.11.
V
,-1
3.4.17. If K(O; r) contains all finite singularities a1, a 2 , ... , an and each K(ak; rk) contains only one singularity, then n
C- ~ Ck '" 0 (mod C",- U {ak}) ,
where
C
=
C(O; r), Ck = C(ak; r).
lienee n
~f(z)dz~" )1 ~ f(z)dz c f::T C~
n
=
-27f:ires(00;f) =, 27f:i )1 res(ak;.f).
f=r
SOLUTIONS
178
3.4.18. (i) res(-I;j) = 2sin2 = -res(oo;j); (ii) res(2;.D = -res( 00;.D = -143/24. 3.4.19. (i) res(O;D = 1, res(I;.D = res( -1;.D = -t; (ii) res(O;f) = -t, res(2i;.D = -ft- i(cos2+isin2),
res(-2i;.D = --ft-i(cos2-isin2), res(oo;j) = +(sin2-2); (iii) res(krt;f) = 0, k = 0, =n, =F2, ... ; (iv) res(krt;f) = -1, k = 0, =Fl, =F2, ... ; (v) res( -1 ;f) = -res( oo;f) = -cos 1; (vi) res(O;.D = 1/2, res(2k7tiln;f) = 1/2k7d, k = =Fl, =F2, ... , (vii) res(O;.D = ofor negative and odd positive n, res(O;f) = (_1)n/2[(n+l)W 1 for even positive n, res(oo;f) = -res(O;.D. 3.4.20. lim (z-a)fo cp(z) = lim f(C) [C-cp(a)] [cp(z)-cp(a) z--,>a i:--,>
(1.)
]-1 =
Alcp'(a).
(-2 1 z-3 + 2! 1 z -4 +... ) [({3 -ct)z+2 1 ({32 -ct 2) z 2+3 1 ({33 z +TI
3 3+ ... ] -_ ... + [({3 -ct)+ 2! 1 ({32'-ct 2) +3f 1 ({33 -ct 3) + -ct)z
... ] z -1 +
... ., h ence
res(O;.D= eP-e"; (ii) Log[I+(z-I)]cos(z-1)-1 = [(Z-I)-
~
(Z-I)2+-} (Z-I)3- ... ] X
X[I-;! (z-1)-2+ 31, (Z-I)-4- ... ] _ ( 1 1 1 1 - ... + -T2! +3 4! -
1 1
561+'"
) -1 z +
.
....
3.5.1. C is a contour containing inside one simple pole -2-1/2 (1+i) with residue (1 +i)/4 yT, hence 1= (i-l)rt/2
V2.
3.5.2. (i) -rtil yT; (ii) -2rti/3; (iii) 3rti/64; (iv) 0 (show that res(oo;.D = 0). 3.5.3. (i) There are two simple poles inside C: al, 2 = t=Fit
yf with residues
--}ak ; (ii) C(O; R) contains inside a simple pole z
= 0, res(O;f) =
1/2rti, and 6
3. COMPLEX INTEGRATION
179
simple poles a v = Vkev, e = exp(7ti/3), on the circles C(O; ilk); res(av;f) .. 1/61ti; (iii) there is one simple pole +(I+i) inside y ... with residue +(l+i)en / 2. 3.5.4. res( 00 ;fl = i-, hence I = - 27ti/3. 3.5.5. There is exactly one root of the equation e--w = 0, i.e. z = Logw, situated in Q. Thus the integrand f has inside Q a simple pole z = Logw with res (Logw;f) = 3.5.6. res(oo;fl
=
(z-Logw)ze Z (e z -w)-l
lim
=
Logw.
z...,.Log w
-(n+1)-1(bn +1-a"+1), hence 1= -27tires(00;fl.
3.5.7. res(00;f)=t(a-b)2, 1= -27tires(00;fl. 3.5.8. We have
[(z-a) (Z-b)]-1/2 = Z-l
(1-_
a+b Z
+ ~)-1/2 = Z2
Z-l +.!.. (a+b)z-2+ 0 (Z-3) 2
'
hence: (i) res( oo;f) = -+(a+b); (ii) res( oo;f) = -1. 3.5.9. (i)
~
[z(z+I)]-1/2z dz
=
-7ti (cf. Ex. 3.5.8 (i»;
C(O; 2)
(ii) 7ti (cr. Ex. 3.5.8 (ii». -1
-2
3.5.10. We have z" (l+z)
-1/2 _
-z"
-1
-2:1 z,n-3 +
1 . 3 -5 2.4 z" - ... ,
hence
res( 00 ;f) = 0 for odd nand ( 'f) res 00,
= (_I)k+l
1·3· .... (2k-l) 2 . 4. .., . 2k
for n = 2k.
3.5.11. Let H(R) be a contour consisting of an arc of C(O; R), R> 1, cut off by the parabola and situated in the right half-plane and of an arc of r contained in K(O; R). There are two simple poles of f inside H(R): =f2-1/2(1 +i) with residues (8
V1+ y2)-1( -1- V2=fi) and hence ~
= T7ti-y 1+
y2
since
H(R)
H(R) is described in the negative sense. Note that the integral over the circular arc does not exceed in absolute value R(R4 _1)-1(R2_1)-1/2 which tends to 0 as R ~ +00, hence ~ =~. H(R)
r
1 (z+z -1 ), sinO 3.6.1. If z = e '6 , then cos 0 = 2: hence
1 (Z-Z-l) , dO = -iz- 1dz; = 2I
180
SOLUTIONS
3.6.2. We have
1= _i ~ [z2(z-a) (Z_PW1(Z2_1) 2dz, 2b C(O; 1) where a, P are roots of the polynominal z2+2ab- 1z+ 1; ifJis the integrand, then res(O;f) = -2ab-I, res(a;J) = 2b- 1(a 2_b 2)1/2, (J is situated outside C(O; 1) since ap = 1; 7t 1= -7J[res(O;f)+res(a;f)]. 3.6.3. (ii) cosnO = t(z"+z-n) for z = e i6 • 3.6.4. We have cosnO
=
rez" for z
=
I=re~ ~ I C(O; 1)
ei9 , hence
(z2+z+1)n z3+3z+1 dz;
the only singularity inside the unit disk is a simple pole Z1 residue 5- 1 / 2 (3_ y'5)n.
=
-}( -3+
Vs) with
3.6.5. Let J y denote the integrals taken over the consecutive sides oRn; then n
= ('
J 1
J
-n
n.
x dx = _ i (' XSIll X dx a-cosx+isinx J l+a 2-2acosx -n
the real part of integrand being an odd function;
n
J3 = -
('
J
(in+x) dx -+ 0 a-e"e ix •
-n
If n Hence
>
loga, there is only one simple pole inside Rn with residue a- 1Ioga.
or
1= 27ta- 1Iog[(1+a)/a], If 0
< "
~
-n
a
< 1, then
l/a
a>
> 1 and this gives
xsinx 1+ a 2 - 2a 1 cosx -dx
=
27talog(1+a),
i.e.
1.
3. COMPLEX INTEGRATION
181
3.6.7. We have 2",
~
e% z-n- I dz = i ~ exp(cosO) exp[i(sinO-nO)] dO
C(O;I)
0
2",
2",
=
~ exp(cosO) sin(nO-sinO) dO+i ~ exp(cosO) cos(nO-sinO) dO
=
27tijn!
o
0
since the integrand has a pole of order n+ 1 at the origin, the residue being equal to 1 jn!. Compare now the real and imaginary parts of both sides. 3.7.2. The inequality sinO >
~0 7t
is a consequence of convexitY'of sine in
(0, +7t); ",
",/2
~ exp(-AsinO)dO
",/2
2 ~ exp(-AsinO)dO
=
o
< 2 ~ exp(-2AOj7t)dO <
0
7tA- I .
0
3.7.3. E.g. (iii) can be solved as follows. The only singularity of e im%(a 2 +z 2 )-2 in the upper half-plane is a double pole z = ai with residue -i-}a- 3e- am (1+am); if R > a, then ' Rl'
~ e imx (a 2+x2)-2dx+ ~ e im%(a 2+z 2)-2dz -R ",
I ~ I<
!}7ta- 3e- am (1+am);
=
r(R) ,
R(R 2_a 2)-2 ~ e-mRsin6dO
r(R)
<
7tm- 1 (R 2_a 2)-2 -+ 0
as R
-+
+00
0
(cf. Ex. 3.7.2). Equating the real parts and making R
-+
+ 00,
we obtain
+ <Xl
~ (a 2+x2)-2cosmxdx
=
T7ta-3e-am(1+am);
(ii) integrate ze i':(z2+ a2)-2 and compare imaginary parts of both sides after making R -+ + 00 • 62
3.7.4. ~J(z)dz = i ~ [b+e(a+re i6)]dO = ib(02-0I)+O(l). Yr
3.7.5. We have:
6,
~ [-R, -r]
=
~
; moreover, IimzJ(z) = -2i, hence .:--..0
[r, R]
as r -+ 0 (cf. Ex. 3.7.4);
I ~ I ::c.:'; 27tR- 1 -+ 0 r(R)
as
R
~
-r(r)
-+
+00;
-+
-27t
SOLUTIONS
182
this implies +00
2 ~ x- 2(1-e 211dx-27t o
Equating real parts we obtain I
=
O.
=
+7t.
3.7.6. Integrate z-2(z2+a2)-1(I_e2miz) over the contour of Exercise 3.7.5 and compare after making R -+ + 00, r -+ 0, the real parts of both sides. The integrand has a simple pole at z = ai with residue (2ia 3)-1(e- 2am _I), ~
-+-27tma- 2 as r -+ 0 (cf. Ex. 3.7.4), hence
-r(r)
+00
2 ~ x- 2(x 2+a2)-1(1-cos2mx)dx = 7ta- 3(2am+e- am -l)
=
41.
o
3.7.7. We have
(sinx)3 = (2i)-3(e ix _e- ix)3 = _(8i)-1[e3ix_e-3iX_3(eiX_e-ix)] -i-sin3x+i-sinx
=
=
im [i-(1-e 3;X)-i-(1-eiX)].
Integrate now J(z) = [+(1-e3iX)-i-(1-ei1]z-3 round the contour of Exercise 3.7.5. We have: limzJ(z) = i- (de I'Hospital rule) which implies ~ -+ -i-7ti -r(r)
z-+O
as r -+ 0 (Ex. 3.7.4), moreover, ~ -+ O. Hence r(R) R
+
~ -[R. -r]
~
R
2i ~ imJ(x)dx
=
[r. R]
=
2i ~ (sinx/x)3dx
r
r
and using the fact that the integral round the contour vanishes, we obtain +00
2i ~ (sinx/x)3dx-i-7ti+o(1)
=
0,
1= 37t/8.
i.e.
o
3.7.8. Integrate then
J round the contour of Exercise
+
~ [-R. -r]
~
-+ -7t/2a2
as
r
< r < a < R,
R
~
=
2 ~ x- 3(x 2 +a 2)-1(x-sinx)dx,
[r. R]
-+ 0
r
(Ex.
3.7.4),
~
-+0 as
R-+oo.
Moreover,
r(R)
-r(r)
27tires(ai;f)
3.7.5. If 0
=
-7ta-4 (a+e- a-l), hence 21-7t/2a 2+o(1)
=
-7ta- 4 (a+e- a-l).
3.7.9. J has inside the rectangle a simple pole z
=
7ti with residue
-e"1tI.
Now,
3. COMPLEX INTEGRATION
183
f(z+27tl) = e 2n1af(z), hence the sum of integrals over the horizontal sides is equal to R
(1_e 27t1a ) ~ (1 +eX)-leaxdx, -R
whereas a corresponding expression for vertical sides tends to 0 as R Thus
--+
+ 00 •
+
(1-e 2n "a) ~ (l+e X)-leax dx+o(l) = -27tiea7ti , -00
i.e. 1= 7tjsina7t. (i) introduce the new variable eX (ii) take t = x", a = mjn in (i).
=
t;
3.7.10. fez) = eaz (1 +ez+e 2Z)-1 has inside the rectangle considered in Exercise 3.7.9 two poles Zl = 27ti/3, Z2 = 47tij3 with residues (iv3t1exp[27ti(a-l)/3J, -(i y3)_1 [exp47ti(a-l)j3]. Similarly as before (1-e 2nia )I+o (1) = 27t3-1/2 [exp27ti(a-l)j3-exp47ti(a-l)/3].
3.7.11. The only singularities off situated inside QN are poles Zk = (2k+ l)i, k = 0, 1, ... , N-l, Zo being a double pole, all remaining poles being simple We have: res(zo;D = (27ti)-1, res(zk;f) = (27ti)-1 (_I)k+1 (1jk-l/(k+l)); If(z) I ,;;;; [(N 2-l) (cosh2}7tN-l)]-1 on vertical sides of QN, whereas If(z) I ,;;;; (4N2_l)-1 on the upper horizontal side. Therefore the corresponding integrals tend to 0 as N --+ + 00 and, consequently,
"
+00
~ f(x)dx = 1+(1-+)- (i--i-)+ ...
=
210g2.
-00
3.7.12. Suppose that 0 < r < t7t < Rand J y is the integral of f taken over an arc indexed by 'V as in Fig. 2, 'V = 1 to 6. If r(r, R) is the contour of Fig. 2, 6
then
~
=
0
=
r(r, R)
L J".
,,=1
We have: R
re(J1+JS) = (1-e a7t) ~ (e 2Y -l)-lsinaydy; r
" limzf(z) ... 0
=
1 --2" l
hence lim J 2 = 47t (cf. Ex. 3.7.4); similarly limJ4 r ....... O
7t-R
J3 =
~ ~
(e'X-e-jX)r12iel"eaxdx,
=
7tean /4;
SOLUTIONS
184
hence reJ3 = (l-e o")/2a+o(I); moreover, reJ6
--+
0 as R
(l-eo")I+t7t(I+~")-(eOTC-l)/2a+o(l)
+00. Therefore,
--+
=
0
which gives l. iR 1t +iR
(6) (5)
(1)
1t +ir
ir r
1t
FIG. 2
3.7.15. (i), (ii): integrate [Z-l_ (sinhz)-l]z-l and (l_e Oi%) (zsinhz)-l, resp., round the boundary of the upper half-disk with radius 7t(n+-!-); the origin is a removable singularity in (i), in (ii) it should be omitted round - r(r); (iii) put x = et (cf. Ex. 3.7.9). 3.8.1. We have
I ~ I ~ (a 2-r2)-1(logr-l+7t)27tr --+ 0
as
r
--+
0: .
-r(r)
I ~ \~ (R 2-a2r 1(logR+7t)2\tR
--+
0
as
R
--+
+00.
r(R)
Hence R
R
~ (a 2+x2)-1(logx)2dx+ ~ (a 2+x2)-1'(logx+7ti) 2dx+o(l) r
r
=
Making r integrals.
--+
0, R
--+
27tires[ai; (a 2+z2)-1(Logz)2] = 7ta- 1 (loga++7ti)2.
+ 00 and separating real and imaginary parts, we find both
R
R
r
r
3.8.2. ~ (l+x 2)-2Iogx dx+ ~ (1 +X2)-2(logX+7ti)dx+o (1)
= 27tires[i; (l+z 2)-2Logz] = {7t( -2+7ti). Making r
--+
0, R
--+
+00 and separating real and imaginary parts, we fllld I
3. COMPLEX INTEGRATION
185
x a(1+X 2)-2; if x < 0, then f(x) = (-x)a(l+ (l_r 2)-2ra+1 7t -+ 0 for -I < a as r -+ 0; ~
3.8.3. If x> 0, then f(x) +x2)-2exp(a7ti);
<
(R 2_1)-2RaH
I ~ 1<
I I
-~~
-+
=
0 for a
<
3 as R
-+
+00. Hence for -I
[1+exp(a7ti)]I+o(1) = 27tires(i;f)
<
a
<
~~
3 we have:
t7t(I-a)exp(ta7ti),
=
or
3.8.4. We have: ~
-+
0 as r
0, ~
-+
r(r)
-+
0 as R
-+
+00; moreover, ~
r(R)
-+
[r, R]
+00 -+
~ (I +x 2)-1Iogx dx =
0 by Exercise 3.8.1. On the other hand the integral
o
+00
over the segment [2-1/ 2R(I+i), 2-1/ 2r(l+i)] tends to -2-1/ 2(1+i) ~ (1 +it 2)-1 X o +00
X (logt+-ii7t) dtasr-+ O,R-+ +00. ThisimpIies ~ (1+it2)-I(logt+ti7t)dt=O. o
Separating real and imaginary parts and using Exercise 3.7.9 (ii) we obtain both integrals. 3.8.5. (i) Cf. Exercise 3.8.1; R
(ii) ~ (l+x 2)-llog[ Jh+X2 +iArg(x+i)]dx+o(l) -R =
make R
-+
27tires[i; (l+Z2)-1 Log(z+i)]
=
7tLog2i;
+ 00 and separate real and imaginary parts;
(iii) by Exercise 3.8.1 (a = 1), +00
~ (1+X 2)-11og(x+x- 1 )dx o
1
~ +
= 7tlog2 =
0
+00
~
1
= 2 ~ (put x = t- 1 ).
l'
3.8.7. Integrate the branch of log(z-a) for which 0
0
<
arg(z-a)
<
27t;
2n:
~
=
C(o •• 1)
~ [logle i6 -al + iarg(e i6 -a)]d(ei6); 0
1
the integral over the lower edge of the slit is equal to -
~
(loglx-al+27ti)dx
a+r 1
whereas an analogous integral for the upper edge is equal to ~ log Ix-al dx; a+r 2n:
~ -C(a; r)
= - ~ (logr+i()ire'6d() o·
=
0(1)
as
r
-+
O.
SOLUTIONS
186
The sum of all four integrals is equal to 0 and the result follows by separating the real part. 3.8.9. (i) Suppose that 0
fez)
< r< =
t
and R > 1 and integrate
{(1+Z)[Z2(1-Z)1/3]}-1,
which is analytic in C",-[0, 1], round the boundary of K(O; R) "'- {K(O; r) u uK(I; r) u [0, I]}. Obviously ~ = 0(1) as R -+ + 00; if f is the branch taking C(O;R)
positive values on the negatiye real axis, then on the upper edge of [r, 1- r] we have: fez) = exp(27ti/3) {(1+x)[x2(1-x)]1/3tl and after describing -C(1; r) the value of f on the lower edge is obtained by multiplying the former value by exp(27ti/3). Hence [exp(27ti/3)-exp(47ti/3)]/+0(1)
=
27tires( -1 ;f)
=
27ti2- 1/ 3
as r-+O, R-+ +00; (ii) integrate fez) = z2n[z(1-r)tl/3 which is analytic in C",-[ -1, 1] round the boundary of K(O;R),,{[-l,l]uK(O;r)uK(-l;r)uK(I;r)}, where R> and 0 < r < t; f takes opposite values on the upper edge of [r, 1-r] and lower edge of [-l+r, r] and vice versa, moreover, the integrals round the circles of radius r tend to 0 as r -+ o. Hence
+
2I[exp(7ti/3)-exp(-7ti/3)]+0(1)-21tires(00;f) Now,
fez)
=
_Z2n-l(1_z- 2)-1/3
=
-rn-l(1+-}-Z-2+
=
!::
o. Z-6+ ... )
and finally 1·3· .... (3n-2) 2/'. 7t _ sm 3 - 7t ~ . 6 . ... . 3n . 3.8.10. After describing -C(1; r) f is multiplied by exp( -27tpi); moreover, the integrals round -C(O; r), -C(1; r) and C(O; R) tend to 0 as r -+ 0, R -+ +00. Hence /[l-exp( - 27tpi)]+0(1) = 27tires( -1 ;f) = 27tie-nPi2p-3p (p-1). 3.8.11. Integrate fez) = (l_zD)-l/n round the boundary of n-l K(O;R)" U [0, exp(2k7ti/n)], R > 1. k=O
After describing the point exp(2k7ti/n) the values on the opposite side of u corresponding slit are multiplied by exp(27ti/n); moreover, the integ~als over
3. COMPLEX INTEGRATION
187
both edges of any slit are equal to [1-exp(27ti/n)]I. Hence n[l-exp(27ti/n)]I = 27tires(00;f) = -27tiexp(i7t/n).
3.9.1. We have: I-zl = I > IF(z)l; hence by Rouche's theorem both equations: -z = 0, -z+F(z) = 0 have the same number of roots in K(O; I), i.e. exactly one. 3.9.2. Consider the variation of argP(z) as z describes in the positive sense the boundary y of K(O; R) n (+, +): Ll[O.R]argP(z) = 0; Ll[IR.O]argP(z) = -arg[l+i(y8+5)-1(-3y 2+7y)] = O(R- S ); on the circular arc LlargP(z) = Ll arg[z8(1 + o(R- S ))] = 47t+o(I). Hence LlyargP(z) = 47t+o(l) = 47t for all R sufficiently large. Now, n(r,O) = (27t)-1Ll yargP(z) = 2.
3.9.3. P has no roots on coordinate axes; it is real on the real axis and attains a positive minimum at x = -!-; on the other hand, reP(iy) = y4+10 ~ 10. Suppose now that z describes the boundary y of K(O; R)n (+; +). We have: Ll[O.R]argP(z) = 0 because P is real and positive on [0, R]; on the arc of C(O; R) LlargP(z) = Llargz4+Llarg[l+z-4(2z 3 -2z+10)] = 4.-!-7t+o(l) = 27t+o(1) as R-+ +00, moreover, on [iR,O] the initial value of argP(z) is arg[I-2i(R4+ + 10)","1 (R+R 3 )] = 0(1) as R -+ + 00, and all the time reP(iy) ~ 10, the final value argP(z) being 0 which means that Ll[iR.O]argP(z) = 0(1). Finally, Llyargp(z) = 27t+o(1) for large R and being a multiple of 27t it is equal to 27t for all R sufficiently large. A similar reasoning can be made for remaining quadrants. 3.9.4. P(iy) = (_I)n[y2n_ia 2y2n-1]+b 2, hence w
=
P(iy)
=
y2n+b 2_ia2y 2n-1
for even n.
Suppose that z ,describes y = a[K(O; R)n {z: rez > O}] and R is large. For z = iy moving on [iR, -iR] the point P(iy) describes an arc with end points W1 = R2n+b 2_ia2R2n-1, W2 = R 2n+b2+ia2R 2n-1 situated in the half-plane {w: rew > b2}, hence Ll[iR._IR]argP(z)
as R
-+
= argw2-argw1 = 2 arc cot a- 2(R+b 2R-2n+1) = 0(1)
+ 00. If z is moving· on the arc of C(O; R), then Ll argp(z) = Ll argz2n +Ll arg(1 +a2z-1+b2z-2n) = 27tn+o(I).
Consequently, Llyargp(z) = 27tn for all R sufficiently large. If n is odd, then AargP(z) = 27tn+o(l) on the arc of C(O; R). If z = iy moves on [iR, -iR], then w = P(iy) = b2_R2n+ia2R2n-1 describes an arc 1 with end points W1 b 2- R2n+ la 2R 2n-1, W2 = b2- R2n_ ia 2R 2n-1, the equations of 1 being: u b2_y2n, V = a ZyZII-l. We have: dargw/dy = a2Iwl-2(b 2+y4n-Z) > 0 which
r
r
SOLUTIONS
188
means that argP(iy) decreases as z moves on [iR, -iR]; if y
+ iv E ( - ; +); WE
if 0 (+; -); if y <
lim argw2
=
Vb,
then
then
+; +); if -). Moreover,
W E (
WE ( - ;
Vb, then w = u+ Vb < y < 0, then
>
lim argw1
=
7t,
R--,+oo
-7t, hence LI[iR,_iR]P(Z)
=
-27t+o(1) and
finally,
n(r, 0)
R...,.+oo
=n-l. 3.9.5. Similar proof as in Exercise 3.9.4. 3.9.6. Izsl = 32 > l-z+161 on C(O; 2), hence the polynomials zS, zS-z+16 have exactly 5 roots in K(O; 2). Moreover, l-z+161 > 14 > IzsJ on C(O; 1), hence the polynomials -z+16, zS-z+16 have no roots in K(O; 1), and consequently, all the roots are situated in {z: 1 < IzJ < 2}. If z moves on [iR, -iR], then P(iy) = 16+iy(y4-1), i.e. LI[iR,_iR]argP(z) = -7t+o(I), whereas on the right-hand half of C(O; R) LlargP(z)
=
5L1argz+o(1) = 57t+o(1).
Hence LI argP(z) = 47t+o(l) = 4n for all R sufficiently large which means that there are two roots of positive real part.
u
FIG. 3
3.9.7. If z = iy moves on [iR, -iR], then W = u+iv = P(iy) = 1-2y2+ +iy(y2-1)(y2_4) describes an arc F1 which can be outlined by means of the following table indicating the signs of u, v as depending on y (Fig. 3): y
u v
R
2
-
+
o
0
o +
0 _2- 1 / 2 +.i\ 0 0
-1 0
-2
+
-R
0
We have: argw1 = -t7t+o(I), argw2 = --t7t+o(1) and Llr,argw = -7t-J-o(l) as R -+ +00; moreover, Llargw = 57t+o(1) on the right-hand half of C(O; R), hence n(r,O) = 2--1-0(1) = 2 for R sufficiently large.
3. COMPLEX INTEGRATION
189
3.9.8. We find the variation of argP(z) as z moves on the contour y consisting of [- R, R], two arcs of C(O; R) contained in D and the segment IR = hi R2 1+ +i, _yR2 I+i]. We have: P(x) > 0 on the real axis and LlargP(z) = 0(1) on both circular arcs, henc"e LI argP(z) = 0(1) on all three arcs. If z = x+i, then P(z) = x 4-6x 2 +3x+4+i(4x 3 -4x+3). The. polynomial 4x 3 -4x+3 has a unique root Xo E (-2, -1) and x~-6x5+3xo+4 < 0 which follows from the inequality: (x 2_2)2 < x(2x-3), x E [-2, -1]. Hence, if R is large and z moves on fR' then at the beginning argP(z) = 0(1) and imP(z) > 0, afterwards P(z) meets the negative real axis for z = xo+i and then for x decreasing we have imP(z) < 0 and at the end point again argp(z) = 0(1). This means ,that LlI~argP(z) = 27t+o(I) and also Llyargp(z) = 27t+o(I) = 27t for all sufficiently large R. 3.9.9. [iz 3 [ = 27/8 < 4 < 81/16-1 ~ [z4+1[ on C(O; f) and by Rouche's theorem both polynomials Z4 + 1, Z4 + iz 3 + 1 have four roots in K(O ; f). The number of roots in (+; +) can be found similarly as in Exercise 3.9.2.
3.9.10. (i) [-5z+II > 3 > IZ41 on C(O; 1), hence q = 1; (ii) 1-4z5 -II > [Z8+ Z 2[ on C(O; 1), hence q = 5. 3.9.11. laz"[ = a > e > eX = [-e"[, z = x+iy E C(O; 1); hence both equations: azn = 0, azn-e'Z = 0 have n solutions in K(O; 1) by Rouche's theorem.
3.9.12. If rez > -f, then Iz+21I1z+I[ > 1, whereas I-e-'Z[ = e-~ ~ 1 for z = x+iy in the right half"plane. Hence on the boundary of K(O; R) n {z: rez > O} we have Iz+2I/iz+I[ > II-e-'ZI. Since (z+2)(z+1)-1 has no roots in the right half"plane, so does (z+2)(Z+1)-1_ e-'Z. 3.9.13. Compare Il-zl, I-e-'Z[ on the boundary of K(O; R)n {z: rez > O} for R> l+1. If z = iy E [-iR, iR], then [l-iYI;;?; l > [-e-:.'[ = 1; if [zl = R, rez;;?; 0, then [l-zl;;?; [zl-l = R-l > 1 ;;?; [-e-'Z[. Now apply Rouch6's theorem. 3.9.14. The case a = 0 is trivial. If [a[-l ~ 2n and Zl, ... , Zn are roots of the ro1ynomial, then [ZlZ2 ... zn[ = la[l ~ 2n, hence IZkl ~ 2 for some k. If la[-l > 2n, then [al- 1 > (2+ <5)n for <5 > 0 sufficiently small and we have: let ·1 (1 +z)[ > [a[-l > (2+ <5)n ;;?; [z[n on C( -1; 1+ <5) which means that the polynomials a-1(1+z), a-1(1+z)"+z" have the same number of roots in K(-I; I+~) for all sufficiently small <5 > 0 the number being equal to 1. Note that
n
d>O
K(-I; 1+<5) c::: K(O; 2).
SOLUTIONS
190
3.9.15. l(z-1)P1 = 1 > a ~ I-ae-"'I = lale- x for z = x+iy E C(1; 1), hence by Rouche's t.heorem the equation (z-l)P-ae-'" has p solutions in K(l; 1). If Zo is a root of order ~ 2, then (zo-l)P-ae-"'o = 0, p(zo-1)p-1+ ae-",o = 0 which means that either Zo = 1: or Zo = -p+1, Le. either eZO = 0, or rezo ~ 0 and this is a contradiction. 3.9.16. tanz is uniformly bounded on the boundaries of squares Qn with corners n7t(=t=FFi), hence for z moving round 8Qn: Llarg(z-tanz)=Llargz+ +Llarg(1-z-1 tanz) = 27t+o(1) = 27t for n sufficiently large. On the other hand, if N is the number of roots and P is the number of poles of z-tanz inside Qn, and n is sufficiently large, then N - P = 1. Obviously P = 2n and this implies N = 2n+1 and after rejecting a triple zero at the origin there are 2n-2 nontrivial Zeros left. The existence of 2n-2 real, non-trivia~ zeros inside Qn is an immediate consequence of the Darboux property. of real continuous functions. .
3.9.17. If Izl = 1, then I(z-a m)/(l-om z)1 = 1, hence IF(z) I = 1 > Ibl on C(O; 1); by Rouch6's theorem withf= F,g= -bthe equationsF = O,F-b = 0 have n roots in K(O; 1). 3.9.18. Apply Rouche's theorem with f= F, g = -F(O). 3.9.19. Suppose that r < R1 < R2 < R. By the argument principle n('YR 2 , a) -n('YR1' a) = Na-P, whereNa is the number of roots of the equationf(z)-a = 0 and P is the number of poles off in the annulus {z: R1 < Izl < R 2}. 'We have Na = P = 0 by our assumptions. 3.9.20. Suppose that D~ = {z: loglf(z) I ~ <5}, <5 > o. If <5 is small enough, then a, Zo are interior points of D~. Since the possible multiple points of 8D~ correspond to zeros off', we can take <5 so that 8D~ = 'Y1 +'Y2+ ... +'Yn, 'Yk being closed analytic curves with positive orientation. If is the image cycle of 8D~ under J, then ncr, a) = Na-P = 0 because r c R(O; e6) for lal >~. Now, P = 1 and, consequently Na = 1.
r
CHAPTER 4
Sequences and Series of Analytic Functions 4.1.1. fn(x) = xexp(-+n 2x 2)
= 0 on the real axis because sup [fn(x) I = -,ie/no x
Suppose In(z) =f(z) in some disk K(O; r). Since fez) = 0 for all z E (-r, r), we have also fez) = 0 for all z E K(O; r) and, moreover, f:(z) =I'(z) = 0 in K(O; r). In particular f:(O) ~ 0 which is obviously false. 4.1.2. If Izl::(; r < I, then Iun(z) I ::(; r n(l_r)-1(I_r 2)-1 = An and LAn is obviously convergent. If Izl ~ R > I, and n is sufficiently large, then lun(z)1 ::(; 2/R"+1. This proves a.u. convergence. Now, un(z) = (l_zn)-1_(I_zn+1)-1+ n
+zun(z) and hence putting sn(z)
L
=
Uk(Z) we
obtain sn(z) = (l-Z)-1-
k=l
< 1 and n ~ +00, then s(z) = limsn(z) =
-(I- z n+2)-1+ zsn(z).1f Izl
(I-Z)-1-
n
-I+zs(z); if Izl
> I, then
s(z) = (l-Z)-1+ ZS(Z) and s(z) can be evaluated in
each case. 4.1.3. Isinz[2 = sin2x+sinh 2y (z = x+iy), hence [yl::(; Olog3, 0::(; 0 implies: 3-n [sinnzl < 3-n(l+exp(nlyl) ::(; 3-n+3-(1-9)n
which proves the uniform convergence in the strip limzl
r. n3n=1 00
1'(0) = 4.1.4. Suppose that fez)
= -~ 2m
rJ
Z E
t·
= f(z)-f(O) = _1_. 2m
C(O; '1)
rJ
zf(i;,) di;, i;,(i;,-z)
C(O; '1)
hence [f(z) I "-( M(r)lzl, where M(r) L.I~zn) has a majorant
=
< Olog3;
= -}(l+r) and 0 < r < 1. We have
K(O; r), r1
f(i;,) di;, i;,-z
n
< 1,
L
=
2(1-r)-1suplf(rleil1)l. This implies that o M(r}r" in K(O; r). 191
SOLUTIONS
192
4.1.5. im(z+n)-1 = -y[(x+n)2+ y 2rl,
I
hence the series
(_I)n+ lim(z+n)-1
is absolutely and uniformly convergent in K(zo; r), since it has a convergent majorant IAn- 2 ; re(z+n)-1 = (x+n)[(x+n)2+ y 2]-1 and from the fact that t(t 2+y2)-1 strictly decreases for t ;> Iyl it follows by Leibniz test of convergence that -1)n+lre (z+n)-1 converges. Now, the rest in an alternating series is bounded in absolute value by the first term omitted, i.e. by Ix+n+ 11 [(x+n+ 1)2+ +y2r1 ~ Ix+n+ll- 1 ~ (n+l-lx ol-r)-1 for large n which proves the uniform
I(
convergence in K(zo; r). 4,1.6. The points -n (n =
=
1,2, ... ) are simple poles of
f and res(-n;f)
(-I)n+\ hence
~f(z)dz
= 27ti
2: (-I)k+ In(y,-k) ,
y
the sum being finite. 4.1.7. We have T(n)
~
n and this implies that the power series on the righthand side is convergent (and also a.u. convergent in K(O; 1). If 0 < r < 1 and z E K(O; r) then
Izn(l-zn)-11
~
r n(1-r)-1
and this proves the a.u. convergence of the series on the left-hand side. Moreover, both sums represent analytic functions in K(O; 1). The identity of both functions OC!
follows from the fact that the double series
OC!
LL
•
zmn is absolutely conver-
n=lm=l gent and its sum does not depend on order of summation.
4.1.8. Differentiate k times the identity (1-Z)-1 = l+z+z 2+ ... 4.1.9. (i) K(O; 1), K(oo; 1);
(ii) the real axis; (iii) C""N 1, where N1 is the set of negative integers; (iv) the annulus {z: q < Izl < q-1}; no domain of analyticity does exist in
(ii).
n [1 + (R/n)2] exists and m
4.1.10. The finite limit H(R)
=
lim m
consequently
n=1
the given series has a convergent majorant
I
R2 H(R) (n+ It2 in K(O; R).
4.1.11. The series is a.u. convergent in C""N where N is the set of all integers, hence it. can be differentiated term by term; the sum to be found is equal to 7t 3 COS 7tz(sin 7tZ)-3. .
4. SEQUENCES AND SERIES
193
4.1.12. Suppose that Go is a subdomain of G such that Go c G and aGo is a cycle consisting of a finite number of contours. If w does not lie on I(aG o), the same is true for I" (aGo) for all n ;;:: N, hence
~ (J" (z)- wtlI~ (z)dz since
I"
~ (f(z)-wtlI(z)dz = 0, 1
-+
oGo
oGo
are univalent. This proves the univalence of f, unless it is a constant.
4.2.1. If ql
=
l/al, q2 = at/a2, ... , q" =;= an_liaR, then la"I- 1/" = (lqlllq21···lqnl)1/"-+ Iql·
4.2.2. (i) {-; (ii) e; (iii) 1; (iv) min(l, lal- 1). 4.2.3. (i) For any e
>
° there exists an integer k such that
la"1 1/n < R 1 1+e,
Ib"11/" < Rzl+e
for all n;;:: k;
hence
la"b" 111" < (RIR2)-1+s(Rl1+Rzl)+e2 for all n;;:: k and consequently (ii) a" = b"(a"/b"), cf.(i); (iii) both series L a"z", L b"z" are absolutely convergent in K(O; R o), hence after multiplication and rearrangement according to increasing powers of z we obtain a convergent series for any z, Izl < Ro. 4.2.4. Take real increments of z and verify by induction that imI(k)(z) for all z E( - 15, b). Note that ak = j
4.2.5. I(reiB)
L: a"r"ei"9,
=
";0
=
°
00
I(rei~
I
=
ii"r ne- i"9. The· product If can be
";0
arranged and written in the following form 00
.z= la" 12r2"+Al (r)e
i9
+Bl (r)e- i9 +
...
"=0
which follows from its absolute convergence. The sum is uniformly convergent in [0,27t] for any fixed r E (0, R) and can be integrated term by term w.r.t. () 2n
which yields the desired result because ~ e i"9d() o
=
°
for any integer n #0 .
4.2.6. We have 2n
(27t)-1
~ II(re f B)1 2 dO
o
2: la"1 r 00
=c
2 2 ":(
,,~o
M2,
SOLUTIONS
194
hence
m
for any
rE
[O,R),
L lan 2R 2n ~ M2, n=O
I.e.
l
"'"
pr
m being arbitrary and this implies
L Ian 12R2n ~ M2.
n=O
m
4.2.7. lanl2r2n~
L lanl2r2n~ [M(r)]2
by Exercise 4.2.6.
n=O ao
4.2.8.
L Ian 12 ~ M2
by Exercise 4.2.6, hence an ~ O.
n=O
4.2.9. Suppose that fez) o;E. z and f(zl)
=
f(z2) , i.e.
ao
'2: an (z1-z1) n=1 Mter dividing by
Zl-Z2
=
0
for
Zl
1= Z2'
we obtain
ao
'2: an (z1- 1 +z1- 2Z 2+ n=2
and hence
... +z1- 1)
=
-al
<
'2:nlanlrn-l n=2
ao
ao
'2:
lall = I an (z1- 1 +z1- 2Z 2+ ... +z1- 1)[ "n=2 which is a contradiction. 00
L
4.2.10., 1 ;> nrn-1/n! = er -1 holds for r ~ log2. However, from the prop, n=2 erties of exponential function it follows that log2 can be replaced by 'It. 4.2.11. Put in Exercise 4.2.3 (i): an = Cn, bn = lIn!; this gives Rl The sequen«e {c n(OR)n} is bounded, hence IclIl (OR)n ~ M(O) ana also
I
CI)
If(z) I ~ M(O)
•
(n!)-l(Jzl/OR)n
=
M(O)exp(lzl/OR).
n=O
4.2.12. We have r
A (r) =
~~ If'(pe i6Wp dp dO K(O;r)
now, f'(z)
=
=
2n
~ pdp
0 1f'(pe WdO);
0
. 0
i6
al +2a2z+3a3z2+ ... and by Exercise 4.2.6 r
ao
('2:
2'1t ~ n 2 1an 12p2n-l) dp o n=1 which can be integrated term by term. A (r)
=
= 00.
4. SEQUENCES AND SERIES
195
4.3.1. The power series expansions of elementary functions known from the real analysis also hold in complex case. The cases (i)-(iv) can be settled by the method of undetermined coefficient~, (i) Z == (z_-}.r+t z3- ... ) (1+<-'IZ+C2Z2+ ... ), hence c1--} = 0, C2--}C 1 + +t = 0, etc. and z/Log(1+z) = 1+-tZ-112 z2+i4 Z3+ ... , R = 1 which is the distance between z = 0 and a singular point z = -1; 4 4 (ii) 1+ 31 z 2+ 45 z - 7.44 135 (1'1'1')
1
- 41 Z2 - %1 Z4 -
Z
6
+ ... ,
R = 1;
23 6 15. 48 Z - ... ,
1 2 5 4 61 6 (iv) 1 + "2 Z + 24 Z + 720 Z + ... ,
1 1 2 (v) Iog2+"2 z +S z (vi) e [1+Z+.r+
R=~' 2'
1 4 192 z + ... ,
~ Z3+ ... J.
R=+oo.
4.3.2. It follows from Weierstrass theorem that
and putting Z = 0 we obtain: an = aO n+aln+ ... 4.3.3. (i) I+z+tz2+-1f-z3+ ... , R .. ) ( II
2
=
1;
R -- l',
5 Z3 1 4 +6 +zz +
z+z ... , (we can apply e.g. Exercise 4.3.2 with un(z) = (n!)-Iz"(I-zrn in (i)).
. I[ 1 1 1 ] 4.3.4. (I) an ="2 l(n-I) + 2(n-2) + ... + en-l)1
1) because k(n-k) 1 n1(1k+ n-k1) ' R
... + n-I
=
=
(_I)n-l n-l
(iii) an
=
(-l)n-12(2n+ 1)-1 ( 1
(iv) (COSZ)2 tlo
(1 +-}+ ... +
(ii) an
1, R
1
="2 (1 +cos2z) , -1- 00;
2n~I)'
=
n1(1 +"21 +
...
1;
cf. (i), R
+ ~ + ... + 2~)' hence a2n
=
=
=
1;
R = 1;
2 2n - 1 (_I)n (2n)!' a2n+l
=
0 (n ~ 1),
SOLUTIONS
196
f
(v) f'(z) = Z-1[1_(1+Z2)-V 2] = -
(-~/2)z2n-1
and hence J(z) = log2-
n=1
- n=1f (2n)-1 (-~/2) z2n, R
1;
=
(vi) J(z) = -}Log(yI+z+yI-zY = -} log2+-}Log(I+YI-z2); the second term expansion can be obtained from (v) by putting iz instead of z and therefore
L (_I)n(2n)-1(-~/2) z2n, 0()
J(z) = -}log2+-} 4.3.5. limsin7tz 2/sin 7tZ =
removable; R
=
+
R = 1. n=1 2n (cf. Ex. 3.4.8) and therefore all singularities are
00 •
4.3.6. If t e (-1, 1) then
, /- , / ,/• /- ( 1 1 1.3 ) V 2re V I+it = (1+v I+t 2)1/2 = V 2 re 1 +2 it + 2.4 t 2_ 2.4.6 it 3+ .... Hence . (,/-2)1/2_./-( 1 t 2- 2.4.6.8 1·3·5 t 4 + ... ) forrea I tee-I, 1) (1) I+vI+t - V2 1+ 2.4 and also for complex t e K(O; 1); put next z = t 2 ; (ii) t(l+yI+t2t1/2 = yiim(l+it)1/2 for t e (-1,1) and similarly as in (i) we obtain:
~ - 2 ~ ~: 6 z+ 2 .I~ .3~ .58·?1O Z2~ ... ). (cO+C1Z+C2Z2+ ... )(1-z-Z2) = 1 and by comparing the coefficients
(1+yl+zt 1/2 = yi( 4.3.7. we obtain:
Co
= 1, -CO+c1 = 0, cn+2-cn+1-cn = 0;
hence C n
n
=
(0) =5-112 [('/5+1 )n+1 +(_I)n ('/5-1 )n+1] J (n) _ = _ V V
nl
0,1,2, ... ; R
2
2
= -}(y5~I).
4.3.8. We have
J(z)
=
~t:~;; + ~
= (-
~~ - ~~ -
... ) :
(Z2+
~~ +
... ) +
~ ~ 0 as z~ 0,
hence the singularity at the origin is removable; J(-z)
=
-e%+I-I +Z-1+~= -_I_+Z-1_~=-J(z). e%-1 2 eZ-l 2
4. SEQUENCES AND SERIES
197
4.3.9. (i) If f is the function of Exercise ~U.8, then -!-zcoth-!-z = l+f(z); (ii) replace z by 2iz in (i). 00
4.3.10. (i) z[Log(sinz/z)]' = zcotz-l = -
L
[22kBk/(2k)!] Z2k by
Exercise
k=l
4.3.9 (ii) and after integrating 00
Log(sinz/z)
=
,L [2 2k -
-
1 B k /k(2k)!]Z2k,
< 7t;
Izl
k=l
(ii) use (i) and the identity Log(sin2z/2z) = Log(sinz/z)+Logcosz, 21z1 hence
< 7t;
00
Logcosz= - ,L[22n-1(22n-l)Bn/n(2n)!]z2n,
Izl
<-!-7t;
n=l
(iii) differentiate (ii): 00
tanz= ,L[22n(22n-l)Bn/(2n)!]z2n-1,
Izl
< -!-7t;'
n=l
(iv) sec 2z
00
=
L
[22k+1(22k+2-1)Bk+1/(k+ 1) (2k) !] z2k, which is obtained by
k=O
differentiating (iii); Izl < -!-7t; (v) tan 2z = sec 2z-1; (vi) subtract both sides in (i) and (ii): 00
Log[tanz/z] = 2)22n(22n-1_l)Bn/n(2n)!]z2n,
Izl
< -!-7t;
n=l
(vii) differentiate both sides of (vi) and replace 2z by z: 00
z/sinz = 1
+,L [(22n-2)Bn/(2n)!]z2n, n=l
Izl
< 7t.
SOLUTIONS
198
4.3.12. yuArctanyu = u--fu 2++U 3- ... , lui < 1. If Izl < -}, then Iz/(I-z)1 hence the radius of convergence R > t. Now, the shortest distance from the origin to the points where p ceases to be analytic (z =.1) is equal 1, hence
< 1;
R=1. 4.3.13. By equating Taylor's coefficients of both sides we obtain a1 = 1, 3a 2 = 2a1, ... , (2n+l)a n+1 = 2na n, ... , hence
2n a n+1 = 2n+l an· 4.3.14. Putting z
u2 in the formula of Exercise 4.3.13 we obtain
=
(l-u2)-1f2Arctanu(l-u2)-1f2
=
(l-u2r1/2Arcsinu
_ 2 3 2·4 5 - u+ 3 u +3":5 u
+ ... ,•
integrate now both sides.
4.3.15. Rn(z) = J(z)-sn(z) = zn+1 pn(z), where Sn is a polynomial of degree at most nand Pn is analytic in K(O; R). We have: Pn(z)
z-n-1 [J(z)-sn(z)]
=
=
~
(27ti)-1
(C -zr 1Pn(C)dC
C(O; ,)
(27t0- 1
=
~
c n- 1(C-zr 1J(C) dC
C(O; r)
by Exercise 3.1.28. 4.3.16. We have
sn(z) = J(z)- Rn(z) =
~
(27ti)-1
(C-z)-lJ(C)dC-(27ti)-1
C(O; r)
=
~
(27ti)-1
~
(zg)n+ 1(C-Zr1J(C) dC
C(O; r)
c-n-1(C-Zr 1(ClI+1_ Zn+1)J(C)dC.
C(O; r)
4.3.17. If Iz-al < R, then J is analytic in K = KC-}(a+z); R--}Iz-al) and the Taylor series with center -}(z+a) has the following form:
Jm =J(-}(a+z))+(C--}(a+z))J'(-}(a+z))+-}(C--}(a+z)Y!,'(-}(a+z))+ ...
Put first C= a and then C= z and subtract both sides. 4.3.18. If an = lin+i(Jn, then
.L (lin+i(Jn) (cosnO+isinnO)r 00
n=l
n
= P(O)-HQ(O)
4. SEQUENCES AND SERIES
199
and consequently co
P(O)
=
L:(cxncosnO-PnsinnO)rn. n=O
Thus by Euler-Fourier formulas 2n
2n
7tCX nr n = ~ P(O)cosnOdO,
-7t{3nrn
=
o
\
o
P(O)sinnOdO,
or 2",
7tan r n = 7t(cx n+i{3n)
=
~ P(O)e- in6 d8 o
and similarly 2n
7tan r n = ~ iQ(O)e- in6 dO. o 2n
4.3.19. lanl ~ (7tr n)-1 ~ P(O)dO by Exercise 4.3.18 and the equation P(O) o 2n
=
(27t)-1 ~ P(O)dO (cf. Ex. 3.2.5) yields lanl ~ 2r-n. Since r E (0, 1) can be aro
bitrary, we have lanl
~ '2.
4.3.20. IJ(z) I ~ I+la11· Izl+la21·lzI2+ ... ~ I+2Izl/(1-lzj) = (1+lzl)/(1-lzl). In order to obtain a lower estimate, note that I/J is analytic in K(O; 1) and satisfies the assumptions of Exercise 4.3.19. Equality holds for real z and J(z) = (1+z)/(I-z). 4.4.1. (i) divergent everywhere on C(O; 1); (ii) divergent only at z = 1; (iii) absolutely convergent on C(O; 1); (iv) conditionally convergent everywhere on C(O; 1) except at z = -1 and
z = +(I=fi}l3). 00
4.4.2. an = (-I)n; lim
L
r-----+1- n=O
anrn = lim (1+r)-1 = r~l-
i.
4.4.3. Zn = ei6 (1-r n expicx n), where Icxnl ~ +7t-b, b
IZnl = (I-2rncoscxn+r,;)1/2 = I-rncoscxn+O(r,;),
> 0; we have: rn = le"6_ zn l.
Note that a necessary and sufficient condition for Zn to be situated inside a Stolz angle is: (cOSCXn)-l = rnlrncoscxn = 0(1). 4.4.4. We may assume e l6 = 1, hence the radius of convergence of l:ak Zk is ~ 1.
SOLUTIONS
200
By Exercise 4.2.3 (iii) the esries L8kZk is convergent in K(O; 1), moreover
00
Thus
L
akz! can be considered as a transform of a convergent sequence {8n } by
k=O
means of a Toeplitz matrix.whose nth row has the following form: l-zn' zn(1-zn), z;(1-zn), ... Now, cf. Exercise 1.1.37 and Exercise 4.4.3. 4.4.5. (i), (ii) The series Z_-}Z2+-tZ3- '" whose sum in K(O; 1) is equal to Log(l+z), is also convergent for all Z E C(O; 1) except for z = -1, which follows from Abel's test of convergence. If z = cosO+isinO and 101 < 'It, then by Abel's limit theorem:
L (-I)n+ln-l(cosnO+isinnO). 00
Log (1 +e1fl) = log (2cos 0/2)+;0/2 =
n=l
Separate now real and imaginary parts. 4.4.6. The seriesz+-tz3+tzs+ .,. is convergent on C(O; 1) except at z = =fl and its sum is equal to tLog[(I+z)/(1-z)] in K(O; 1). By Abel's limit theorem:
tLog[(1+ei fl)/(1-e1fl)] = tlogl cot (0/2) I+i-i'lt = eifl+-te3ifl++eSifl+ ...
= cosO+-tcos30+tcos50+ .. , +i(sinO+-tsin30+ ++sin50+ ... ),
0
<0<
'It.
4.4.7. The series is an alternating series of the form L (-I)Run with Un = 22n(n!)2/(2n+l)!. From Stirling formula for asymptotic value of n! it follows that Un= O(n- I/2 ), hence Un -. O. Obviously Un decreases, hence the series is convergent. Use now Abel's limit theorem and Exercise 4.3.13 with ~,' -1. 4.4.8. The series is a convergent alternating series; put z = i and apply Abel's limit theorem for -(Arcsinz)2. ' 4.4.iO. Use the expansion 1-1 1 1 1·3 Arsinhz = Log(z+ V l+z 2) = z-32 z3 +S 2.4 zs- .. ,
and Abel's limit theorem for z = 1. 4.5.1. For example
00
00
n=l
n=l
L n- 2 zR+ L n- 2 z-n;
the regular part is convergent in
K( 00; 1). The principal part is convergent for Iz I > lim Ia_n 11/n = r and divergent
K(O; 1) and the principal part is convergent in
4.5.2.
4. SEQUENCES AND SERIES
201
for Izl < r, whereas the regular part is convergent for Izl < (limlanI1/n)-1 = R and divergent .for Izl > R. Hence the annulus of convergence is not empty iff r
4.5.3. (i) an = (-t)n+1, ao = -t, a_ n = (_2)"-1; (ii) an = ao = a_1 = 0, a_ n = (_I)n(3 n- 1 _2n-1); n = 1,2,3, ...
4.5.4. [z-t(a+b)]--fi-(a-b)2[z-t(a+b)tl+ ... 4.5.5. (i) (a-b)-1{a- 1[alz+(alz)2+ ... ]+b- 1[I+zlb+(zlb)2+ ... (ii) z-2+(a+b)z-3+(a 2+ab+b2)z-4+ ... 4 • 5•• 6 (1.)
Z -2 -z-4+ Z -6 -
...
1 -2
+ TZ 1 2 - g1Z 4 -
n;
... ,.
ex>
l:
(_I)n(2n-l_l)z-2n. n=2 1 -4+ 3Z 1 -6+ 4 .5.7. z -2+ -ZZ (ii)
...
7r
4.5.8. (i) an = (7t)-1 ~ exp(2cosO)cosnOdO; o ex>
(ii) an
L [k!(n+k)!]-l,
=
n ~ 0;
a_ n = an'
k=O
4.5.9. We have 7'1:
~
an = (27ti)-1
exp[tz(C-C-1)]C-n-1dC = (27t)-1 ~ exp[i(zsinO-nO)]dO
C(O; 1) 7r
-7r 7r
= (27t)-1 ~ cos(zsinO-nO)dO+i(27t)-1 ~ sin(zsinO-nO)dO -7r
-7r
7'1:
= (27t)-1 ~ cos (nO-zcosO)dO; -7'1:
ex>
an =
L
(_I)k[k!(n+k)trl(zI2)2k+n which is the coefficient of cn in the pro-
k=O
duct exp (zC 12) exp( - zC- 1/2).
4.5.10. fez) = 1+(z+I)-1+2(z-2)-1+3(z-2)-2; ao = ex>
L zn; n=O (ii) _(Z-1)-1+ 2: (-I)n(z-l)n; (iii) - L z-n.
4.5.11. (i) Z-l+
ex>
n-2
i, an =
(3n-l)2- n- 2.
SOLUTIONS
202 00
4.5.12.
L
00
c_nz- n+
n=1
L
cnz", where
n=O
L (-,!!2r200
n= -2nc_ n,
C
Co =
2- 1/2i(1 +
m] •
m=l 00
4.5.13. =j=[z-t(a+b)+
L c_nz-n] for
Izi
> max(lal, IbJ), with
n= 1
+00
4.5.14. The sum of the series
L
n=
(z-n)-2 is an analyticfunctionh in C",N.
-00
Suppose F is a compact set such that F (IN = 0. If bn
=
inf 11- z /nl then bn
-+
1
ZEF
as n -+ =j=oo and consequently Iz=j=nl- 2 < +n- 2 for all n sufficiently large which implies a.u. convergence and analyticity of h. Obviously z = n is a pole of second order with principal part (z-n)-2. If t = z-n, then 7t 2 cosec 2 7tz- (z-n)_2 = 7t 2 cosec 2 7tt- t- 2 = [7t 2t 2_(7tt-(7tt)3/3!+ ... )][t 2(7tt-(7tt)3/3!+ .. Yr 1 = 0(1) as t -+ 0 which means that 7t 2 /sin 2 7tz- h(z) has a removable singularity at z = n .. 4.5.15. Evidently both h(z) and 7t 2 /sin 2 7tz are periodic with period 1. Since g(z) = 7t 2/sin 27tz-h(z) has removable singularites at zEN, and has period 1, it is bounded in the strip limzl < 1. Suppose now that limzl ;;: 1 and 0 < rez < 1. We have: 00 00 1 00 00 2 2 In-zl- = (x-n)2+y2 .< 2 (n +y2)-1 < 2 (n 2 +I)-1.
nboo
nboo
b
k
Hence h is bounded in {z: limzl:;;:' I}. The same is tr~e for 7t 2 /sin 2 7tz and consequently g is bounded in C, i.e. g = const = 0 which follows from Exercise 4.5.14. 4.5.16. The series is a.u. convergent, hence it can be differentiated term by term and this gives the equality of Exercise 4.5.15. Note that 7tcot7tZ-Z- 1 -+ 0 asz-+O. 4.5.17. (i) 2z/(z2-n 2) = (-2z/n 2) (1 + (z/n) 2 +(z/n)4+ ... ) and consequently 7tcot7tZ = z-1-2s1Z-2s2Z3_2s3ZS- ... Hence (27t)2kBk= 2(2 k)!Sk; (ii) 3z- 1 +2z- 3+2z- s + ... -2(sl-I)z-2(s2-1)z3-2(s3-I)zs- ...
4. SEQUENCES AND SERIES
203
4.6.1. We have Icot7tzl ~ M for each Z E frQN and each N, where M does not depend on' N, (cf. Ex. 2.7.3, 2.7.7). Hence
I ~ J(z)cot7tZdZI ~ M oQ N
~ IJ(z)lldzl ~ 8MICNIIJ(CN)I, oQ N
where sup IJ(z) 1= IJ(CN) I· Hence ~ J(z) cot 7tzdz -+ zefrQN
oQN
°asN
-+
+
00.
If a1, ... , am
are inside QN, then
~ J(z)7tcot7tzdz oQN
= 27ti {
N
m
n=-N
k=1
L res[n; 7tJ(z) cot 1tz]+ L res [ak; 7tJ(z) cot 7tz]} ° -+
as N
-+
+00.
Note that res[n; 7tJ(z)cot7tz] = limJ(z)coS7tz(7tz-7tn)/(sin7tz-sin7tn) =J(n). z->-n
(z2+ z+1r 1 satisfies the assumptions of Exercise 4.6.1, hence s = -7t(b1cot7tal+b2cot7ta2) where a1 = -(I+q/3)/2, a2 = a1 are poles of J and b1 = res[al;!] = i/]l3,b 2 = res[a2;J] = -i/]l3. Hence 4.6.2. The function J(z)
s
=
3- 1 / 27ti[cot7t(I+i]l3)/2-cot7t(I-i]l3)/2] = -3- 1 / 227titani7t ]13/2 = 27t3-..1 / 2tanh 7t ]13/2,
=
because cot ( ; +cx) = -tancx. 4.6.3. (i) If a =F 0, =j=i, =f2i, ... thenJ(z) = (Z2+ a2)-1 satisfies the assumptions of Exercise 4.6.1 and therefore
L 00
(n 2 +a2 )-1
= -(2ai)-1 7t [cot 7tai-cot(-7tai)] = a- l 7tcoth7ta;
11==-00
(ii) if a4 =F 0, -14, _24, ... thenJ(z) = (Z4+ a4)-1 satisfies the assumptions of Exercise 4.6.1; it has 4 simple poles a. = 2- 1/ 2 a(=fl =fi) with residues b. = -{-a- 3 a. and therefore 4
~
L n~-~
(n 4 +a4 )-1
=
-}7ta- 4
L a. cot 7ta. '=1
= 2- 3 /2a-3 7t [(1 +i)cot 7ta(I+i)/~/2+ (l-i)cot 7ta(I-i)/t!2]
= 7ta-- 3 2- 1 / 2 (sin 7ta]l2+sinh 7ta~2)/(cosh 7taJl2-cos7taV2) hy Exercise 2.7.16 and the identity zcotz ,,- -zcot(-z);
SOLUTIONS
204
4
L 00
2s =
-i- 7t
n 2J(n 4+a4) =
n==-oo
a;lcot 7tay
v=1
where a y are the same as in (ii); hence by a1a2 2s
L
_a 2 we obtain:
=
(2a2rl7t(a2cot7ta1 +a1 cot7ta 2) = r 3 / 2a- 17t[(1+i)cot( -1 +i)a7tjV2+( -1+i)cot(1 +i)a7tjV2] =
and the result follows similarly as in (ii); (iv) if a i= ni Y (v = 1, ... ,4; n = 0,1,2, ... ), then
L 00
(n 4_a4)-1 =
-i- a- 47t
4
L
a y cot7ta y ,
\1=1
n=-oo
where a y = ai Y , v = 1, ... 4; --}a-47t(a1 cot 7ta1 +a 2cot 7ta 2)
=
--}a- 3 7t(cot7ta+icot7tai)
=
-ta-37t(cot 7ta+coth 7ta).
4.6.4. The function fez) = (z-a)-2 has a double pole· at z = a; moreover, res[a; 7t(z-ar 2cot7tz] = --7t 2 jsin2 7ta, hence by Exercise 4.6.1
I
00
(n-a)-2 = 7t 2 jsin2 7ta.
n=-oo
4.6.5. fez) = (z-a)-l(z-b)-l satisfies the assumptions of Exercise 4.6.1. We have: res[a; 7tf(z)cot7tz] = 7t(a-b)-lcot7ta and consequently s = -7t(a-b)-l X X
(cot7ta-cot7tb).
4.6.6. Putting a = i, b 2s+1
=
-i in Exercise 4.6.5 we obtain:
=
7ticot7ti
7tcoth7t
=
or
s
=
-}7tcoth7t--}.
4.6.7. If Z E frQN' then Isin 7tzl- 1 ~ 1 for each N, hence
I~
(sin 7tz)-lf(z)dz/
~ ~
oQ N
If(z)lldzl
~ 81 CNf(CN) I =
0(1)
oQ N
as N ~ +00 (CN being the point yielding the maximum of If I on frQN). If all a, are inside QN, then \ (sin 7tz)-l7tf(z)dz = 27ti{
". oQ N as N
~
~ res [n; ~(Z)] + ~ res [ak; ~f(Z)]} L.J SIll 7tZ L.J SIll 7tZ k=l
n=-N
+00. Note that res[n;
~f(Z)l
Sill 7tZ
=
fen)
cosn7t
=
(-I)nf(n).
=
0(1)
4. SEQUENCES AND SERIES
205
4.6.8. E.g. (iv). The function J(z) = Z2 /(Z4 +a4 ) satisfies the assumptions of Exercise 4.6.7. It has 4 simple poles ak = 2- 1/2 a(=j=I=fi), k = I, ... ,4 with residues (4ak)-1. Hence 4
1 "'., ( . )-1 = - 21 7t [alslll ( . 7ta1 )-1 + ( . )-1] , S = - 4 7t L...J ak Slll7ta k a2S1117ta2 k=l where a1, a2 are situated in the upper half-plane. Now, a l a 2 = -a 2 and therefore s
(2a2r17t(a2 sin 7ta2 +a1 sin 7ta1)J(sin 7tal sin 7ta2)
=
and the result follows. 4.6.9. We have: Isin az/sin 7tZ12
=
(sin 2ax+sinh 2ay)(sin 27tx+sinh 27ty)-1,
x+iy; therefore Isinaz/sin 7tzl ~ 1 on vertical sides since sin 27tx = 1 sin 27ta, sinh27tY ~ sinh 27ta. Moreover, Isinaz/sin 7tzl ~ 0 uniformly as Iyl ~ +00. Hence ~ J(z)dz ~ 0 as N ~ +00 (f(z) = 7tsinaz(z3sin 7tZ)-1). Now,
z
=
~
iJQN
~
=
27ti
iJQN
f
res[n;f],
res[O;j] = a(7t 2-a 2)J6,
res[n;j] = (-I)nn- 3sinna,
n=-N
n = =fl, =f2, ... and consequently N
a(7t2-a2)/6+2L(-I)nn-3sinna~ 0 n=1 (ii) put a = 7t/2 in (i).
as
N~ +00;
4.6.10. Similarly as in Exercise 4.6.9. we verify that Icosaz/sin 7tzl ~ 1 on v~r tical sides of QN, whereas Icosaz/sin 7tzl ~ 0 uniformly as limzl ~ + 00. Hence IN ~ 0 as N ~ +00. IfJ(z) = 7tcosazcosec7tz/(x 2_z 2), then res(n;f) = (-I)n X x cosna/(x 2-n 2) , res(x;j) = -7tcosax/(2xsin7tx) = res(-x;j), and consequently for all N sufficiently large R
IN
=
27ti{-7tCosax/(xsin7tx) +
N
L
(-I)ncosna/(x2-n2)} ~ O.
n=-N
4.6.11. If J(z) = z(sinh 7tazsin 7tZ)-l, then for real a "" 0 and z on the vertical side of RN we have: IzJ(z) I ~ (N+i-)2(I+a-2)/sinh7tlal(N+i-) --+ 0
as
N ~ +00;
for z on the horizontal sides of RN we have: IzJ(z) I < (N+t)2(l+a-2)/sinh7tlal-1(N+i-) ~ 0
I'his implies IN
~
O.
as
N ~ +00.
SOLUTIONS
206
(i) There are following singularities of f inside RN: simple poles -N, -N+l, ... , -1,0,1, ... , N, with residues (-I)"n/(7tsinh 7tan) , n i= 0, as well as simple poles -Ni/a, ... , -i/a, i/a, ... , Ni/a with residues (-I)mm X X [7ta 2sinh(7tm/a)tl; res(O;f) = 1/7t 2a. Hence IV
IN
=
27ti [(7t 2 atl+2
I
(-I)mm(7tsinh 7tamt 1
m-l N
+22.".:(-I)mm(7ta2sinh(7tm/a))-1]~0
as
N~+oo.
m=l
Note that each sum has a finite limit. (ii) Put a = 1 in (i).
°
4.6.12. Suppose that h,O are arbitrary real numbers satisfying: h > 1, < 0 < 7t/2 and D(h, 0) is the bounded, closed domain whose boundary consists of: 1 the segment I on the straight line rez = h such that -0 < argz < 0, 0
20 the circular arc y arising from I by inversion, 3° two segments joining the end points of I and y.
FIG. 4
Obviously a E D(h, 0) implies a- 1 each z E D(h, 0) and therefore
E
D(h, 0). If b = h-1coS 20, then rez ~ b for
1m/sinh 7tmzl ~ 1m/sinh 7tmbl ~ Am- 2 for all sufficiently large m and all Z E D(h, 0). This implies that both series or Exercise 4.6.11 (i) are uniformly convergent in D(h,O) and represent analytic functions of a in the right half-plane. By Exercise 4.6.11 (i) the difference or both sides vanishes identically for real a E [h-t, h], hence being analytic in the right half-plane, it vanishes there identically. An analogolls identity for I he
4. SEQUENCES AND SERIES
207
left half-plane is obtained by a change of sign. On the imaginary axis both series are divergent. 4.6.13. If N+iexceed
> lxi, the sum of integrals over the horizontal sides does not
"4(N+i-) [I +sinhlal(N+i-)] V2(N+i-) [sinh 7t(N+-iW t [(N+t)2_lxI 2]-t which tends to 0 as N -+ + 00. The parametrized integral over the vertical right-~
d
hand side (0
=
argz
E
[-i-7t, i-7tJ) has the form ~ + -/j
rr/4
~ + ~ , where the first -rr/4
/j
integral can be made arbitrarily small since the integrand is bounde~ independently of N and in the remaining intervals Isinazl/lsin7tzl-+ 0 uniformly as N -+ +00. This'implies that the integral over the vertical right-hand side (and also over the left-hand side) tends to zero. Hence N
res(x;f)+
I
res(n;f) -+
o.
n=1
Note that •
res(x;f) = -7tsinax/(2sin 7tX),
_
res(n,J) - (-I)
n
nsinan X
2
-n
2·
4.6.14. (i) Put a = 7t/2, x = 2z in the formula of Exercise 4.6.13; (ii) replace Z by iz in (i). 4.6.15. cot7tZ and coth7tz are uniformly bounded on frQN' where QN is the square of Exercise 4.6.1. Hence
~ f(z)dz -+ 0
as
N -+ +00.
iJQN
The integrand has following singularities inside QN: 2N simple poles on the real axis, 2Nsimple poles on the imaginary axis with res(n;f) = res(ni;f) = n- 7 coth 7tn and a pole of order 9 at the origin. In order to evaluate res (0 ; f), we find the Laurent series expansion of the integrand near z = 0 by using the formulas: 7tzcot7tZ = ,1-Bt (27t)2z 2/2!-B2(27t)4z4/4!+ ... ,
7tzcoth7tz = 1+Bt (27t)2z 2/2!-B2(27t)4z4/4!+ ... . Hence res(O;f) = 7t-t(27t)8[(B2/4!)2_BtB3/6!-2B4/8!] = 287t7(-19)/(6!7!). ()hserve that N
4~n-7coth7tn+reS(O;f)-+O
as
N-++SO.
SOLUTIONS
208
2>n
4.7.1. Suppose F is a compact subset of G and an arbitrary convergent series with decreasing positive terms. Suppose {An} is an increasing sequence such that b
I ~ W(z, t)dtl :;:;:; Un for any b > An and any ZE F . . An OC> An+l +00 The series 2..: ~ W(z, t)dt is uniformly convergent on F, hence ~ n=1 OC>
+ 2..:
An
a
A, ~
+
a
An+l ~
n=1 An
is analytic in G.
4.7.2. Suppose F is a compact subset of the right half-plane. We have: rez ;;?: b
> 0 for all z E F; if s > 0, arbitrary, and 2d- 1 e- M < s, then
B
I ~ e-ztdtl
< s
b
for any z E F and any b, B half-plane. Evidently
>
A, hence the integral is a.u. convergent in the right
A
hence
~ e-ztdt = _Z-l e- Az +z - l ~ Z-l o
A ~ +00,
as
4.7.3. The integral is a.u. convergent in C and can be differentiated under the sign of integral. +OC>
4.7.4. Put g(z)
=
~
(x-z)-2q;(x)dx and verify that
-OC>
Ig(z)- [f(z+h)-f(z)]jhl
=
O(h)
as
h
~
0.
4.7.5. Suppose that Zo is a pole of F and Zo is not an integer. Since F(zo+n) = (zo+n-l) ... (zo+l)zoF(zo), also zo+n would be a pole of F which is impossible for F is analytic in the right half-plane. 4.7.6. Consider G(z)jF(z). 4.7.7. Put x"
=
t.
4.7.8. If z = Re iO and s = a+ir, we have:
le-zzs-11 ::s;; Ra-lexp( - RcosO); hence by Exercise 3.7.2:
I. ~ I::s;; y(R)
re/2
Ra
~
0
exp(-RcosO)dO::S;;
_}7tR<7-1
-->
0
209
4. SEQUENCES AND SERIES
as R
~
+00
< 1). Similarly
(for a
rr./ 2
I ~ I ~ (j" ~ exp(-(jcosO)dO ~ -~:,(j" ~ 0 0
y(6)
<
as (j ~ 0 (for 0
. 7t a). Here yet) = te'6, 0 ~ 0 ~ T' Now, +00
~
0
=
~ e-xxs-1dx -
=
OD(6,R)
0
+00
~ e-iYexp [(s-l) (Iogy+ i7t/2)] idy
0
and finally
res)
+00
=
exp(7tis/2) ~ e-iYyS-ldy. o
4.7.9. Put s-1 We have:
=
a in Exercise 4.7.8 and separate real and imaginary parts. . +00
Iimr(l-a) cos (7ta/2)
=
0<->1
Iimr(s) sin (7ts/2)
=
7t/2
~ x-1sinxdx.
=
8->0
0
4.7.10. Putting t = sin 2 0 we obtain the integral 1
-i ~ t
P/
2- 1 (l-t)Q/2- 1 dt =
fB(P/2, q/2)
o
which is equal to +r(p/2)r(q/2)/r(f(p+q)). 4.7.11. (i) Put p = a+l, q (ii) put p = 3/2, q = 1.
=
I-a in Exercise 4.7.10;
4.7.12. (i) +7t- 1 / 2 r 2 (+); (ii) 27t 3 / 2 [r(+W 2 • 4.7.13. If e hence
=
exp(27ti/n), then (z-I)(z-e)(Z-e 2 )
(l-e)(I-e 2 )
...
(l-en -
1) =
...
(Z-e n- 1)
lim(zn-l)/(z-l)
=
Z----7'-l
=
z"-1 and
J
n. .
n sin(k7t/n)
n-l
Now, [1-e k [
=
2sin(k7t/n) which gives the identity
=
n2 1 -". Use
k=1
now the formular(z)r(l-z) = 7t/sin7tz with z = 7t/n, 27t/n, ... , (n-l)7t/n. 4.7.14. If f is nonnegative and decreasing in (0, 1] then 1
~ f(x)dx ,,10
1-10
n- 1 fJ( I /nHI(2/n)+ ...
+f(n/n)]'(
~ 0
f(x)dx
SOLUTIONS
210 1
which implies that the improper integral ~ f(x)dx exists, iff the finite limit o
lim n- 1 [f(1ln)+f(2In)+ ... +f(n/n)] exists and both are equal. Note that logr(x) n-+<Xl
is nonnegative and decreasing in (0, 1] and use the equality of Exercise 4.7.13. 4.7.15.
We
have:
I'(a) = logr(a+1)-logr(a) = loga, = +log27t.
therefore
l(a)
= aloga-a+b; b = 1(0+)
4.8.1. If I is an arc in the z-plane through zo', III its length and 11 is the spherical image of f(l) , then p (zo,f) = lim III I / III
4.8.2. p(O,f)
=
as 1 shrinks to Zo. 2 2Ial/(1+lbI ) is unbounded.
4.8.3. Note that any normal family in which the case of a.u. dive'rgence to is excluded, is necessarily compact.
00
4.8.4. Suppose that K(zo; 2r) c: D and M is the common bound of If I in this disk. Then for z E K(zo; r) by Cauchy's formula:
If'(z) 1= i (27ti)-1
~
(C-z)-2f(C)dCi ~ 2Mr- 1 ;
C(zo;2r)
hence f' are locally uniformly bounded and form a compact family: Marty's condition shows that n(z2- n2) form a normal family, whereas their derivatives do not. 4.8.5. If H is a compact subset of D, then If(z) I ~ R for all z E H, f E§". Hence IFof(z) I ~ sup IF(w)1 for all z E H. Hence {Fo!} is a compact family. Iwl";;R
4.8.6. {az} form in Izi > 1 a normal family by Marty's condition. Consider the sequence f.(z) = F[(2n+ )z]; we have:
+
fn(7t) = exp[(2n+-})7t] ~
f.(27t) =
+00,
°
which shows that {Fo!} is not a normal family. 4.8.7. By Marty's criterion {llf} is a normal family. Hence llf. is either a.lI. convergent in D and the limit function 1/g is analytic which makes g(z) = 0 impossible, or 11fn tends a.u. to 00 which gives g = 0.
°
4.8.8. By Ex. 4.8.7, g(z)-g(a) is either never in D"",a (which means that is univalent since a is arbitrary), or g(z)-g(a) == 0, i.e. g = const.
°
4.8.9. The sequence n- 1 (e nz -1) is equal to at z hence the family considered cannot be normal.
=
°and tends to
00
~
at z ' 1/2.
4. SEQUENCES AND SERIES
211
4.8.10. Let g = logf(z)/z with f E To be this branch which is equal to 0 at the origin. Since g does not take the values =j=27ti, so {g} is normal and even compact (cf. Ex. 4.8.3), since g(O) = O. By Exercise 4.8.5 To is compact. 4.8.11. The functions gn form a normal family (gn i= =j=27ti). Since gn(O) = 0, contains a subsequence gnk a.u. convergent in K(O; 1) whose limit g is analytic and g~/O) -+ g'(O) = (3 with finite (3. Now, g~k(O) = -1/IY. nk -+ 00 which is a contradiction.
gn
4.8.12. Obviously we can assume IY. = O. Since a family is normal, iff it is locally normal, we can also assume that D = K(O; 1). Now, 10g[f(z)/f(0)] = g(z) are analytic in K(O; 1) and omit the values =j=27ti; hence {g} is normal, and also compact (Ex. 4.8.3; g(O) = 0). By Exercise 4.8.5 {f(z)/f(O)} form a compact family, as well as {f(O)/f(z)}. Hence there exists a finite mer) such that If(z) I ~ If(O)lm(r) and also If(O)1 (m(r)tl ~ If(z) I in K(O; r). From this the normality readily follows. 2~
4.8.13. (J(zo)Y
=
(27t)-1 ~ f2(zo+re'8)dO, by Cauchy's formula: hence o 2~
If(zo)1 2 ~ (27t)-t ~ If(zo+re iO )1 2dO. o
After multiplying by r and integrating W.r.t. 'r over [0, R] we obtain 7tR2If(zoW
~ .. ~~ K(zo; R)
which shows that
If(z)1 2dxdy
~ ~~ If(z) I2dx dy ~ M D
CHAPTER 5
Meromorphic and Entire Functions 5.1.1. In what follows H denotes an arbitrary entire function. <Xl
(i) F(z)
=
L z2[n(z-n)rl;
H(z)+
"=1 00
(ii) F(z)
=
L
H(z)+
z/(z-a");
"=1 00
L Z2 /[n(z- yn)];
(iii) F(z) = H(Z)+Z-1+
"=1
(iv) we have n(z-n)-2 = n- 1[1+2(z/n)+3(z/n)2+ ... ], Izl < n, and smce In(z-n)-2- n- 11~ (n- )/71)-2 in K(O; yn), therefore
L z(2n-z)/[n(z-n)2]; 00
F(z)
H(z)+
=
"=1 00
(v) F(z)
=
H(z)+
L
[n 2(z-n)-2-1-2z/n+(z-n)-1+1/n];
"= 1 00
(vi) F(z)
=
H(Z)+Z2
L
00
[n(z-n)]-1 +Z2
"=1
L
[(z+n)r 1
"=1
00
L
(vii) F(z)
=
[n(z2-n 2)r1; "=1 H(z)+7tcot7tZ (cf. Ex. 4.5.16);
(viii) F(z)
=
H(Z)+Z-1_ Z
=
H(z)+2z 3
00
L [n(z+n)]-1. "=1
00
5.1.2. F(z)
H(z)+
L
(-l)"/(z+n). "=1 5.1.3. (Z-W)-1 is the principal part at a pole wand =
I(Z-W)-1+ W-1+ZW-21 ~ 21z1 2 1wl- 3
for
Izl <-kIwi.
There are 8k poles on the boundary of the square with vertices k(=Fn':i) and 212
5. MEROMORPHIC AND ENTIRE FUNCTIONS
213
grouping together all corresponding terms, we obtain after dropping a finite number of initial terms an absolutely and uniformly convergent series. Hence F(z)
=
H(Z)+Z-l+
'2:' [(z-m-ni)-;-1+(m+ni)-1+ z(m+ni)-2]. m.n 00
5.1.4. We have: e- t (z-l
L
=
(-1)nt n+z- 1/n!and integrating term by term, we
n=O
obtain the desired result. 1
+00
5.1.5. From the equality r(z)
=
~ o
e- t t z - 1 dt = ~
+
+00
~1 and the
equality of
0
Exercise 5.1.4 we obtain
'2: (-l)n[(z+n)nW\ 00
r(z)
=
H(z)+
n=O +00
where H(z)
=
~ e- t t Z - 1 dt
is an entire function.
1
5.2.1. From 2° and 3° it follows that for all n sufficiently large the points 0, z lie inside en. If ep(t;) = [t;(t;-Z)r1J(t;), then ~ ep(!;)dt; --+ 0 as n --+ +00 which en
is a consequence of 1°,4° and 5°. On the other hand, the integral can be evaluated by means of the theorem of residues. Since res(O; ep) = -J(O)/z, res(z; q;) =J(z)/z, res(a n; ep) = An[an(an-z)r\ we have
'2: An [anCa n-z)]-1} 00
27ti{ -J(O)+J(z)+z
=
O.
n=1
5.2.2. The function g(z) = J(z)-G(l/z) is analytic, at z = 0, hence we can apply Exercise 5.2.1 for g and then add G(l/z) to both sides. 5.2.3. (i) If Qn are squares with vertices (n+i)(=r-1 =r-i), then IZ- 1-7t/sin 7tzl :.;;; 7t+2 for all positive integers n and all z E frQn (note that Izl ~ i). Moreover, ". = 0 is a removable singularity of z-1- 7t/sin 7tZ, hence by Exercise 5.2.1 :
'2: (-1)n[(z-n)-1+n-1] m
7t/sin7tz-z- 1 = lim
m-+C()n=_m
(ii), (iii) can be solved in an analogous manner.
5.2.4. By Exercise 5.2.3 (iii) we have
L (_1)n+1/(n2_z2); 00
=
2z
"=1
SOLUTIONS
214
and hence
~ [(mx+,Brl-(nIX+IX-,B)-I] = next we put IX
=
3, ,B
:
tan~(I- 2:) ~
; cot 7t:;
1.
=
5.2.5. Using the results of Exercise 5.2.3, the equality of Exercise 4.5.16 as well as the identities of Exercise 2.7.15 we obtain: 00
7t/sinh 7tZ
=
Z--:l +2z 2: (-I)n(n2+z2)-I; n=1
7t/cosh 7tZ
=
22: (-I)n(n+ 1 )[(n+ 1)2 +z2rl;
00
n=O 00
7ttanh 7tZ
=
2z 2: [(n+ t)2+Z2]-1; n=O
I
00
7tcoth7tZ
(n2+z2rl. n= 1 5.2.6. The function fez) = (sinzsinhz)-I- z-2 has a removable singularity at the origin and we may assume f(O) = O. On the boundaries of the squares Qn with vertices 7t(n+ t)(=f1 =Fi) we have If(z) I :s;; 1+4/7t 2 since Izl ~ 7t/2; f has 4n simple poles m7t, m7ti (m = =F 1, =F2, ... , =Fn) inside Qn with residues: res(m7t;f) = (_1)m /sinhm7t, res(m7ti;f) = (-I)m- 1 i/sinhm7t. Hence =
z-1+2z
00
fez)
=
2: (-I)'"4m7tz 2 [(z4- m47t4)sinhm7t]-I. m=l
5.2.7. We have: coshz-cosz
=
cosiz-cosz
~
-2sin,CsiniC
2sin((l+i)z/2)sin((l-i}z/2)
= =
-2isinCsinhC,
where C = (l + i)z /2. Hence (coshz-cosz)-l
=
1 i(sin CsinhC)-l;
use now Exercise 5.2.6. 5.2.S. fez) = (sinz-zcosz)-lzsinz-3/z = O(z) as z _ 0 so that we may assume f(O) = O. Furthermore, fez) = (tanz/z-1)- l tanz-3/z is uniformly bounded on frQn, where Qn are squares with vertices n7t(=Fl=Fi) and has simple poles at An with residues 1. Hence 00
00
f(z)= ) ' [(Z-A n)-I+I/An] = 2z 2: (Z2_ A;)-I. "=-00 "=1
5. MEROMORPHIC AND ENTIRE FUNCTIONS
5.2.9. The functions tanz,
(COSZ)-l
l15
are uniformly bounded on frQ., hence
~ -+ 0 as n -+ + CXJ. The integrand f has the following singularities inside Q.: oQ. a simple pole z with res(z;f) = sinzlzcos 2z, as well as double poles (m+i-)7t with residues
0()
Since
2..:
(_I)m (m+i-)-2 = 0, the desired result follows from the theorem of
m=-co
residues.
5.2.10. fez) = eaZ (e"-1)-1_ z-1 -+ a--t as z -+ 0; moreover f is uniformly bounded on fr Q., where Q. are squares with corners 7t(2n+ 1)( =f 1 =f i) and has at z = 2m7ti (m = =fl, =f2, ... ) simple poles with residues exp(2am7ti). Hence by Exercise 5.2.1:
L' exp (2am7ti)[(z-2m7ti)-1 + 1/2m7ti] 0()
fez)
=
a- i- +
m=-co
L CD
=
(z2+4m27t2tl(2zcos2am7t-4m7tsin2am7t) "
m~l
since
L(m7t)-lsin2am7t 0()
-a+ t
=
0< a
for
< 1.
m=l
5.3.1. If m
p(z)
=
fez)
IT (r 2- a z )[r(z- a k
•
k)]-1
k= 1
IT r(z-b
k)
(r2-bk z)-1,
k= 1
then p is analytic in K(O; r) and has.no zeros. Moreover, [pi = [fl on C(O; r). Since log[p[ is harmonic in K(O; r), we have by the Gauss formula (cf. Ex. 3.2.5): 2n
(27t)-1 ~ log [f(re
2n i8 ) [
dO = (27t)-1 ~ log [p(re i8)[ de = log [p(O)[
o
0
which gives Jensen's formula. 2n
5.3.2. It is sufficient to prove that ~ log [1- ei8 [ de
=
O. This can be done e.g.
o
hy integrating (izt I Log (1-- z) round the boundary of K(O; I) with an indentation al z ~. I so as to leave z = I outside. [f the indentation Yr has the equation:
SOLUTIONS
216
z = l-rei'P, CPl ~ cP ~ CP2, and E = E(r) = Argzl for its upper end point Zl'
then 2n-£
P2
~ Log(l-ei6 )dO+ ~ (l-rei'P)-lLog(rei'P)rei'Pdcp • 'P.
Since the latter integral tends to 0 as r 5.3.3. If f(z)
=
z).F(z), then 10gl!1
-t
=
O.
0, the result follows.
= Alogr+loglFl and
2"
(21t')-l ~ log I!(rei6 ) IdO o =
Alogr+logIF(0)I+log[r m /la l a2
5.3.4. (i) lall = I, la21 = la31 =
=
am l]-log[r"/lb l b2
.. •
bol].
}if, therefore
I
for
r E (0, I],
110g3+log r
for
r E [I,
for
r ~
Iog 3
if>(r)
.. •
310gr
V3],
}i3;
(ii) by Exercise 5.3.3 we have: if>(r) = (2n+I)logr-2nlog1t'-210gn!
for n1t'
~
r ~ (n+I)1t';
(iii) we have for cosz: if>(r)
=
2nlogr-2nlog1t'-210g[t·
f .. · (n-t)],
(n- t)1t' ~ r ~ (n+ t) 1t'; now subtract both expressions for if> for cosine and
sine. 5.3.5. Obviously logx = log+x-Iog+(llx). Since the zeros of of 1If, we have only to show that r
10g[r Ilb l b2 O
...
bol]
=
~ t-ln(t.!)dt. o
r
~ t-lnk(t)dt o 00
Observe now that
I
k=l
nk(t) = n (t,f).
= log[rllbkl].
f are poles
5. MEROMORPHIC AND ENTIRE FUNCTIONS
217
5.3.6. f is analytic and 1= 0 at the origin and moreover n(r,!) = nCr, llf). 5.3.7. n(r,f)
= 2[r/7t+1/2]; nCr, 1If) = 2[r/7t].
5.3.8. Since n(r,f) == 0, we have 12.
T(r,f) = m(r,f) = (27t)-1 ~ rcos()d() = r/7t. -12
5.3.9. P(z) = lalrnexpi(oc+n())+O(r n- l ) where a = lalei .. ; reP(z) = lal rncos(oc+n())+O(r n- 1).
Thus 27t
T(r,f) = m(r,f) = (27t)-llalrn ~ {cos(oc+n())}+d()+O(r n- l ) = lalr n/7t+O(r n- l ), o
where {f}+ = -}(f+lfD. Hence r-nT(r,f)
lal/7t.
--+
5.3.10. loglf(reifl) I = reLogj(re ifl) = rsin()/(1+r 2 -2rcos()) which is ?: 0 for o ::( () ::( 7t. Hence 7t
m(r,f) = (27t)-1 ~ (1+r 2 -2rcos())-lrsin()d() o
O
+ ... , ICAI
CAZ A
> 0, then r
which shows that
5.3.12. The integral} t-1",(t)dt is finite (cr. Ex. 5.3.1) and this implies that also o
r
I
} ",(t) dt is finite. Since ",(t) is increasing, we have lim (1- t) ",(t) = O. Observe 1/2
' ...... 1-
now that -logr'" I-r as r --+ I. 5.3.13. (i) By Exercises 5.3.11,5.3.12 finite limits lim
=0
r-+l-
exist. Using this and Exercise 5.3.1 we see that also a finite limit lim logla 1a2'" ani n-++aJ
exists.
4.18
SOLUTIONS
(ii) iloglanll/(I-Ianl) --+ 1 because lanl --+ 1. Now; the series 2)ogla nl is absolutely convergent by (i) and this implies the convergence of L (I-Ian I).
5.3.14. The function F = (f_g)Z-A is analytic and bounded in K(O; 1) and I does not vanish at the origin. Ityanishes at z = ail and if we hadf-g 1= 0, then the series L (1-la nl) would be convergent by Exercise 5.3.13 (ii). 5.4.1. We have 0 < A ~ Iz- 2[Log(1 +z)-z]1 ~ B for z ciently small r > O. Therefore
Alunl2+un ~ Log(1+un) ~ Blunl2+un =
for all
E
K(O; r) and suffi-
n ~ no.
5.4.2. The series L Un is convergent as an alternating series, whereas L I/n is divergent (cf. Ex. 5.4.1).
L u~
5.4.3. (i) P n = PIP2 ... Pn = (n+I)/2n; (ii) P n = -;-[1 I/n(n+ 1)];
+
(iii) P 2k = 1,
P 2k + 1 = (2k+2)/(2k+ 1).
m
544 • ..
D. (1 +z2n) = n=O
1+Z+Z 2 +
'" +Z 2 m + 1 - l ~ '
5.4.5. (i) The finite plane; (ii) K(O; 1/e); (iii) the finite plane; (iv) E"'--.N1 , Nl is the set of all negative integers. 5.4.6. Put un(z) = (z-a n)/(z-a;l)-1. If 0 < r < 1, and z E K(O; r), we have lun(z)1 ~ (1-la nI2)/(1-r) = A.n. Since. LAn < + 00, the product (1 + Un (z)) represents a function F analytic in K(O; 1). If z 1= an, then all the factors are different from 0 and consequently 'F(z) 1= 0 which follows from the definition of convergence for infinite products. Moreover,
n
l(z-an)/(z-a;l)1
=
la nIICz-a n)/(1-anz)1 < lanl < 1
K(O; 1) and this implies W(z) I < 1 in K(O; I). 5.4.7. We have: 1(1-z)e -11 = IzI 21(1-1/2!)+(1/2!-1/3!)z+(1/3!-]/4!)z2+ ... 1 ~ IzI 2[1-1/21+(1/2!-1/3!)+ ... ] = Izl2
for
Z E
Z
for z E K(O; I). Suppose now that z and this implies
E
K(O; R); if n > R, then Izl/n < R/n < 1
Iun(z) I = 1(1 +z/n)e-z/n-II ~ Iz 2 [/n 2 < R2/n 2. 5.4.8. We have: n
hn(z)
=
zexp[z(1+1/2+ ... +I/n-Iogn)]
Il (/+z/k)e- z/
k•
k=1
5. MEROMORPHIC AND ENTIRE FUNCTIONS
219
Note that Yn = 1+1/2+ ... +l(n-logn -+ Y = 0.5772 ... (y is the so-called Euler's constant) and use the result of Exercise 5.4.7. 5.4.9. (i) With the notation of Exercise 5.4.8 we have: hn(z+l) = z- l hn(z)(z+n+I)/n and therefore F(z+I) = zF(z); (ii) hn(l) = I+l/n-+ I, hence F(I) = I; now apply (i). 5.5.1. (i) q = 2; (ii) no finite genus exists; (iii) q = 2; (iv) q = 3; (v) q = 4; (vi) q = 3.
n co
5.5.2. (i) zexpg(z)
(I-z/n)e zfn ;
n E(z/log(l+n), n), n=l (vi) zexpg(z) n [E(z/n, 2)]n. "=1 co
(ii) zexpg(z)
mn
=
n;
co
5.5.3. If F is an entire function with zeros a1 , a 2 , ••• , an, ... (0 < Iall la21 ~ ... ) written as many times as their order shows, then F'!Fis a meromorphic function which has at an simple poles with principal parts (z-a n)-1 (for a zero of order k, we have k terms of this kind). Now, according to MittagLeffler formula, ~
{I 1[ z-a an
(an
F'( ) -~= ~ - - + - I+~+ ... + ~)mn-1]} +h'(z) <Xl
F(z)
L...J
n=1
n
an
where h is an entire function. If G is a simply connected domain such that 0, z and all an belong to C,"", G and y c G joins 0 to z, then
f F'(t;)
J
E
G
F(z)
F(Cr dt;
=
log F(O)
o
=
fIOg{(l-
n=1
~)exp[~ + ... + ~n (:nf n]}+lo gexP(h(Z)/h(O»,
where log denotes a single-valued branch of logarithm in G. The term by term integration is admissible by uniform convergence on y. 5.5.4. If bo = 0, b 1 , b 2 , ••• , bn , ••• are poles of/and h is the Weierstrass product for {b n } , then H = /h is an entire function. 5.5.5. Suppose that 0 < la 1 1~ la21
Kn ,., K(O;
~ lanl)
and
~
... ,
Mn
=
~~f..I-W'(;n)~;~a") I·
220
SOLUTIONS
Choose qn so that 2- qn M n :s;; 2- n. If N > n, then the Nth term of (A) is dominated by 2-N in Kn and this implies a.u. convergence of (A). If Z ~ an, then the only term with a removable singularity at an tends to 'Y}n' while all the remaining ternis tend to zero because they contain the factor w(z). 5.5.6. We can take sinz instead of w in Exercise 5.5.5. We have Isin 7tZI K(O; n/2) hence e.g.
< exp(n7t/2) in
F(z) =
~ (_I)n ~ n=1
(n-I)!sin7tz 7t(z-n)
(~)n2 n
5.6.1. The genus of the sequence of zeros is equal to 2, hence
IT (l-z/n)e 00
sin7tz = 7tzexpg(z)
00
zfn
= 7tzexpg(z)
"=-00
I1 (l-Z2/n2). n=1
Taking the logarithmic derivative of both sides and using Exercise 4.5.16 we see thatg=O. 5.6.2. (i) Replace z by iz in Exercise 5.6.1; (ii) use Exercise 5.6.1 and the identity: cos iz-cosz = 2sin[(1 +i)z/2]sin[(I-i)z/2]; (iii) a particular case ,)" (iv); (iv) replace z by (a-b, "''(,, in (i). 5.6.3. Cf. Exercise 5.6.1. 5.6.4. (i) (coshz7t V2-cosz7t V2) [27t 2z 2(I+z 4)t1 (cf. Ex. 5.6.2 (ii)); (ii) (sin 7tzsinh 7tz)[7t 2z 2(I-Z 4)tl; take the limits of both sides as z ~ 1. 5.6.5. (i) If 'Y} =
1- (1 + i y'3),
then (l-z2'Y}2/n 2) (l-z 2i?/n 2)
=
I+(z/n)2+(z/n)4-
and using Exercise 5.6.1. we obtain 00
I1 [1 + (z/n) 2+ (z/n)4] =
(7tz)-2sin7tz'Y}sin7tzrj;
n=1
Oi) if
T
= +(V3+i), then
I+(z/n)6 = [1+(z/n)2] [I-(Tz/n)2] [I-(Tz/n)2] and using Exercises 5.6.1, 5.6.2 (i), we obtain that the value of the product is equal to (7tzF3sinh 7tZ sin 7tTZ sin 7tTZ; (iii) (7tz)-3sin 7tZ sin 7t'Y}Z sin 7trjz.
5. MEROMORPHIC AND ENTIRE FUNCTIONS
5.6.6. (i), (ii). Put z
=
221
1 in Exercise 5.6.5 (i), (ii).
5.6.7. Integrate both sides of the formula of Exercise 5.2.8 similarly as in Exercise 5.5.3. 5.6.8. If z
C+t, then by Exercise 5.6.3 we have:
=
sin7':z
---:-:----:- 7':z(l-z) -
4cos7':C 4n°o n 2+n ( Z-Z2) I+ 2 7':(1-4C ) - -;- n=1 (n+f)2 n+n2
n (n 2+n) (n+i-)-2 which gives the desired 00
making z
--+
0 we obtain I
=
(4/7':)
n
1
result. 5.6.9. Taking
~he
logarithmic derivative of the right-hand side we obtain:
~,[( 3 -~)-2(---~~ n7':+3z n7': n7':+z
n=-oo
__ I n7':
)+(~I +_1)] z-n7': n7':
=
3cot3z-cotz
which is the logarithmic derivative of the left-hand side. The two expressions differ from each other only by a constant factor which shows to be equal 1 (put z = 7':(2). 5.6.10. Integrate both sides of the equation of Exercise 5.2.1: 00
F'(z) F(zf
=
F'(O) F(O)
~
+~ n=1
(1 1) z-a + a::
(An
=
1).
n
5.6.11. The function F(z) = sin 7':(z+a)/sin 7':a vanishes at z = -a-n, n being an integer, moreover its logarithmic derivative F'(z)/F(z) = 7': cot 7':(z+a) is uniformly bounded on squares with corners (n+t)(=f1 =fi)-a. Hence by Exercise 5.6.10: F(z)
J~l (1+ a~n)exp[-z/(a+n)]
=
exp(z7':cot7':a)
=
(1+~)expz(7':cot7':a-a-1) a
=
oo
(1+ :)expz{ 7':cot7':a-a- 1 -
X
Jl(l+
IT '(I+_z_)exp(_~+~ __ z_) a+n n n a+n
n=-oo
nt~ [(a+n)-l- n-ll} X
a_tn)exp(-z/n).
Nole that the expression {... } vanishes identically.
SOLUTIONS
222
5.6.12. F(z) = cos(7tz/4)-sin(7tz/4) = y2sin[7t(I-z)/4]; F'(z)/F(z) = - t7tcot[7t(I-z)/4] is uniformly bounded on the boundary of squares with vertices (4n+2)(=t=I=t=i)+l, moreover F(z) = 0 for z = 4n+l,
where n is an integer. Hence F(z)
=
exp(-7tz/4) (l-z)ez(l+z/3)e-z/3(I-z/5)ez/5 ...
=
(l-z) (l+z/3) (I-z/5) ...
because 1-1/3+1/5- ...
=
7t/4.
5.6.13. We have
[r(z)r(l-z)t1
=
(_Z)-1 ze YZ
=
[-zr(z)r(-z)t1 00
00
"=1
"=1
IT (I +z/n)e-z/"( -z)e- Yz IT (l-z/n)e z/"
n (I-1/2n)e 1/
=
sin7tz/7t.
00
(i)
5.6.14.
[r(-t)t1
=
-te-y/2
,
=
now,
2 ";
r(-t) (-t)
n=1
r
V;,
=
and therefore 00
IT (I -1/2n) e
y~Y /7t;
1 / 2" =
"=1
n (l-z/2n)e z/ 00
(ii) [r(-z/2)t1
=
-tzexp(-yz/2)
2 ",
"=1
00
=
t(z-l)exp[y(z-I)/2]
IT (l+z/(2n-l)) (l-1/2n)exp[(I-z)/2n] "=1
and hence
il (I + 2n~1)(1(Hi) putting z
;n)
=
Z~I~:)
[r(-z/2)r(z-I)/2)r 1 ;
1/2 in (ii) and using the equality - {- r( -{-)
=
=
obtain: (l+t)(I-{-)(I+~) ...
=
y;[r(tW 2 •
5.7.1. M(r)
=
expr\ hence loglogM(r)/logr
=
k.
5.7.2. M(r)
=
expe r , hence loglogM(r)/logr
=
r/logr
-+
loglogM(r)/logr
=
5.7.3. (i) M(r)
coshyr, hence
=
logM(r)
+00.
=
v'r+o(I),
or
{-+o(l);
r(~) we
5. MEROMORPHIC AND ENTIRE FUNCTIONS
(ii) if P(z)
223
az"+ ... +ao , then
=
10gM(r)
= loglal+nlogr+o(l) = nlogr(l+o(1))
and consequently 10glogM(r) = logn+loglogr+o(l) = o(logr). 5.7.4. Suppose that there exists a positive integer k and a sequence {rn}, + 00, such that
rn ~.
10gM(rn,f)/logrn < k,
i.e.
M(rn,f)
< r!.
From Cauchy's formula for coefficients it follows that ak+1 = ak+2 = ... = 0, where am are Taylor's coefficients off at the origin. This means that f is a polynomial of degree at most k. 5.7.5. log M(r, f)
= klogr+logM(r,
g)
= 10gM(r, g)[l+o(I)]
by Exercise 5.7.4.
5.7.6. If f is a polynomial, then pC!) = pCf') = 0; if p(f) > 0, then p(zf') = pCf') by Exercise 5.7.5. Using the formula (5.7A) we easily verify that p(zf') = p(f). 5.7.7. We may assume (cf. Ex. 5.7.5) thatf(O) =f. O. Then for r sufficiently large r
2tt
~ t- 1m(t)dt
-loglf(O)1 + (2,,)-1 ~ loglf(re i9 ) IdO
=
° (8
>
8
only) since loglf(re i9 ) I < 10gM(r)
for r sufficiently large. Moreover, ~ t- 1m(t)dt
°
2r
2r
m(r)log2
=
mer) ~ t- 1dt
< ~
since m(t)' is increasing. Hence mer)
<
< (3 <
C(,
Alanl P, or lanl-a
<
<
Kr P+'
2 P+'r P +' K and consequently 2r
t- 1 m(t)dt
< ~
t- 1m(t)dt
< 2P +'r P+ K B
0
r
r
5.7.8. If p
Kr P +'
0
0 is arbitrary, K depends on
2r
~ n
<
<
(log2)-12 P +'r PHK.
< Ar P and putting r = lanl we obtain mOan!) (A/n)aIP for all n sufficiently large.
then mer)
<
5.7.9. It is sufficient to apply Weierstrass factorization theorem and take 00
an integer m such that m
< p < m+ I. Then by Exercise 5.7.8 the series L lanl-m- 1 n=l 00
is convergent and therefore the product
•
TI E(z/a n , m)
is convergent.
n=l
5.7.10. (i) -nlogn/loglcnl = 1 for all n;:?: 2, hence p larly: (ii) p •.c a; (iii) p = 0; (iv) p = I.
=
1 (cf. 5.7A); simi-
CHAPTER 6
The Maximum Principle 00
6.1.1. If III has a local maximum at a ED and I(z) =
2.: An(z-a)n,
then
o
I/(a) I = IAol and I/(a+re i6 12 :0:;; IAol2 for all real () and all r sufficiently small.
Hence 2~
~ I/(a+re i6)1 2d()
=
21t(IAoI2+IAlI2r2+ ... ) :0:;; 27tIAo12
o = A z = ... = o. 6.1.2. Consider II! and cf. Exercise 6.1.1.
and consequently Al
6.1.3. D is a compact set on the sphere, hence III attains in D its lower and upper bounds m, M. If Zo ED, then the equalities I/(zo) I = m, I/(zo) I = Mare impossible.
6.1.4. If we had I(z) Exercise 6.1.3.
=1=
0 for all zED, then I/(z) I would be. a constant by
6.1.5. Each component is bounded by a system of curves where III = A, hence by Exercise 6.1.4 there is at least one zero of I in each component. 6.1.6. A corollary of Exercises 6.1.3, 6.1.1. 6.1.7. e > 0 being given, choose the integers p, q so that q is positive and 1 < (M l /M 2 )Q(r1 /r 2 )P < I+e. Then the absolute value of (J(z))qz-p attains its maximum on C(O; r1), i.e. [M(r)]qr- p < M~rlPandhence 10gM(r) < 10gMl'I + (p/q)log(r/r1)· Now, 0 < qlog(M1/M 2 )+plog(r1!r 2 ) < 10g(1+e) < e and COil· sequently p/q < [(log(M2 /M1)+q-1e] : log(rz!r1). Thus
10gM(r) < logM1 +log(r/r1)[log(M2 /M1)+q-1 e]: log(rz/r1) and by making e
~
0 we obtain our result.
6.1.8 .. After a substitution C= Z-1 we obtain a function Fanalytic in K(O; 1)', 0 with a removable singularity at the origin. Note that F cannot have a maximulll at C = 0 and use Exercise 6.1.6. 224
225
6. THE MAXIMUM PRINCIPLI!
6.1.9. By Exercise 6.1.8 the maximum of [z-np(z) [ in K(oo; 1) is attained on C(O; 1), i.e. [z-nf(z) [ ~M for zeK(oo; 1). 6.1.10. The rational function g(w) = p(t(w+w- 1») is analytic in K(oo; 1) and satisfies [g(w) [ ~ M on C(O; 1). Hence by Exercise 6.1.9 we have: Iw-np(t(w+w-l»)1 ~ Min K(oo; 1). The image domain of the annulus {w: 1 < [wi < R} under t(w+w- 1) is an ellipse H with semiaxes a, b = t(R=fR-1) slit along [-1,1], hence for zeH we have: [P(z)[
=
IP(t(w+w-l»)1 ~ MRn
=
M(a+b)n.
6.1.11. Take a positive integer n such that 2rt/n < (J-rx; if z approaches C(O; 1) within the angle rx+2krt/n ~argz ~ (J+2krt/n, thenf(ah), and also qJ tend uniformly to O. Hence lim M(r) = 0, i.e. q; = 0 by Exercise 6.1.6. r-->1
6.1.12. Iq;(z) I < expJ.IY[lexps(x 2-y2+2ixY)1 ~ exp(sa 2+J.IYI-ey2)
if Iy[ is large enough. Making e
~
< C,
0 for a fixed z, the result folldws.
6.2.1. z-lf(z) is analytic in K(O; 1) since it has a removable singularity at the origin, hence the maximum of Iz-Y(z) I in K(O; r) does not exceed l/r and making r ~ 1 we obtain Iz-Y(z) [ ~ 1 in K(O; 1). If Iz-lf(z) [ = 1 for some z e K(O; 1), then z-Y(z) == eiCt:.
6.2.2. The function [f(z)-f(O)][I-f(O)f(z)]-l satisfies the assumptions of Exercise 6.2.1. 6.2.3. From Exercise 6.2.2 it follows that 11'(0)[ ~ l-lf(0)[2, hence 11'(0)[ ~ 1. Moreover, if 11'(0)1 = 1, thenf(O) = 0, thus f satisfies the assumptions of Exercise 6.2.1 and If(z)/z[ attains a maximum at z = 0, hence it is a constant. 6.2.4. By Schwarz's lemma for f-l(W) we have [1'(0)[-1 then fez) = eirxz by Exercise 6.2.3.
< 1;
if 11'(0)1
=
1,
6.2.5. F(z) = M[f(z)-a o][M 2-aof(z)r l satisfies the assumptions of Exer1. cise 6.2.1, hence lim IF(z)/zl z-+o
6.2.6. Put Ihe function
OJ =
<
exp2rti/(n-m), F(z)
=
zn-2mf(z) and
which satisfies the assumptions of Exercise 6.2.5.
C= zR- m and consider
SOLUTIONS
226
6.2.7. The function [w(z)- e<][I- e<w(zW l satisfies the conditions of Exercise 6.2.1, hence !(w(z)-e<)/(I-e<w(z)! ,:::;; Izl or l(w-e<)/(W-e<-l)1 ,:::;; re<, where r = Izl and W = w(z) (cf. now Ex. 1.1.25). 6.2.8. If w(z) satisfies [w(z)- e<]f[1- e<w(z)] = ze iP with real fl, then its values taken on C(O; r) cover C(zo; p). Suppose now that WI' W2 E C(zo; p) and WI = Wl(Z) , W2 = W2(Z). If 0 < A < 1, then Q(z) = AWl(Z)+(1-A)W2(Z) takes for suitably chosen WI, W2' A each value from K(zo; p), for some z E C(O; r). 6.2.9. The set Q r is a bounded, convex domain whose boundary is of circles C(zo; p) of Exercise 6.2.8 with r fixed and e< ranging over [0, 1]. The envelope consists of the left half of C(O; r) and two circular arcs emanating from z = 1 and symmetric with respect to the real axis which intersect at a right angle the imaginary axis at =Fir. 6.2.10. Put P(z) = zQ(z); K(O; 1) c H, where H is an ellipse with semiaxes V2, b = 1 and foci =F1. It follows from Exercise 6.1.10 that IP(z)1 ,:::;; M(1 + +V2)n+l in H and also in K(O; 1) and by Schwarz's lemma
a
=
IP(z)1
=
IzQ(z) I ,:::;; IzIM(I+V2)n+l
in
K(O; 1).
6.2.11. P(z) = (z-1']) Q(z) = (z-1']) (aO+alz+ ... +anz") = -ao1']+(ao-1']al)z+(al-1']a2)z2+ ... +(an_l-1']an )z"+anz"+I.
Hence laol :':( M, lao-1']all:,:( M, i.e.
lall:S;; la ol+l1']a 1 -a ol :S;;2M, la 21:S;; 3M, ... Similarly lanl:S;; M, lan-II':::;; lanl+lrjan-1-anl ,:::;; 2M, lan- 21:S;; 3M, If n = 2k+l, then
IQ(z)1 ,:::;; (Ja ol+lanJ)+(Ja 1 1+lan-lJ)+ ... +(lakl+lak+lJ) ,:::;; 2[M+2M+ ... +(k+l)M] if n
=
=
i-(n+l) (n+3)M < i-(n+2)2M;
2k, then IQ(z)1 ,:::;; 2(M+2M+ ... +kM)+(k+l)M = i-(n+2)2M.
6.3.1. If fez) = F(w(z) and Iw(z)1 < r1 in K(O; rl), then also Iw(z)1 < r2 ill K(O; r2) by Schwarz's lemma for any r2 < r1 which means that f -
=
I.
F(rle i 'P) with rl ,,; r and conse-
227
6. THE MAXIMUM PRINCIPLE
6.3.4. Note that w = F- 1 of satisfies the conditions w(O) and is analytic in K(O; 1).
=
0, Iw(z)1 ~ Izl,
6.3.5. The function F(z) = (2/7t)Log[(l+z)/(I-z)] maps conformally K(O; I) onto {w: Irewl < 1} and F(O) = 0, hence f -< F; (i) follows from Exercise 6.3.2; (ii) follows from Exercise 6.3.3. 6.3.6. Suppose that F(z) = a-ar(a-rz)/a(r-az), r < lal (F maps K(O; I) onto the outside of K(a; r». Obviously w = F- 1 of is analytic in K(O; 1) since possible poles of f are removable singularities of w. Moreover w, satisfies the conditions of Schwarz lemma, hence Iw'(O)1 ~ 1 which implies the deliired result. 6.3.7. If Inl = I, then p -< F, where F(z) = F'1(z) = (l +nz)/(l-nz) and also p -. 6.3.8. We have (l-lzj)/(l+lzj) ~ Ip(z) I ~ (1+lzj)/(I-lzj), -2arctanlzl
~
Argp(z)
~
2arctanlzl,
which are precise estimates of w and Argw in K" r
Izl (cf. Ex. 6.3.7).
=
l-lzl ~ () 1+lzl I· ()I 21z1 639 . . . 1+lzl '--" rep z ~ I-Iii' Imp z ~ 1-lz12
W
h· h .. IC are preCIse estI-
mates of rew, im w in K r • 6.6.10. We may assume that f(O) > o. The function M/f(z) is analytic and does not vanish in K(O; I), hence we may consider a single-valued branch, of logM/f(z); obviously [logM!.f(z)]: [logM!.f(O)] Er!J> and hence
(1-lzj)/(l+lzj) ~ (logM/lf(z)j):(logM/f(O») by Exercise 6.3.9 which gives the desired estimate of If(z)l.
6.3.11. From Exercise 6.3.10 it follows that Ifn(z) I ~ If.(0)1 1/ 2M1/2 i.e.
If.(z)1 2 ~ MIJ.,(O)I
for
ZEj{(O;
i-).
6.3.12. There exists a single-valued branch of log[a-f(z)][I-lXf(z)] in K(O; I) and obviously q;(z) ... -]og[a--f(z)][I-lXf(zW 1 : (-Ioga) EflJ, hence 'P
-<
(I I =)(1
=)
I
and consequently 1q;'(0) I . 2 (Ex. 6.3.2).
228
SOLUTIONS
6.3.13. cp -< F, hence Icp' (0) I :( IF'(O)I = 1 and using the equality f'(a) = 1'( -a) we obtain (1-laI 2)1f'(a)1 :( 1. 6.3.14. Put w = CIZ+C2Z2+ ... ; from Exercise 6.2.5 as applied to w(z)jz we obtain Ic 21:( 1-lclI2. From the identity /= Foro we obtain: al = A1Cl, a2 = A2ci+A1c2' which implies
6.4.1. If K(zo; r) c D, there exists a function/= u+iv analytic in this disk; obviously the absolute value of exp 0/ i.e. exp 0 u has a local extremum i= 0 at zo, hence u = const (cf. Ex. 6.1.1).
6.4.2. u being continuous on the Riemann sphere attains a lower bound at Zo and an upper bound at Zl. If u i= const then Zo i= z 1 which means that u has a local extremum at a point i= 00 which is a contradiction (Ex. 6.4.1). 6.4.4. U K(C; p(lln») = Bn is an open set and Hn = (C"'-B n) n D a closed ~
set whose boundary points are interior points of D; moreOVer u(z):( M + lin in Hn. We may assume that p(1ln) :( lIn for all C and hence D = UHn and u(z) :( M for all ZED. Note that actually u(z) < M by Exercise 6.4.1., 6.4.5. If Ci= 0 is a pole of F, then h ~ +00 as z ~ C; if Izl ~ 1, then limh(z) ? 0 and if z ~ 0, then h(z) ~ log IAI. Now, z = 0 is a removable singularity of /_1 0 F hence either log IAI > 0, i.e. IA I > 1, or else log Jzl = log If- 1 0 F(z) I, i.e. F(z) == ei!T./(z). 6.4.6. The function
h(z)
=
A(rl)(IOgB):(IOg~)+A(r2)(10 B):(log!..2..)-u{Z) r g r1 rl 2
'2
is harmonic in the annulu~ rl < Izl < r 2 (R 1 < r 1 < r 2 < R 2 ) and nonnegative on its boundary hence it is nonnegative on C(O; r) and
. A (r 1) (log 6.4.7. We have:
;J:
(log
~~ )+A(r2) (lOg ~ ): (lOg ~: )-A(r) ?
:e argf(rei8) = re[zf'(z)I/(z)] ? 0 for z
E
O.
C(O; r). Since j
is univalent,/(z) i= 0 for z i= 0 and therefore re [zf'(z)lf(z)] is harmonic in K(O; R), nonnegative on C(O; r) and positive in K(O; r). This implies for any p
< r.
:0 argf(pei8) > 0
CHAPTER 7
Analytic Continuation. Elliptic Functions 7.1.1. The former series represents log2+Log[1-t(1-z)]
Log(l+z) in
=
K(I; 2), whereas the latter series represents Log(l+z) in K(O; I)
7.1.2. Both series represent (1-zr1 in K(O; I) and K(i; K(O; l)nK(i;
y'2),
c
K(I; 2).
resp.;
y'2) =I 0.
7.1.3. The former series represents -Log(1-z) in the disk K(O; 1), while the latter one represents 'lti-Log(z-l) in K(2; 1), both disks being disjoint. The function element (-Log(l-z), {z: imz > O}) is a direct analytic continuation of both series which is easily verified for z approaching 0 through the upper half-plane. 7.1.4. Suppose that D1
11(Z)
=
=
C",(-; +), D2
=
loglzl+iargz,
--'--'It
= loglzl+iargz,
0
C",(+; -), and
< argz < t'lt,
while 12(Z)
< argz < t'lt.
12 in (+, +), while 11 =I 12 in (-, -). 7.1.5./(z) = -z-1Log(1-z), z eK(O; 1); if this element is continued along C(1; 1), then after encircling the point 1 we obtain the element
Obviously 11
=
11(Z)
= -[=f2'1ti+Log(1-z)]Z-1,
z e K(O; 1)",0
(the sign depends on the sense of encircling). 7.1.6. No radial limits at points z of C(O; 1) do exist.
=
exp(2'1tim2-n) which form a dense subset 00
7.1.7. Suppose that the point z
=
1 is a regular point of I(z)
=
) .......'
n=O
the Taylor series of I with center h:
00
2: n-O
129
(z-h)nJ
anzn. Then
SOLUTIONS
230
must be convergent in K(h; r), where 0 < h < 1 and h+r> 1. Obvicusly Ipn>(he iO ) I ~ pn>(h) and this implies that also the series
L (z-heioypn>(heiO)/n! 00
n=O
is convergent in K(he iO ; r) for any real O. This implies that f is analytic in K(O; h+r) which is a contradiction. 7.1.8. The functionfis analyticin H+and in H_ which is easily proved by using the M-test. Suppose that there exists a disk K(xo; r), xo real and a function F analytic in K(zo; r) and such thatf = Fin H+n K(zo; r). There exists a rational number Wk E K(zo; r) and by our assumptions lim f(Wk+iy) = F(Wk). However, Y--"O+
as
y
-+
0+
which is a contradiction. 00
7.1.9. E.g.
L 2-nz2n ; the derivative cannot be continued beyond the unit disk n=l
and the same necessarily holds for the function itself. 7.2.1. After reflections w.r.t. the real axis we obtain fez) = fez) and after reflections w.r.t. the imaginary axis we obtain f(Z) = -f(-z)· which gives f(z)+f(-z)
=
o.
7.2.2. After reflections we obtain a function meromorphic in the extended plane, i.e. a rational function. 00
7.2.3. If fez) =
L
An (z-ay, then
n:::::-oo 00
feZ)
=
L
An (z-a)n.
n:::::-oo
7.2.4. We have: z
=
a+R 2 /(Z*-a), b = a+R2 /[b*-a);
z-b = R 2 (b*-z*) [(Z*-a) (b*-a)rl
and consequently
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
231
where g is analytic in some neighborhood of b *. Observe that res[b*; ck(z*-a)\a-b*)k R- 2k (z*-b*r k] = (-l)kkck (b*_a)k+1 R- 2k . 7.2.5. Take f1
1
=
2
[f(z)+f(z)] , f2 =
zi1 [f(z)-/(z)].
7.2.6. f is a rational function with poles fez)
=
Zl'
l/z1, hence
(AZ2+Bz+C) [(Z-Z1) (I_Z1Z)r1
for Z1 i= 0 (note that f is analytic at
or fez) = Az+B+Clz for Z1 = 0;
00),
moreover, fez) =f(llz) which implies: C
=
A, B is real.
7.2.7. After reflections we obtain a function meromorphic in the extended plane which has two poles 0, 00 and maps C onto a two-sheeted w-plane; hence fez) = az+blz+c; 1'(=[=1) = 0 because the angles with vertices at these points are doubled. Moreover f(=[=I) = =[=2 which gives a = b = 1, c = O.
7.2.8. f is a rational function which has a double pole z = 1 (the angles with vertices at z = 1 are doubled), moreover, fez) == f(llz) which implies f(O) =/(00) = 0 and finally fez) = Az(1-z)-2. f(-I) = gives A = 1.
-+
7.2.9. Ai'(Z-Z1) (Z-Z2) ... (z-ze) [(I-z\z)(I-z 2z) ... (l-z/z)rl, k+1 = n, IAI = 1. 7.2.10, 11. After reflections we obtaina function f analytic and univalent in C such that 00 ~ 00, which means that fez) = az+b. 7.2.12. After suitable rotations of D and K(O; 1) round the origin we can achieve that the real axis in the w-plane is the axis of symmetry of D and/, (0) > O. Suppose now that cp =f- 1 and consider cp(!(z» = '!j!(z); thus '!j! is analytic and univalent in K(O; 1). Moreover '!j![K(O; 1)] = K(O; 1), '!j!(0) = 0, '!j!'(O) > 0, hence '!j! must be identity which implies fez) == fez), i.e. f is real on (-1, 1). 7.2.13. If the branch f2 arises from f1 by an analytic continuation along an arc I, then the arc I (with both end-points in the upper half-plane H+) intersects the real axis an even number of times. To each intersecting of the real axis there corresponds a reflection with respect to some boundary arc of D, hence /2 arises from f1 by an even number of reflections. Observe that two reflections can be replaced by one linear transformation. 7.2.14. We have: F' = (ad-bc)f'(c!+d)-2,
'·fence
and also
F"IF'
=
f" ff'-2cf'(cf+dr 1.
SOLUTIONS
232
and moreover, (F" JF') 2 =(f" J/)2 +4c2f' 2 (Cf+d)-2_4cj"(cf+ d)-1 which implies {F, z} == {t, z}. 7.3.1. Suppose (j, D) is a function element of the given global analytic function and K is a disk contained in {w: rew < O} such that z = expw is univalent in K and exp(K) cD. The function element (f0 exp, K) can be continued along any arc situated in the left half-plane and hence it determines a single-valued function F(w). Note that F(w) = fo expw, WE K, or fez) = F(logw) with suitably chosen branch of log. 7.3.2. Suppose g(zo) = Wo, Zo being arbitrary. Since g'(zo) i= 0, there exists a branch of g-1 in a disk K(wo; r) = K, say (j, K), such that f(w o) = zoo By our assumptions the element (j, K) can be continued arbitrarily in C, thus it defines a single-valued inverse function g_1 in C. "Hence g as an analytic, univalent function mapping C onto itself must be a similarity transformation.
7.3.3. Sincef(z) i= 0, we havef= exph, where h is an entire function .. Hence f' = h' exph and f'(z) i= 0 implies h'(z) 0 for all z E C. Suppose h does not take a finite value a. This means that fez) i= ea , 0 which is impossible for nonconstantfby Picard's theorem. Thus h assumes all finite values, whereas h'(z) i= 0 for all z, and consequently, h(z) = az+b (cf. Ex. 7.3.2).
*"
7.3.4. Let z = z(t), 0( ~ t ~ f3, be the equation of Y and ~ = dist(y, C ",D). Let 0( = to < t1 < ... < tn = f3 be a partition of [0(, ,8] such that for m = 1,2, ... ... , n-l botp. arcs Ym-1, Ym of Y corresponding to [t m_1, tm], [t m, tm+1] are situated inside Km = K(zm; ~), where Zm = z(tm). Put Fm(z) = ~ f(C)dC+C m , [Zm. Z ]
where Cm are such that Fm_ 1 = Fm in Km- 1 ("\ Km and Co = O. Evidently Fo admits analytic continuation along y, Fm being functions of the chain. Moreover,
~
=
Fm(zm)-F,n_1(Zm_1)'
Ym
Hence n
~
=
L ~ = FnCzn) = F(Z). m=lY m
7.3.5. Take K(zo; r)
(
~
=
KeD
and
continue
mdC, K) along Yi which gives (Fi' K(Z;
the
function
~), j = 1,2. Now Fl
element
=
F2 =F
[zo ... ]
by theorem of monodromy since Yl' Y2 are homotopic w.r.t. D. However,
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
thus both integrals have the
s~e
233
value.
7.3.6. Let y be an arc situated in D and r = /(y). If argw is a continuous branch of argument of r, then 10glf(z)l+i argf(z) is analytic and single-valued in any sufficiently small disk K(z; rz ), z E y. This implies that any initial element of logf(zo), Zo ED can be continued "along any arc y cD starting at Zo and defines a single-valued branch of log / in D.
= g(y). Since
7.3.7. Take any finite Zo and any arc y. starting at Zo and let r
r omits 0, 1, with each CE y, we can associate a disk K, such that r
0 g is analytic in K" r being a branch of ./l-1, and moreover the corresponding elements (rog, K,) are identical for C sufficiently close to each other. This procedure defines a single-valued entire function G(z) with reG(z) > 0, which evidently implies G = const. 7.3.S. If fez) = a, b, then g = (f-a)/(f-b) is an entire function which omits the values 0, 1 and therefore g = const. 7.3.9, 10. It follows from the definition of ./l and from the reflection principle that ).(r+2) = ./l(r); moreover, im(Iogw/ni) > 0 in Iwl < 1. Hence Q is singlevalued and analytic in K(O; 1). An open segment (-!5, (5), 0 < () < 1, is mapped under Q onto another segment (-n, n) of the real axis, hence Q(w) = A 1 w+ +A2w2+ ... and all Ak must be real. Ai > 0 since the angle of local rotation at w = 0 is equal to O. An element of the inverse function w = Q-l(W), 0 = @-1(0), can be continued on IWI < 1, and its values cover a part of K(O; 1), hence Id~/dWlw=o < 1, which implies Ai > 1.
7.3.11. The function element (Q-l 0/, K(O; (5»), where !5 is sufficiently small, can be continued along any arc situated inside K(O; 1) and determines a singlevalued analytic function ()) with 1())~z)1 < 1. Hence / = Q 0 ro, i.e. / -< Q. 7.3.12. A consequence of Exercises 7.3.11, 6.3.2 and the remark given in Exer-
cise 7.3.10. t
7.4.1. If !(t)
=
A~
n (r-b,,)fJkn
1
dr+B maps {t: imt
> O} onto the inside
Ok=l
of D and z
=
(b n -t)-1, then
t-bk = (bn-bk)z-l(Z-XIc), k = 1,2, ... , n-l, n-l
dw/dz
=
dw/dt· dt/dz
=
t-bn
CIT (z-X")',,.-l. k_1
=
-z-1,
SOLUTIONS
234
7.4.2. w X3
=
= co; IXI
z \
o
C- 2/3 (1-0- 2/3 dC;
a = (27t)-13 1/ 2 [r(+W;
Xl
= 0,
X2 = 1,
+.
= 1X2 = 1X3 = z
7.4.3. w = ~ C,,-l(1-C 2r"dC; a
=
i-[r(IX)COS(IX7t/2)r1 r2(-tlX).
o
7.4.4. We have dw = z-l[z/(I-z 2 )]"dz, hence argdw = const on the circle z = ei8 , as weII as on the line z = iy, Y > 0; z = i corresponds to the center of the rhombus. 7.4.5. The image domain is bounded by the polygonal line with interior angles ~7t, f7t, f7t, -}7t, hence z
W=
~ [(C 2- r')2)/(C 2-a2)]1/2dC, a a
h
~ [(t 2- r')2)/(a 2_ t 2)]1/2dt,
=
k
2 ~ [(r')2_ t 2)/(a 2_t 2)]l/2dt.
=
o
If r') ~ 0, then k ~ 0, h ~ a, w ~ vz2-a2; in the limiting case the image domain is H+~(O; ia]. 7.4.6. Put Xl = -1, x 2 = 1, IXl = -}, 1X2 = -} which gives· w = 2i7t- l X x(yl-z 2 -Arcsinz), the branch being the principal branch of arc sin taking on (-1, 1) the values belonging to (-7t /2, 7t /2). z
7.4.7. w
=
A~ (1+C- I )I-odC+B, where A, B are real constants which can be 1
determined from the equality: 7r
7ti
=
Ai ~ (1 +e- i8 )I-oei8 dfJ+B o
by separating real and imaginary parts. If r') ~ 0, A(o) ~ 1, B(r')) ~ 2 and the limiting function w = z+ I+Logz maps conformally H+ onto H+ slit along the ray: im w = 7t, re w :(; O.
+
7.4.8. Applying the reflection principle to the mapping w = z+ 1 Logz we can continue the mapping through the positive real axis, the image domain = C which gives the desired mapping: being the given domain; then we put w = C2 +] +2LogC.
yz
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
235
7.4.9. The mapping of H+ onto a polygonal domain with interior angles rr/n, 2rr(l-I/n), rr/n has the following form: S
fn(z)
=
\
A
J
(C_a)1-2 /n (C2-1)1 lIn dC+B= A
o
r z
J
C-a C2-1 dC+B
0
= -}A[(I-a)Log(z-I)+ (l+a)Log(z+ 1)]+B. We may expect that after a suitable choice of parameters we obtain the desired mapping function. In fact, the function v Log (z-I)+ ' v -;\ 1~ ) Log(z+ 1)+ B,
where Vl
Vl
rr-Vl
-B=-log--+log--+log2, rr rr-vl rr satisfies our conditions. If Vl = rr/2, then B = 0 and fez) branch of logarithm corresponds to argz E (0, 2rr).
= -}log(z2-1); the
7.4.10. The mapping can be obtained as a limiting case of a mapping of H+ onto a polygonal domain with interior angles rr/2, rr/n, (2-1/n)rr, as n -.. +00. w
=
2C/(a 2-1)+Log[(I+C)/(l-C)],
C= {z;
h = 2a/(a 2-1)+log[(a+ 1)/(a-l)]. z
I
7.4.11. K(O; ~)'" (i~, 0] being simply connected, the integral C- 1F(C)dC So
does not depend on the curve of integration and this implies that W(z) is singlevalued. If z = te lIJ (0 = const) we see that arg (dW/dt) = argF(telll) = const for 0 = 0, rr and this means that the images of (-~, 0), (0, ~) are situated on straight lines. Since W -.. 00 as z -.. 0, these are necessarily infinite rays. Assuming argF(z) = 0 for (- ~, ~) we obtain that the distance d between the rays is equal to im[W(-r)- W(r)] = limim[W(-r)- W(r)] ,->0
n
= im lim ~ i[F(O)+o(I)]dO = rrF(O). r->O
0
In the general case d = rrlF(O)I. 7.4.12. Suppose that F(z) = (z_b)-1(z_a)-1/2. Obviously argF(z) = - i- rr ror all z < a. We have argH'(z) = -rr:/2 on (-00,0), (0, a) hence H(z) carries thesc intervals into two parallel rays inclined to the positive real axis at the angle
SOLUTIONS
236
-1t12. The distance between the rays is equal to we obtain: w
=
1tlb Va.
Putting Z
=
z-b,
"
~ [1J(1J+b)J/ 1J+b-ar1d1J "0
and a similar reasoning shows that the images of (a, b), (b, +(0) are infinite rays parallel to the negative real axis, the distance between them being 1tlb Vb-a. _ _ _ _ _ _ _--,H (aJ
H(b)=OO-
H{oo)
~ H(O) = 00 FIG. 5
If 2a = b, both distances are the same. If Zo = a, two perpendicular rays of the boundary and the negative coordinate axes in the w-plane coincide.
7.4.13. d
=
1ta-1 {3.
7.4.14. After the transformation W 1 = w- 1 we obtain the domain bounded by the rays [0, +00), (--!-i-oo, -+i] and the segment [0, -ti] which contains Q1. This domain is carried under W 2 = 4 W 1 + i into a domain of similar type as considered in Exercise 7.4.6 and being the image of H+ under W 2 = 2i7':-1 X X (J/ l-z2+Arcsinz). z
7.4.15.
HI =
7':-1
~
(1+C)IlC- 1 dC+i (cf. Ex. 7.4.11).
-I
7.4.16. The mapping obtained in Exercise 7.4.15 after a reflection w.r.t. thc positive real axis in the z-plane whose image is the whole real axis in the IV-planc gives the desired image domain. Put now z = Z2. 7.4.17. The mapping z = i(l-t)/(1+t) carries H+ into the unit disk in the t-plane; if (t = ck) ~ (z = Xk), then Z-Xk = 2i(t-c,,)[(1-Ck)(1-t)r1;
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
237
moreover, n
dw/dt = dw/dz' dz/dt = 2i(I-t)-2 A
IT (Z-Xk)lXk-1 k=1 n
=
Al (l_t)-2(l-t)IX,+I%2+ ... + lXn- n
IT (t-C
k )"'k- 1
k=1 n
=
Al
IT (t- Ck)"r 1. k=1
since
1X1 +1X 2 +
...
+lX n
= n+2.
7.4.18. If Ck = exp(27tik/n) and F carries the sector 0, Co, C1 into the triangle 0, ao, a1 (ak being the vertices of a regular n-angle)' then after reflections F can be continued on the whole disk, which implies that the preimages of vertices ak also form a regular n-angle (if F(O) = 0). We have: t
F(t)
= A~ o
with
IXk =
n
IT (T-Ck)lXk-ldT+B k=1
1-2/n. If A = 1, B = 0 then t n t
F(t)
= ~ o
IT (T-Ck)-2/ndT = k=l
~(I-Tn)-2/ndT. 0
The radius Rn of the circumcircle is equal to
hence the perimeter In = 2nRnsin(7t/n) = 2r(l/n)r(I-2/n) sin (7t/n) [r(I-1/n)tl =
27tr(I-2/n) [r(I-1/n)t2.
7.4.19. The alternate interior angles of the n-pointed star are equal: 7t(I+A), 7t(I- A-2/n) and a symmetry reasoning shows that the preimages on It I = 1 form a regular 2n-angle. Hence t
W
= F(t) = ~ (1 + Tn)J.(I_ T n)-J.-2/ndT,
0 < A < 1- 2/n.
o
A.
7.4.20. Particular cases of Exercise 7.4.19: (i) take A = 2/5, n = 5; (ii) take 1/3, n - 6.
SOLUTIONS
238
7.4.21. 1° If () C(O; 1); in fact,
argz then dw/d()
=
=
const on both arcs (f{Jk' "Pk), ("Pk> f{Jk+1) of
l-zk Z = -2iexp [i- i«()-f{Jk)] sint «()-f{Jk) ' 1- CkZ = -2iexp[i-i«()-"Pk)] sini- «()-"Pk) '
hence n
arg(dw/d()) = arg(ie i6dw/dz) = const+()+ tn()-i-() (n+
L (3k) = const, k=l
if there are no Zb Ck on the arc considered; hence the image of any arc (f{Jk> "Pk), ("Pb f{Jk+ 1) is a half-line (note that z -+ Ck implies w -+ 00); 2° 11 arg (dw /dz) = -7t when z describes a small semicircle with center at Zk which means that the rays lk corresponding to circular arcs with a common end point Zk coincide (note the continuity of w(z) at Z = Zk); 3° 11 arg(dw/dz) = (1 + {3k)7t when z describes a small semicircle with center at Ck which means that the angle between lk and Ik+1 is equal to (3k. If the rays lk do not intersect each other, the mapping is 1: 1 which may be verified by the argument principle. The univalence may be proved either by argument principle, or by using the remark given in Exercise 7.4.24. z
7.4.22. The mapping IP(z)
~ (1 +t 4 )-1/2dt carries K(O; 1) into a square
=
o
which is a convex domain; we have: IP'(z)Jw'(z) = l-z4 which has a positive real part in K(O; 1). Hence the mapping is univalent (cf. Ex. 7.4.24). The images of arcs (k7tJ4, (k+l)7tJ4) of C(O; 1) can be found similarly as in Exercise 7.4.23. The points i" correspond to w = 00, whereas the points (1 +i)2- 1 / 2 i" correspond to the vertices of right angles. Hence 1
a=
V2 ~ (1+X )-1(1-X )-1/2dx. 4
4
o
7.4.23. Cf. Exercises 7.4.1, 7.4.17. The domain being convex, we have 0 < rY.k < 1 for the finite vertices; if 1 < rY.k < 2, then the image of z = e- i6k is the point w = 00; r 1
f(z) 1
< IAI~ (l-p)-2dp =
If'(0)lr(l-rt1,
o 1
f'(z) 1 < IAI
n
IT (l-r)-
=
If'(0)1(1-r)-2.
k=l
7.4.24. (i) If G = IP(K) and g =fo IP-l, then g'
=
(I'IIP') 0 IP- 1 = po IP-1,
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
239
where p is a function of positive real part in K. By Exercise 3.1.14 g is univalent in G and also f = go t[J must be a univalent mapping. (ii) In view of Exercise 7.4.23 it is sufficient to prove that n
re
IT (l-=-zkz)!(I-Ckz) ~ 0
on C(O; 1)
for
Z =1= Zk> Ck'
k=l
We have on C(O; 1): n
re
n
IT (l-zkz)!(1-Ckz) = re IT (Z-Zk)!(Z-Ck)' k= k=l 1
Now, arg[(z-zk)!(Z-Ck)] = --}(V'k-Pk) if Z is situated outside the arc (Zk; Ck), hence for Z on an arc (Ck; Zk+l) we have n
n
L arg[(z-zk)!(Z-Ck)] = --} L (V'k-Pk) = -7t/2. k=l k=l Moreover, for Z situated on the arc (Zk; Ck) we have arg[(Z-Zk)!(Z-Ck)]
= 7t--}(V'k-Pk),
whereas for the remaining arcs arg[(z-z,)!(z-C,)] = --}(V'.-p,), hence n
n
L arg[(z-zk)!(Z-Ck)] = 7t--} L (V'k-Pk) = 7t!2. k=l k=l n (l-zkz)!(I-Ckz) of arcs (Ck; Zk+1) k=l n
Consequently, the images under p(z) =
are situated on the negative imaginary axis, whereas the images of arcs (Zk; Ck) are situated on the positive real axis, moreover, P(Zk) = 0, P(Ck) = 00. In view of the maximum principle and the relation p(O) = 1 we have rep(z) > 0 in K(O; 1).
7.4.25. The integrand has in K(O; 1)",-0 the Laurent expansion:
Z-2_
n
_Z-l
L
k=l
ct kCk+ ... , hence it does not depend on the line of integration joining n
in K(O; 1)"'-0 the point 1 to z, iff
L ctkCk =
k=l
n
0, i.e. iff
L
k=l
ctkCk =0. Then W(z)
is meromorphic in K(O; 1) and has a simple pole at the origin. Moreover, W(z) is continuous in K(O; 1)",-0 since all ctk > -1. Hence W(e i9 ) is a closed curve. If ()k = arg Ck> 0 ~ ()1 < ()2 < ... < ()k < 27t and Z = e;9, then n
dW!d()
= iz- 1
IT (l-ze- i9k )
SOLUTIONS
240
n
arg(dW/dfJ) = const-O+
Locd-(O-Ok) = const k=l
and consequently L is a polygonal line. If L has no self-intersections and Wo ¢ L, then the index n(L, wo) = 0, -1 and since there is one simple pole inside K(O; 1), we see by using the argument principle that the equation W(z)-w o = 0 has at most one root, i.e. W(z) is univalent in K(O; 1). Moreover, if Wo lies in the exterior of L, then n(L, wo) = 0 which means that W(z) necessarily takes in K(O; 1) the value woo z
n 3
3
3
7.4.26. ~ t- 2 (t-'f/k)l%k- l dt; L OCk = 5, L (ock-l)1Jk = O. 1 k=l k=l k=l 7.4.27. Cf. Exercises 7.4.26, 1.1.10. z
7.4.28. w
= A~
t- 2(t n _l)21 dt, where
1
n
A = 2- 2nnsin(7t/n) [~(sinlP)2/ndlPrl. o
7.5.1. Put x =
ty'2.
7.5.2. Put t = ~ and verify that y'1-z 2
=
sn V2(u-K).
7.5.3. Put t = -Vl-k2u2/k' in the integral 11k ' K'(k') = ~ [(t 2-1) (l-k,2 t2w 1/2 dt 1
which gives K(k) = K'(k'). 7.5.4. (i) If w = sn u, then u = u(w) = w+t(l+k2)w3 + 4~ (3+2k 2+30)w5 + ...
which can be found from the definition· of u(w) by expanding the integrand into a power series and term by term integration; hence (i), (ii), (iii) can be obtained by the method of undetermined coefficients; R = K' since iK' is a singularity nearest the origin. nl2
7.5.5. From the equality K(k)
=
~
(l-k2sin20)-1/2dO we obtain after ex-
o
panding the integrand according to the binomial formula and term by term integration:
241
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
and hence K(k) ./ +00,
K(k')
=
K(V1-P) \. 7t/2
.
as
k ~ 1. .
7.5.6. The product snusn(u+iK') is an analytic, doubly periodic function, hence it is a constant k- 1 (make u ~ K). Consequently sn(u+iK') = k- 1[u- 1+f(1+k 2)u+ ... ]. 7.5.7. Periods: 2K(k), 4K'(k)i; zeros 2mK(k)+2niK'(k); poles: (2m+l)K(k) +2niK'(k)(m, n are integers). 7.5.8. The l.h.s. has periods 2K(k), 2iK'(k) and possibly poles 2mK+2inK'; in view of Exercise 7.5.4: sn- 2(u, k) = u- 2 [1+i-(1+k 2)u2+ ... ] and hence the l.h.s. is equal to 1+ (higher powers of u), near the origin; hence it is identically 1.
7.5.9. Put k
=
1/-V-l so that K(k) feu)
=
=
K'(k)
=
K;
snKusnK(u+l+i)
is a solution which is unique up to a constant factor.
7.5.10. If z describes COAB, then v describes the boundary of (+; +) and (l-v)/(1+v) describes the boundary of the lower half of K(O; 1); note that w = [(1-v)/(1 +V)]1/2. 7.6.1. Grouping the factors corresponding to the pairs (m; n), (-m; -n), we obtain a(z)
=
zIT (l-z2/Qi)exp(z2/Q;);
observe that every factor has the form l-z4/Qt+ ...
7.6.2. If z = x, group the factors mW 1 =fnw2; if z = iy, group the factors =fmw l+ nw 2' 7.6.3. A consequence of Exercise 7.6.3 and the reflection principle .. 7.6.4. C'(U+Wk) = C'(u) since f.J is periodic; hence C(U+Wk) = C(u)+2C(fwk) (put u = -fWk and use the fact that C is odd). Thus a'(u+wk)/a (U+Wk) = a'(u)/a (u)+2C (fWk)
and the result follows by integration since a is odd. 7.6.5. (i) By Exercise 7.6.4 a(2u)/a 4(u) is doubly-periodic and has poles of order three at mWl +nw 2 • Moreover it is equal to 2u- 3 +Au+ ... near the origin,
242
SOLUTIONS
hence fp'(u)+a(2u)/a 4 (u) = 0 since the l.h.s. has only removable singularities; (ii) take logarithmic derivatives of both sides. 7.6.6,....7. Use the definitions of t; and fp, as well as the identities
L
, L (a+n)-2 = 7t cosec 7ta. 00
OCJ
(a 2+m2)-1
=
a- l 7tcoth7ta,
m=-oo
2
2
n=-oo
7.6.S. (i) Follows from Exercise 7.6.2 by differentiation; (ii) fp (Z) = fp (z), hence fp (x+ib) = fp (x-ib) = fp (x+ib) = fp (x+ib) , i.e. fp(x+ib), as well as fp(x+(2n+l)ib) is real; by (i) also fp(x+2nib) is real. A similar reasoning shows that fp (na+iy) is real; (iii) fp is real on the sides of R, moreover fp' =F 0, hence arg fp' = const on each side; fp' = 0 has simple zeros at three corners of R hence the image of aR""- {p}, p being the pole, is the open real axis. The univalence easily follows by the argument principle (p is omitted along a circular arc).
7.6.9. fp strictly decreases on (0, w) from
+00
to e1 because fp'
< 0; if
iv E (0, w'), then ,f.J'(iv) strictly increases from -00 to e2 • Now, the order of
boundary points is preserved, hence e2 < e3 < e1' We have: e1+e2+e3 = 0, hence e 2 < 0 < e1' Conversely, if e2 < e3 < e1 and e1+e2+e3 = 0, then the function u =
t
w
~ [(t-e1)(t-e2)(t-e3)]-1/2 dt maps im w
> 0 onto a rectangle with
00
sides a, b. After reflections we obtain a function w = w(u) meromorphic and doubly periodic with periods 2a, 2ib which can be identified as fp because it has double poles at 2ma, 2nib and the principal part at the origin is u- 2. 7.6.10. If one factor has a pole (which is double) then another one has a double zero, hence the l.h.s. is analytic in C and must be constant. Now put u = w'. 7.6.11. 2K(e1 -e 2 )-1/2 ,
since fp is homogeneous of degree - 2. Hence both sides have periods 2K, 2iK'. Moreover, principal parts of both sides at u = 0 are equal to (e1-e2)u-2, hence their difference reduces to a constant which can be found by putting u = iK' which gives e2 • 7.6.12. If z is replaced by z+wk> t;(z) being an odd function increases by 2t;(-}w k ); after integration we obtain 2'1}lW2-2r12w1 = =[=27ti because res(O; t) = 1; the sign depends on the orientation of parallelogram of integration.
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
243
7.6.13. (i) this is obvious; (ii) if 'l}1 = C(j-) then by Exercise 7.6.4: 1f(z+l)
(iii) if 'I}
=
=
exp (-'I}lZ2-2'1}lZ-'I}1)a(z+ 1)
=
-exp(-'I}lZ2-2'1}lZ-'I}1)a(z)exp(2z+1)'1}1
-1f(z);
C( r /2), then by Exercise 7.6.4:
1f(z+r)
since r'l}l-'I}
=
=
1ti
=
exp(-'I}lz2-2'1}lrZ-'I}lr2)a(z+r)
=
-exp (-7h z2-2'1}1 rZ-'I}l r2) exp(2z+r)'I}a(z)
=
-1f(z) exp [-m'(2z+r)]
(cf. Ex. 7.6.12).
7.7.1. The angles (on the sphere) at u = 0 are doubled, hence u = 0 is a double pole, the only singularity in: Ireul < 2a, limul < b; fis an even function, hence after a suitable choice of A, f-Ap is analytic and hence it reduces to a constant B. Putting u = 2a, ib, 2a+ib we obtain for f(u) the values 0, -h, _h- 1 and this enables us to find A, B, h. 7.7.2. Put z
=
10gC/Q, a
=
7.7.3. Put w = ei6 , 0 < () < # O.
10gQ, b 1t,
=
1t
in Exercise 7.7.1.
and verify that du/d()
=
ie i9du/dw is real and
7.7.4. Use Exercise 7.7.3; note that the image of [K, K+-}iK'] is the segment [Vk, 1]; horizontal sides of the rectangle are mapped onto the upper and lower half of C(O; 1), whereas vertical sides correspond to the slits [-1, -y'k], [y'k, 1]. 7.7.5. Under the mapping w = Vksn(u, k) the rectangle R: Ireul < K, limul < -}K' is mapped onto K(O; 1)",,[(-1, - y'k] u hlk, 1)]. On the other hand; z = csn 1tu/2K maps R conformally onto an ellipse with foci =fc slit along [-a,-c] and [c,a]; if (u=-}iK')--(z=ib), i.e. b=y'a 2 -b 2 sinh1tK'/4K, or 1tK'/2K = log a + bb -, we obtain the given ellipse H. The slits can be removed
a-
by reflection principle.
7.7.6. w = y'ksn(u, k) maps 1: 1 conformally the rectangle Q: Ireul < K, 0< imu < -}K', onto K(O; 1)"'-{[-yk, y'k] u [0, i)}; u = 2Ki1t- 1 Logz maps A"'-( - R, -1] onto Q; the slits can be removed by reflection principle. 7.7.7. Consider W cise 7.7.6.
= [w(Y~)r2,
where w(z) is the mapping obtained in Exer-
SOLUTIONS
244
7.7.S. If K'(k)/K(k)
=
7t/logQ, then w = sn(Klogz/logQ, k).
7.7.9. Find k from the equality of cross-ratios: (al, a2 , bl , b2) = (-k-l, -1,1, k- l ) and put (W, a2 , bl , b2 ) = (w, -1, 1, k- l ), where w is the mapping of Exercise 7.7.8 continued by reflections. 7.7.10. Use the reflection principle and show that fez) has the same periods
and singularities as cn(z, 1/v':2). 7.7.11. After reflections the mapping function can be continued over the whole plane, has simple zeros at ~+ifJ+2ma+2inb, -~-ifJ+2ma+2inb and simple poles at ~-ifJ+2ma+2inb, -~+ifJ+2ma+2inb; it is moreover, doubly-periodic with periods 2a, 2ib. The function .
a (z-~-ifJ) a (z+~+ifJ) [a(z-~+ifJ) a (z+~- ifJ)tl has analogous properties and is an elliptic function of order 2 (cf. Ex. 7.6.4). Both functions are identical up to a constant factor whose absolute value is equal to 1 (put Z= 0 and use Exericise 7.6.2).
CHAPTER 8
The Dirichlet Problem 8.1.1. The inverse mapping would be a bounded, entire function, thus a constant by Liouville's theorem. 8.1.2. The sets of values taken by g on (- 00, -1), (-1,0) coincide. Moreover, if u = -vcotv and vcosecv = exp(vcotv) which holds for. some vE(21tn+1tt4, 21tn+1t/2) and all n large enough, then g(ll+iv)=g(-I). Thus we need to verify that g takes every finite value in C. Suppose that weW - a 1= 0 for Some a and all w. Then also (w +21ti)ew - a 1= 0 and the quotient I+21tie w /(we W -a) which is an entire function 1= 0, 1 must be a constant. This is an obvious contradiction. If h(z) = 4z(I-z)-2, then h[K(O; 1)] = G and 1= {- g 0 h has the desired properties. 8.1.3. Suppose h is a (many-valued) conjugate of g and the increment of h over a small circle with center at Zo and positive orientation is equal to A; then 1= exp o 27tA-1 (g+ih) has all th~ desired properties. Univalence follows from the argument principle; since A is real and negative, I(zo) = O. 8.1.4. The mapping It oli 1 carries the unit disk onto itself; the mapping is a homeomorphism in K(O; 1) and after reflections it becomes a 1: 1 conformal mapping of the extended plane onto itself and consequently it is a linear mapping with 3 fixed points, i.e. an identity. 8.1.5. Both mappings I(z), (f(z*)*, where the stars indicate corresponding reflections carry Zk into Wk (k = 1,2,3), hence they are identical. This means that I(Yl) remains unchanged after a reflection w.r.t. Y2' i.e. Y2 = I(Yl)' 8.1.6. Arcs of: C(I; 1), reA =
t,
C(O; 1) situated in the upper half-plane;
A((I+iV3)/2) = (I+iV3)/2, A(i) = 1/2, A(1+i) = 2.
8.1.7. Suppose h is a positive, continuous and nowhere differentiable function with the period 27t and let D be the domain whose boundary D has the equation r = h«()), 0 ":( () ~ 27t, in polar coordinates. If I maps 1: 1 conformally K(O; 1) onto D, then C(O; 1) is obviously a natural boundary of f
246
SOLUTIONS
8.1.8. If the function f/J mapping D onto K(O; 1) is such that 0 = f/J(a), then all the functions with this property have the form eilXf/J(w); hence q?(w)
= eilXf/J(w)/f/J'(a)
and
r(a; D)
=
1/1f/J'(a)l.
8.1.9. In view of Exercise 8..I.8. r(a; G)
=
Idw/dCJ~=o,
w = J[(C+zo)/(l+zoC)].
where
8.1.10. (i) R- 1(R 2_laI 2); (ii) 2h; (iii) 4d. 8.1.11. If Do = J[K(O; 1)], ao = J(zo) , then r(ao; Do)
= (l-l zoI2)1f'(zo)1
and
r(a; D)
= (l-lzoI2)1f'(zo)IIq?'(ao)l.
8.1.12. Suppose that z = f/J(w) maps Do onto K(O; r). Then q? conformally K(O; r) onto D and IDI
~~ I(q?
=
0
0
f/J-l maps
f/J-l),1 2dxdy
K(O;r)
and since q?
0
f/J-l (z)
= z+ A2Z2 + ... , we have IDI
=
1tr2(l+2IA212r2+ •.. ),
(cf. Ex. 4.2.12). Hence IDI is a minimum if q? i.e. f/J = q?
0
f/J-l is an identity mapping,
8.1.13. We may assume that 00 Ern' We apply Riemann mapping theorem to the domain C"", rn and the mapping function carries G into th~ unit disk minus n-l continua. After a suitable linear transformation one of the boundary continua, say Yn-l will contain 00. We again apply the Riemann mapping theorem to C"",Yn_l and obtain an image domain whose boundary already contains two analytic Jordan curves, C(O; 1) and the image curve of a circle under the latter mapping function. After n analogous steps we finally obtain as an image domain of G the unit disk with removed n-l interior domains of closed analytic Jordan curves.
2n:
8.2.2. J(z)
=
(21t)-1 ~ U«()) (Rei6 +z)/(Rei6 -z)d()+iv(O) , where v(O) is an aro
bitrary real number. 8.2.3. This is a consequence of (8.2A), the mean-value property: 2n:
u(O)
=
(21t)-1 ~ u(Rei6 )d() o
8. THE DIRICHLET PROBLEM
247
and the inequality R-izi R2_lz12 R+lzi :::;; IRe'8_ z I2
8.2.4. u(z)
=
R+lzi
~ R-izi .
-}re(1+z2) (cf. Ex. 3.6.3 (i)).
8.2.5. 21t'u(z) = ~ (1-lzI2) IC-zl- 2dO = ll(Cl-Z)/(C-z)ldO because IC1 -zliC -zl ~
=
~
l-lzI2; moreover, by considering the triangles z, C, C+L1Candz, C1 , C1 +L1C1
FIG.
6
(cf. Fig. 6) we have dO/IC-zl = dOdIC1 -zl, hence 21t'u(z) = ~dO = length 1
of y. 8.2.6. If KeD, where D is the domain where the functions Un are defined, then the formula (8.2A) may be applied, hence the limit function is harmonic, in K and also in D. 8.2.7. In view of Exercise 8.2.6 the function u(z)
= u(re i"') = -}a o +
ao
L (r/R)n(ancosrtq;+bnsinnq;) n= 1
is harmonic and tends to U(q;) as r formulas we have:
-+
R- (cf. Ex. 4.4.4). By the Euler-Fourier
2",
an
=
(21t')-1 ~ U(O)cosnOdO, o
2",
bn
=
(21t'r 1 ~ U(O)sinnOdO, o
248
SOLUTIONS
and hence 2,...
u(re''If) = rc- 1 ~
co
[t + L (r/R)nCos n(p-O)] U(O)dO
OJ
n=1
2,...
=
co
(2rc)-1 ~ re[I+2L(zm n]U(0)dO o
n=1
2,...
=
(2rc)-1 ~ re[(C+z)/(C-z)] U(O)dO, o
.
2,...
8.2.8. u(z) = u(rei'P) = (2rc)-1 ~ re[(z+O/(z-C)]U(O)dO, where C = Re w, o
8.2.9. u(rei'P) = C(rc-1+2~-1
co
L n- (r/R)nsinnr1..cosnp. 1
n=1
8.3.1. A function harmonic in 0 < Izl < 1 and continuous in K(O;I) is harmonicinextensiontoK(O; 1) and hence it is determined by its values on ceO; 1). 2,...
If U(O) =1= (2rc)-1 ~ U(e i8 ) dO , the Dirichlet problem has no solution. o
8.3.2. (i) u(z) = A(1og(lzl/R2»:(log(R1/R2»+B(log(lzl/R1»: (log(R2/R1»; (ii) the coefficients an, bn, Cn, dn may be evaluated from Euler-Fourier formulas: 2,...
ao+bologR k = rc- 1 ~ Uk (0) dO , o 2,...
anR~+bnR"kn
=
rc- 1 ~ Uk (0) cosnOdO , o • 2,...
cnR~+dnRkn = rc- 1 ~ Uk (0) sinnOdO , o
n
=
1, 2, "', k
=
1, 2.
8.3.3. The mapping function
8. THE DIRICHLET PROBLEM
8.3.5. ro(z)
249
= imLog [(z-b)!(z-a)] ;
.
{O
hmro(z) = JI-+~
if if
7t
CE {(-<Xl, a) u (b, +<Xl)}, CE (a, b).
n
=
L
ukimLog[l-Llxk!(z-xk_l)] ' LlXk = Xk- Xk_l· k=l 8.3.7. The continuous function U can be uniformly approximated by a sequence of step functions; in the limiting case we obtain from Exercise 8.3.6: 8.3.6. u(z)
7t-1
+00
u(z)
=
+00
7t- 1y ~ [(x-t)2+y 2r 1U(t)dt
=
im7t-1 ~ (z-tr1U(t)dt.
-00
-00 +00
8.3.8. The integralf(z) =
7t- 1
)
(z-t)- l U(t)dt is a.u. convergent for imz
>
0,
-00
hencefis analytic and u(z) = imf(z) is harmonic in the upper half-plane. Suppose that B is an arbitrarily chosen positive number and () is such that IU(t)-U(to)1
< B for
It-tol
< ().
Put ~-8
U1(Z) =:=
7C- 1y
+00
~ Iz-tl- 2U(t)dt+7t- 1y ~ Iz-tl- 2U(t)dt+ -00
10+8 '0+ 6
+7t- 1y ~ Iz-tl- 2U(t o)dt. 10- 8
If z --. to, the two initial terms tend to 0 and the third tends to U(t o) by Exercise 8.3.5. Moreover, '0+ 8
lUI (z)-u (z)1
:::;:; B7t- 1y ~ Iz-tl- 2dt
10- 6
and the existence of a finite limit limuI(Z) = U(t o) implies limu(z)-limu(z) %-+to
z~to
,Z---?to
:::;:; 2B which shows that limu(z) exists and is equal to U(to).
'0 .
%.....
8.3.9. The mapping z = -}(w+w- l ) gives the annulus 1 < Iwl hence u(z) = loglz+y'z2-1//log (2+y3) by Exercise 8.3.2 (i).
<
2+y3,
8.3.10. The mapping z = -}(w+w-,l) gives the upper half-plane imw > 0 with the negative and positive real half-lines being the image lines of the slits. Hence u(z) = 2h[-}-7t-1Arg(z+vz2-1)],
im(z+Jlz2-1)
> O.
8.3.11. After the transformation z = }c(w+w- 1) the problem is reduced to
SOLUTIONS
250
v'
an analogous problem for the outside of K(O; R) (c = a 2-b2 , a = -}c(R+K1), b = -}c(R-R- 1), R = (a+b)lc > 1). The solution of the latter problem is: 10g(JwIIR)+A. Hence u(z) = loglz+ yz2- c2 1-log(a+b)+A; if A = log [(a+b)/2] , then u(z)-loglzl follows from the maximum principle.
=
0(1) as z --.
+00.
The uniqueness
8.3.12. The mapping w = 2(Z2+-}) carries the given domain into that considered in Exercise 8.3.10. Hence u(z) = 1t'-limLog(2z2+1+2zyz2+1) = 1t'- l imLogF(z). We have
Ux
=0
on the real axis, hence u, = aulan = 1t'- l IF'(z)1 = 21t'-1(x 2+1r 1/2.
8.4.1. re[(l+z)/(l-z)] = (l-lzI 2 11-zl-2 • 8.4.2. (i) Circular arcs with end points a, b situated in the upper half-plane (cf. Ex. 8.3.5); (ii) circular arcs with end points eilJ1 , e11J2 , situated inside K(O; 1).
8.4.3. £O(z; /2' G) = oc-1Argz, £o(z; f2' G) = l-£O(z; fl' G). 8.4.4. If O(z) = imLog[(z-l)/(z+l)], then £O(z; T, G) = 21t'- l O(z)-1; £O(z; 'Y, G) = 2 [1-1t'- 10 (z)]; circular arcs with end points 1, -1 situated inside G. 8.4.5. If z = x+iy, then £0 (z;
'Y, G) = 21t'-1 [1t'-0 (z)] = 21t'-1[arctan(yl(r-x»)+arctan(yl(r+x»)]
= 4Y(1t'r)-1(1+0(1»). 8.4.6. The mapping z = w2 carries Go. 'Yo into G, 'Y and due to the conformal invariance of £0 we have: £O(z; 'Y, G)
=
21t'- l Arg[(z-1)/(z+1)]
=
21t'- l Arg[(w2-1)/(w2+1)]
8.4.7. We have £O(r,.O)
= A(l-logrflogR) +
L n=l 00
[(anrn+bnr-n)cosnO+(cnrn+dnr-n)sinnO] ,
where
=
[1t'n(1-R- 2n)rlsin21t'nA,
an
= [1t'n(l-R 2n)r 1sin21t'nA,
bn
Cn
=
dn = [1t'n(1-K2n)rl(I_cos21t'nl).
[7tn(1-R 2n)r 1(1-cos27tnA),
8. THE DIRICHLET PROBLEM
8.4.8. If O(z)
251
Arg[(z-b)/(z-a)] and CPk
=
w(z; Yb G)
=
(k, 1= 1,2, k i= f).
[O(Z)-CP1]/(CPk-CP1)
=
O(C), CE Yb then
8.4.9. (i) w(z; Y2' G)-w(z; Yl' G) is a bounded harmonic functions with nonnegative boundary values, hence it is nonnegative in G; (ii) consider w(z; Y, G2)-w(z; y, G1), z E G1; again the difference has nonnegative boundary values. 8.4.10. The function
h(z)
=
10g/f(z)l-w(z;
0(,
G)logm-w(z;
p, G)logM
is harmonic except possibly at zeros Zk off and bounded from above. If z ~ Zk, then h ~ - 00; if z ~ CE frG, limh(z) ,;;;; 0 except possibly for end points of 0(. Hence h(z) ,;;;; 0 in G by the maximum principle.
fJ
8.4.11. If Z = x+iy is fixed, then by Exercise 8.4.10 (with m = 1,0(= (-r, r), r) (") H+ and G = K(O; r) (") H+) and by Exercise 8.4.5 we have:
= C(O;
log/f(z)/ ,;;;; w(z; p, G)logM(r) = 4Y(7tr)-110g M(r) (1+0(1)). Suppose that lim r- 1IogM(r) = (],;;;; 0; then there exists a sequence {rn}, r~+co
rn ~ +00, such that r;llogM(rn) ~ (],;;;; 0, and consequently loglf(z)/ ,;;;; 4Y7t- 1 (]
,;;;;
0,
i.e.
If(z) I ,;;;; 1.
8.5.1. If h(z, zo) is the solution of the Dirichlet problem with boundary values 10gIC-zol, then
g(z, Zo; G)
=
h(z, zo)-loglz-zol.
8.5.2. Suppose t =f(z) maps 1:1 conformally G onto K(O; 1). Then
g(z, zo; G) 8.5.3. If dist(w, frG w) if w ~ wo, then
~
=
10g[ll-f(zo)f(z)lIlf(z)-f(zo)l].
0, then dist(z(w), Gz )
~
0, hence g(w, wo; G)
~
0;
g(z(w), z(wo); G2 ) = -loglz(w)-z(wo)I+O(I) =
-loglw-wo/-log[jz(w)-z(wo)l//w-wol]+O(I)
= -loglw-wol+O(1). 8.5.4. u+iv = 10g
r
r.
SOLUTIONS
252
og/on = da/ds = I«P' (z) I, because the ratio of arc length elements da on ceO; 1) and ds on is equal to I«P'I.
r
8.5.5. g(z, zo; K(O; R)
= log[IR2- zzoIlIR(z-zo)l] , hence by Exercise 8.5.4:
og/on, = R-l(R2_lzoI2)IC-zol-2 = R-l(R2_r2) (R2 +r2-2Rr cos (O-/f)t 1 ,
C = Rei9 , Zo
=
rei,,; note that ds
=
RdO.
= 10g[lz-zolllz-zol]; og/on, = 2YoIC-zol- 2 = 2Yo[(x-xo)2+y2rl; ds = dx.
8.5.6. g(z, zo; G)
8.5.7. (i) loglzl; (ii) loglzl+logll+yl-z- 21; (iii) -log(a+b)+loglzl+logll+yl-(a2 -b 2 )z 21; (g(z; G) = logl «P(z) I, where «P: G -4 K(oo; I) is such that «P(oo) = (0). 8.5.8. (i) If argw increases by 2k1r:, the increment of (27ti)-1(1ogw=Flogc) is an integer, hence F(w) remains unchanged by Exercise 7.6.13 (ii) and consequently F is single-valued; (ii) F has zeros at ch 2n and poles at c- 1h2n (n = 0, =FI, =F2, ... ), hence there is one zero w = c and no poles in A; (iii) we have f}(Z) == f}(z), hence If}(Z) I = If}(z) I and this implies
IF(eifl) I = If} [(27t)-1 (O+ilogc)]1 : 1f}[(27t)-1(O-ilogc)]1 = I; moreover,
IF(he '9 )1 where z
=
= 1f}(z)l: If}(z+ or) I,
(27t)-1(ilogh-ilogc+O) and consequently IF(he "9) I = lexp[7ti(2z+or)]1 = lexp(1ogc+Oi)1 = c.
8.5.9. In order to obtain Green's function g(w, c; A) we add to -logW(w)I a term lloglwl such that the sum is equal to 0 on fr A. This gives
g(w, c; A) = 10gcloglwl/logh-logIF(w)l, where F is the function considered in Exercise 8.5.8. 8.5.10. We have: 10g«P(w) = g(w)+ih(w), where I«P(w) I = c on C(O; h). Thus og/on = 8h/os = c- 1o(carg«P)/os = iei9 c- 1d«P/dw = I«P'(w)/«P(w)i ; Now, «p" = wlOgC/lOgh/F(w), hence
og/on = Iw- 110g c/log h-F'(w)/F(w) I where F' /F = w- 1 { -1J17t-210gc+(27ti)-1[C«(27ti)-110g(w/c))-C«(27ti)-110gcw)]J.
8. THE DIRICHLET PROBLEM
253
8.5.11. We may assume that G = K(O; 1) since both p and g are invariant under conformal mapping. Moreover, we can take Zo = O. Then p(z, 0) = i-Iog[(1+lzJ)/(I-lzJ)], g(z,O) = -loglzl, hence g = -logtanhp. 8.5.12. The function 10g1C-Zll is harmonic w.r.t. Zl' hence g(z, Zl) is harmonic in Zl by (8.5A); therefore g(Zl, Z2)- g(Z2' Zl) = h(Zl) is also harmonic in Zl and for Zl ~ frG we obtain limh(zl) :::;;; 0 which means that h(Zl) :::;;; 0 for a fixed Z2 E G and all Zl E G. Similarly H(Z2) = g(zt. Z2)- g(Z2' Zl) ?: 0 for a fixed Zl and all Z2 E G. Hence g(zt. Z2)-g(Z2, Zl) == o. We used the fact that a bounded harmonic function necessarily has a harmonic extension to isolated singularities. 8.5.13. A consequence of the extremum principle for g(z, Zl; G)-g(z, Zl; Go), Z,Zl EG O· 8.5.14. We have
Wk(ZO)
=
(27t)-1 ~ og(C, Zo; G)/ oncds rk
=
_(27t)-1 ~ oh(C, Zo; G)/oscds
=
rk
r
C E k ; og/on
-oh/os by Cauchy-Riemann equations.
=
8.5.15. log€l>
_(27t)-1L1 rk h(C, zo),
=
g+ih, hence for w describing C(O; 1):
L1log€l> = iL1h = -27tiw(c; C(O; 1),A) = -27ti(1=
:::~)
L1logwlo.gc/logh-L1logF(w),
i.e. L1logF(w) = 27ti, similarly for w describing C(O; h) we have L1logF(w) = O.
8.5.16. D is a convex domain, hence there exists a half plane H containig D and such that c E C",,-H. In view of Exercise 8.5.13 we may assume that D = H. Therefore we may take c = 0, D = {z: rez > O}. By Exercise 8.5.6 g(a, b; H) =
log
::~!\
.
If the points a, b are rotated round the origin, la-bl remains
unchanged and la+bl :::;;; lal + Ibl with the sign of equality in case a, 0, -b are collinear. Hence supg(a, b; D) = 10g[(lal+lbJ)/la-bl] in case c = 0; in the general case: D
s~pg(a, b; D) = log
la-cl + Ib-cl la-bl .
SOLUTIONS
254
8.5.17. If g(z) = g(z, 00; G), then Ag(Z)+B q;(z) = { B
for for
z EG, ZEH.
The uniqueness follows from the maximum principle in the usual manner. 8.6.1. If f, g
E
L 2(G) , so does al+bg for any complex a, b because lal+bgl 2 :0::;; laI2IfI2+lbI2IgI2+labl(lfI2+lgI2).
8.6.2. Ilgl :0::;; fClfI 2+lgI 2), hence ~Ug is finite. 8.6.3. Suppose q;:
c"'-.r--* K(O;R)
R = r(C; C"'-.T)
is such that q;'(C)
=
1; then
~~ 1q;')1 2dxdy = 'rCR 2
and
C,,",r
(cf. Ex. 8.1.12) is finite. Now, G c
c"'-.r and
hence q;'
E
L2(G).
8.6.4. If dist(z, C"'-.G) = p, then II(z) I :0::;; (M/7tp2)1/2. 8.6.5. I/(a) I :0::;; (M/7t)1/2(1-lal)-1. 8.6.6. The family of functions IE L2 (G) such that I( C) = 1 and 11I1 t :0::;; y~R, where R is defined as in the solution of Exercise 6.8.3, is compact and nonempty. 8.6.7. If e > 0, 0:0::;; () :0::;; 27t, then f* We have Ilfoll :0::;; Ilf*ll, hence
=
fo+ee i9 g
E
L 2(G) for any g
E
L2(G).
(fo,Jo) :0::;; (fo,Jo)+2ere[e i9 (g,Jo)]+e 2(g, g), i.e. 0:0::;; 2re[ei9 (g,!0)]+e(g, g) for any e > 0 and any () (g,!o) = O.
E
[0,27t] which implies
8.6.8. Any function analytic in G has a primitive, hence we need to find q; analytic in G and such that q;(C) = 0, q;'(C) = 1 and ~ ~ 1q;'12 has a minimum. In view .
G
of Exercise 8.1.12 the extremal function fo = q;' is equal to w'(z)/w'(C); if R = r(C; G), then IlfoW = ~ ~ 1q;'l2 = 7tR2 = 7tlw'(nl-2 by Exercise 8.1.8 'and hence G
k(z, C) = 7t- 1Iw'(C)1 2w'(z)/w'(C) = 7t- 1w'(z)w'(C). 8.6.9. (i) 7t- 1R 2(R 2-zC)-2; (ii) 7t- 1 (Z-C)-2. 8.6.10. w'(z) = V7t/k(C, C)k(z, C). 8.6.11. If g(C) = 0, then (g,!o) = 0 (Ex. 8.6.7) and also (g, k) = 0, I.e. g(C) = ~ ~ f(z)k(z, C)dxdy = 0; G
255
I. THE DIRICHLET PROBLEM'
if I(C) =F 0, then g(z) = l(z)IfW-/o(z, C) vanishes for z = C, hence ~ ~ [/(z)If(C)-/o(z, C)]/o(z., C)dxdy
=
°
G
and consequently
~ ~/(z)/o(z, C)dxdy = I(C) 11/0 112. G
8.6.12. Put 1= k(z, C) in Exercise 8.6.11 and use Schwarz's inequality. 8.6.13. Suppose that I(C) = (f, k 1 ) = (f, k 2); then (J, k 1 -k 2) = 0 for any leL2(G) and also for 1= k 1 -k 2. This implies IIk1-k211 = 0, i.e. kl = k 2 • n
8.6.14. If Sn =
L akf/Jk>
k=O
n
Gn =
L bkf/Jk,
k=O
then from the minimal property of n
Fourier coefficients and from the density property of II/-snil
0, IIg-Gnil
-+
-+
o.
L Ckf/Jk
k=O
it follows that
Hence n
(/-sn,g-Gn) = I(ig)- ~akbkl,::;;; II/-snll·llg-Gnll-+.O.
8.6.15. From Exercise 8.6.11 it follows that k(C, rJ) = (kq, k,), where k, = k(z, C), kq = k(z, rJ); moreover, an = (k" f/Jn) and hence by Exercise 8.6.11: an = (f/Jn, k,) = f/JiC); similarly bn = (kq, f/Jn) = f/Jn(rJ). Using Exercise 8.6.14
we obtain: 00
(kq, k,) =
2:
f/Jn(rJ) f/JnW = k (C, rJ)· n=O 8.6.16. Using polar coordinates we easily verify that {f/Jn} form an orthonormal system; the completeness can be proved by verifying Parseval's identity, the integral /112 being expressed by Laurent's coefficients.
H
+00
8.6.17. k(z, C) = (27tlog(1/h)t 1 (zC)-1+7t- 1
L
(l-h 2n )-ln(zC)n-l.
"=-00 00
8.6.18. Both series
00
L (n+ lr1lbnl2, L
n=O gent, cf. Exercise 4.2.5.
n=2
(n-lrllb_nI2h-2n should be conver-
8.6.19. If z = z(t), C = z('r), then after the change of variables in the formula of Exercise 8.6.11 we obtain: I(Z(T)) = ~ V(z (t))k (z (t), Z(T)) 1z'(t)1 2dudv Gl
and due to uniqueness of kwe obtain: kl (t, T) = v'k(z, z) Idzl.
=
k(z, C) Idz/dt/ 2. Hence ykl (t, t) Idtl
CHAPTER 9
Two-Dimensional Vector Fields 9.1.1.
W
=f'(z);
Iwl
= 1f'(z)l.
9.1.2. g = ux-iuy, G = Ux-iUy are analytic in D and we may assume that g and G are not identically 0; then g, G do not vanish on D""-.H, H ED being an isolated set. If z E H ny, then the normal vector of y has components ux , uy, or Ux , Uy, and consequently arg((Ux-iUy)/(ux-iuy)) = 0, hence the quotient
G/g must reduce to a constant (which is real). 9.1.3. The lines of flow are circles im(ilz) = const, hencef(z) = kilz; hence IW11/1w21 = 1/2. 9.1.4. The lines of flow can be represented in the form im(iz-2) fez) = kiz- 2 (k is a real constant); IW11/1w21 = 2- 3 / 2.
=
= f'(z) ,
W
const, hence
9.1.5. Circles ceO; r). 9.1.6. (i) Equipotential lines : straight lines x = const; lines of flow: y velocity: w = a; (ii) y = C1 , X = C 2 , W = -ai; (iii) X2+y2 = C1x, X 2+y2 = C 2y, W = Z-2; (iv) Iz-bl/lz-ci = C 1 , circles through b, c,
IV
const;
(b-c)[(z-b)(z-C)r1;
=
(v) x 2_y2 = C1, xy = C 2 , (vi) X2+y2 = C 1 , Y = C 2i, (vii) y = C 1x, X2+y2 = C 2 ,
IV
=
W
= 2z.;
W =
W
=
alz; -ailz.
9.1.7. If W is the velocity and f = u+ iv is the complex potential of flow, then and this means that u x , U y are components of w. Hence
= f' = ux-iuy
~f'(z)dz = ~(ux-iUy)(dx+idy) y
=
y
=
~uxdx+uydy+i~ uxdy-uydy y
~ wsds+i ~ wnds = r+iQ. y
y
(wn denotes the normal component of w). 256
y
9. TWO·DIMENSIONAL VECTOR FIELDS
257
9.1.8. The points -=t=(1 =Fi)/J/2: are sources of intensity 21t, the points 0, sinks of intensity 41t.
00
are
9.1.9. Equipotential lines: equilateral hyperbolas through =Fa with center 0; lines of flow: lemniscates with foci =Fa, r = -41t. 9.1.10. Q = 61t, I'_= O. 9.1.11. J(z) = Qi1t- 1Arcsinz; according to Bernoulli law the pressure
P = C-tlw'12 = C_Q2/[21t 2(I_x)2],
C
= const =
Poo'
9.1.12. J(z)
= woo (Z+Z-l), w(i) = 2woo , w(2)
=
fw
oo ,
p(2)-p(i) = ~w!.
9.1.13. The complex potential of flow Ji outside the circular cylinder lei > R has the fotmJl(C) = w(e-icrC+R2ei R}, R = t(a+b), is mapped onto the outside ofthe given ellipse so that (dz/dC)oo = I. Hence J(z) =Jl(C(Z), where C(z) = t(Z+J1Z2_ C2). 9.1.1~. (i) Map the outside of the wedge, or segment onto the outside of K(O; R) so that (dC/dz)oo = I; then w(z) = Ji (C(z) similarly as in Exercise 9.1.13.
9.1.15. J(z)
= (21t)-lQlog(z-a).
= (21t)-lQlog(z+I)/z). 9.1.17. J(z) = (21t)-lQlog(I+4z- 4 ). 9.1.16. J(z)
9.1.18. J(z) = -p/1tZ; circle,s C(O; r).
= (21ti)-l rlog(z-a). 9.1.20. J(z) = woo z+(21t)-lQlog(Z2+ a2); w(O) =
9.1.19. J(z)
WOO'
9.1.21. (i) J(z) = -1t- 1[p(z-ai)-1+p (z+ai)-1]+w oo z; (ii) J(z) = (21ti)-1 rIog(z-ai)/(z+ai)+ (21t)-lQlog(z2+a2)+w oo z. (21ti)-1 rIog{ (z-ai)(z+i/a) [(z+ai) (z-i/a)r 1}.
9.1.22. J(z)
=
9.1.23. J(z)
= (21t)-lQlog(z+z-1- a- a-l).
= (21tW l rIog(z-a)/(z-a- 1). 9.1.25. We have w = ie i6 [(21tR)-1 r-2asinO] at z = Rei6 , hence C(O; R) is a line of flow; moreover, w(iR) = 0, if a = r/41tR = w; the force F = -iwoor. 9.1.24. J(z)
9.2.1. Suppose that the charged wire coincides with the OZ axis of the rectangular system OXYZ of coordinates and the unit charge is placed at the point a in the XY-plane. The force acting at a due to two linear charges qdZ situated at
SOLUTIONS
258
Z, -Z has the direction Oa and its magnitude is 2q(Z2+r2)-lcosctdZ = 2qr(Z2+ r2)-3/2dZ, where r = IOal and ct = 1:: (aDZ). Hence the magni+co
tude of the resulting force is equal to 2rq ~ (Z2+r2)-3/2dZ
=
2q/r.
o
9.2.2. The components of the force ware equal to
hence the (real) electrostatic potential "P(z) tential J(z) = - 2qilogz .
=
-2qloglzl and the complex po-
#
9.2.3. J(z)
=
-2iqlog(z+h)(z-h); lemniscates with foci
~h.
9.2.4. J(z) = 2Miz- 1 if the positive charge is placed at h > 0;. equipotential lines: X2+y2-C1x = 0, lines offorce; X2+y2-C2y = 0; W = -if'(z) = 2Mz- 2. 9.2.5. Consider a portion .1 of the surface of the conductor corresponding to the arc ds on its intersection with the plane of reference and let S be the closed surface being the boundary of a solid formed by shifting .1 both sides along the normal. According to Gauss theorem:
~~ wndS = 1.11 Wn
=
41t1L1la,
s
hence . a = (41t)-lW n = -(41tt1o"P/on.
9.2.6. q
=
~ ads
r
=
_(41ttl ~ (0"P/on)ds.
9.2.7. The total charge on the outer coating qo = ~ aods = _(41t)-1 ~ (0"P/on)ds, ro ro
similarly ql = _(41t)-1 ~ (0"P/on)ds. r,
The Green's formula with h
=
I gives the result.
9.2.8. The electrostatic potential of a condenser has the form
hence c = [41t("Pl-"PO)tl ~ ("Pl-"PO) (-ow/on)ds = -(41tt1 ~ w(ow/on)ds r, ~+~
9. TWO.DIMENSIONAL VECTOR FIELDS
259
since 0) = jon rj,j = 0, 1. The application of Green's formula gives the result (O)xx+O)yy = 0). 9.2.9. After a change of variables: 0);+0); =
(O)~+O);)a(x, y)/a(u,
v).
9.2.10. If 1Jlo = 0, 1Jll = 1, thenf(z) = (ilog(z/R)): log(r/R); introduce polar coordinates and apply Exercise 9.2.8. 9.2.11. Suppose that the cross-sections of wires are disks K(=fi-b; a); after the linear transformation w = (z-d)(z+drl, where =fd = =fVb 2-4a2/2 (=fd are symmetric w.r.t. either circle) the problem is reduced (cf. Ex. 9.2.9) to Exercise 9.2.10.
v'3
9.2.12. If d = 7-4 and w cylinder 1 ~ Iwl ~ 2+y3; hence fez)
=
=
(z+d)(1+zd)-l, we obtain the concentric
-(ilog(w/(2+y3))): log(2+r3)-1),
c
= --}[log(2+V3)tl.
9.2.13. The complex potential can be evaluated similarly as in Exercise 9.2.12.; 10'1 = (47t)_11f'(z)l, hence . (967tlog2)-1 ~ 10'1 ~ 3(327tlog2)-1.
9.2.14. fez) = 2iAlog[(z+ih)/(z-ih)] , 0' = 7t- l Ah(x2+h2)-1 (z-plane is intersected by both wires at a right angle, the intersection points being =fih). n
9.2.15. If the density of charge per unit length of
r = U r k is equal
to 0'( 'z),
k=l
then the real potential 1Jl(z) = 2 ~ 0'( .)log Iz- .I-lds. r
= 210glzl-1 ~ 0'(.)ds.+2 ~ 0'(.)logll-./zl-1 ds. = 2Aloglzl- l +o(l) r . r as z -+ 00. We have, moreover, g(z)-log Izl = 1'+0(1) since a bounded harmonic function with an isolated singularity has a harmonic extension to this singularity. We have also 2Aloglzl- 1 +2Ag(Z)-2Ay = 0(1),
or
1Jl(Z)+2Ag(Z)-2A'Y = 0(1).
The left-hand side has a constant value on r and vanishes at 00, hence it is identically O. Therefore 1Jl(z) = -2Ag(z)+2A'Y, which gives 1Jl = 2Ay on
r.
9.2.16. By Exercise 8.5.7 (iii), g(z) = loglzl-log(a+b)+log/l+yl-(a 2-b2)/z2/,
hence g(z)-loglzl+o(l) = -log[(a+b)/2] = y.
CHAPTER 10
Univalent Func.tions b
b
10.1.1. If ~ f dg = 0, the inequality is obvious. If ~ f dg 1= 0, take real oc such a
a
b
b
b
b
a
a
II
II
that e'cz ~ f dg = I~ f dgl. It is easy to verify that c ~ f dg = ~ cfdg for any complex c, hence b
b
a
a
b
!~fdg! =e'''~fdg= ~eiCZfdg a b
b
re {~ei"fdg}
=
a
= ~ re [e i '1] dg ~ V:(g)maxlfl. a
t
10.1.2. Obviously f is continuous, hence by a theorem of Helly (cf. [23], p. 252): b
b
~fdgn -+ ~fdg. Moreover, a
a
b
!~ (fn-f)dgn! ~ maxlfn-fl' K -+ 0 a
as n -+
+ 00,
if V:(gJ ~ K for all n. Therefore b
b
a
II
!~ fndgn - ~ f
b
dg! <;
b
II
II
!~ (f,,-f)dgn! + !~ f dgn - ~f dgi-+ O. II
t
10.1.3. ftn(t)
b
•
= (27t)-1 ~ u(Rnei6)dO is a continuously differentiable function o
of t
E
[0, 27t], hence for any
Z E
K(O; Rn)
Since ft~ > 0, ftn is strictly increasing for any fixed n, moreover ftn(O) = 0, ftn(27t) = u(O) = 1. By Helly's selection principle (cf. [23], p. 241), we may choose a subsequence {ftnk} convergent in [0, 27t] to ft which satisfies ft(O) = 0, ft(27t) = 1. 260
to.
UNIVALENT FUNCTIONS
261
Obviously (Rneit+z)/(Rne't-z) tends uniformly in [0, t] to (Reit+z)/(Reit-z) for any fixed z E K(O; R). Apply now Exercise 10.1.2. 10.1.4. The linear transformation w = (1+z)/(I-z) maps in a univalent manner K(O; 1) onto {w: rew > O}, and the image of C(O; 1) is the imaginary 0 and the Schwarz formula gives O. We can take 1'(0) = 0, axis. Hence u(e i6 ) p(t) = 1 for t E (0, 27t], R = 1, in Herglotz formula.
=
10.1.5. If WE convr, H being continuous we can find tk E [a, b], k = 1,2, such that W = AH(t 1)+ (1- A)H(t2), 0:;:;:; A:;:;:; 1, t1 < t 2 . Take now p(t) as a step b
function with jumps A, 1- A at t1, t2 resp., which gives
W
= ~
H(t)dp(t). Con-
a b
n
versely, each integral sum of ~ H(t)dt has the form:
Wn
=
L H(tk)iJPk
with
k=1
a
n
iJPk ~ 0,
L iJPk =
1, hence
Wn E
conv Hand conv H being closed the result
k=1
follows. 10.1.6. en
=
2"
~ 2e- int dp(t); the circle,C(O; 2) is the curve o
r
of Exercise 10.1.5
for any positive integer n. 10.1.7. The curve r of Exercise 10.1.5 is the circle (1 +ze- it )/(I-ze- it ) , 0:;:;:;t<;27t, whose diameter is the segment [(I-r)(I+r)-I,(I+r)(I-r)-I]. Therefore convr = K(zo; p), where Zo = (l+r 2)(I_r 2)-I, P = 2r(l-r2)-1, r = Izl. 10.1.S. q;(z) has the form: (w-wo)/(w+wo), where rew > 0, rewo > 0, hence Iq;(z) 1 < 1. Moreover, q;(0) = 0, hence by Schwarz lemma 1q;'(0)1 :;:;:; 1 which gives
10.1.9. We have: d '6 -loglf(re')1 dr
=
'61 d '6 If(re)'I--If(re')1 dr
hence :, 10g{lf(reiB)I/"P(r)}:;:;:; O. A a finite limit as r
-+
R-.
nonnega~ive,
decreasing function has always
SOLUTIONS
262
10.1.10. By the inequality of Exercise 10.1.8:
1f'(rei6)l/lf(rei6 )1 :(; :, 10g[(I+r) (l-r)-l],
hence lim If(re i6 ) I(I-r) (l+r)-l
=
t
r--+l-
lim (l-r)lf(re i6)1 r--+l-
exists for any real (). 10.1.11. Suppose that e > 0 is arbitrary. Choose r5 > 0 such that f-l«()o+ r5) - f-l«()o - r5) < t e. Split now the integral 1 of the Herglotz formula into 3 integrals: 60-6
~ o
2"
60+6
, ~ , ~ ; the integrand in two initial terms is 0(1) as z
-+
e i60 , hence
60 +6 60-6
both integrals multiplied by I-r tend to 0 as r -+ 1-. The third integral 13 has for z = rewo an upper estimate 2(I-r)-lVZ~:!:~{f-t) = te(l-r)-1, i.e. (1-r)13 < Finally (I-r) III < for all r sufficiently close to 1.
teo
e
10.1.12. Suppose that h(t) = f-l«()o-) for 0 :(; t < ()o, and h(t) = f-l«()o+) for ()o < t :(; 27t and h«()o) = f-l«()o). Obviously y(t) = f-l(t)-h(t) is continuous at t = ()o and increasing in [0, 27t]. Hence 2"
J(z)
~ (eit+z)(eit-z)-ldy(t)+h(ei60+z) (e i60 _zr1.
=
o
Similarly as in Exercise 10.1.11 we show that 2"
(1-r) ~ (eit+rei60) (e it -rei60 )-ldy(t)
-+
0
as
r
-+
1-
o
and
clearl~
lim (l-r)h(ei60+rei60) (ei60_rei60)-1 = 2h. r--+l-
10.1.13. If f-l is continuous at 27t-OC and v(q;) = f-l(q;+27t)-f-l(27t--oc) for -oc:(;q;:(;O and v(q;)=f-l(q;)+I-f-l(27t-oc) for o
J(z)
=
~
(ei'l'+z) (e i'l'-z)-ldv(q;) ,
-IX
hence lim (1-r)J(r)
=
2[v(0+)-v(0-)]
=
2(J-t(0+)+1-f-l(27t-)].
r--+l-
10.1.14. The function f-l(t) is increasing and hence it has at most a countable
to.
UNIVALENT FUNCTIONS
263
number of discontinuities (h (=f. 0, 27t) with jumps hk ; we have rx(Ok) as for 00 = 0, 27t we have
2hk , where-
=
Hence
L rx(Ok) 00
[2: hk+ fl(27t)- fl(27t-)+ fl(O+)- fl(O)] :;:;:; 2. k=l 00
=
k=O
2
For 0 =f. Ok we have rx(O)
=
0 by Exercise 10.1.1l.
10.2.1. logf(z) = 10glf(z)l+iargf(z) is analytic in the annulus slit along a radial segment. If z = re i8 , then after differentiating both sides w.r.t. 0 we obtain:
. i8'!'/,'! -- 7fO a Iog If I+17fO . a arg f', Ire now compare the imaginary parts of both sides. 10.2.2. If y(r) is the image curve of C(O; r), 0 then the index n(y(r),O)
=
(27ti)-1 ~ w-1dw y(r)
=
< r < R, and p(z)
(27ti)-1
~
z-lp(z)dz
=
=
zj'(z)/f(z) ,
1.
C(O; r)
Hence f/J = argf(reiB) is a strictly increasing function of 0 which increases by 27t as 0 ranges over [0,2] and therefore y(r) has a representation in polar coordinates: R = R(f/J), rx:;:;:; f/J:;:;:; rx+27t. 'If Wo = Roelo is not on y(r) and Ro < R(f/Jo), then the segment [0, wo] does not meet y(r) and consequently n(y(r), wo ) = n(y(r), 0) = 1 which means (argument principle) that the equationf(z) = Wo has 1 root in K(O; r). Similarly, if Ro > R(f/Jo) then the ray with origin at Wo being the prolongation of [0, wo] does not meet y(r), hence n (y(r), wo ) = n (y(r), 00) = 0 which means that the equation fez) = Wo has no roots in K(O; r); f being univalent in K(O; r) for any r < R is univalent in K(O; R). 10.2.3. (i) We have re[zfUfl] = re[(1+2z)/(I-z)] > 0 in K(O; i). In fact, the image curves of C(O; r) under (1+2z)/(I-z) are Apollonius circles with limit points - 2, 1 situated in the right half-plane for r < t; (ii) re(zf'lf) > t(2-rx) in K(O; 1). 10.2.4. From the solution of Exercise 10.2.2 it follows that the conditions are sufficient for Df to be starlike. We now prove the converse. Suppose Df is starlike, f being univalent with/CO) = 0,/'(0) = 1. We first prove that also f[K(O; r)] is starshaped for any r E (0, 1). Obviously w(z) = /-1 (if(z» satisfies the conditions
SOLUTIONS
of Schwarz's lemma for any 0 < t < 1. Hence lro(z) I ~ Izl. If Wl Ef[K(O; r)] and Wl =f(Zl)' IZll < r, then lro (Zl) I = Irl(twl)1 ~ IZll < r. Therefore tWl = f(Z2) willi Z2 E K(O; r) which means that f[K(O; r)] is starshaped. Hence f[c(O; r)] is a curve starlike w.r.t. the origin and so argf(reiIJ ) increases with () which yields re[zf'(z)lf(z)] ~ 0 as in Exercise 10.2.1. The case re[zf'(z)lf(z)] = 0 in K(O; 1) is excluded by the maximum principle. 2"
10.2.5. !,(z)lf(Z)-Z-l = Z-l [P(z)-I] = 2 ~ (e!IJ-zrldft«()
and the integra-
o
tion gives 2"
fez)
= zexp{-2 ~ Log (I-ze-i9)dft«()}. o
10.2.6. If(z)/zl ~ exp{210g(I-lzJ)-l} = (I-lzJ)-2, hence If(z) I ~ Izl(I-lzJ)-2 and similarly (1+ IzJ)- 2Izl ~ If(z)l. Hence using the estimates
(I-lzJ)(I + IZJ)-l ~ Izf' (z)lt(z) I ~ (1 + IzJ)(1-lzJ)-l, cf. Exercise 10.1.7, the estimates of!' easily f(jIlow: (1-lzJ)(I+lzJ)-3 ~ 1f'(z)1 ~ (1+lzJ)(I-lzJ)-3. 10.2.7. Log[zlf(z)]l/2
=
2"
~
Log(I-ze-ilJ)dft«() by Exercise 10.2.5. Using Exer-
o
cise 10.1.5 and Exercise 2.6.8 we see that Log[zlf(z)]l/2 ranges over a convex domain being the image of K(1; r). under Log. Hence [Zlf(Z)]l/2 ranges over K(I; r) for varying f E S* and fixed z, Izl = r. . 10.2.S. If f, z range over S* and K(O; 1) resp., then [z/f(zW/ 2 ranges over
U
K(I; r) = K(1; 1), i.e. ff(z)/Z]l/2 ranges over {w: rew
> t}.
rE(O.l)
10.2.9. Suppose that IZll ~ IZ21 < rand Wk = f(Zk) , k = 1,2; by Schwarz's lemma I1J'(z) I ~ Izl and for z = Z2 we have
Ir l [twl +(1-t)W2] I ~ IZ21
< r;
if Zo =f- l [tw+(1-t)W2], then IZol < r andf(zo) = tWl+(I-t)w2 which means tlaat Br is convex since t E (0, 1) can be arbitrary. 10.2.10. If z!' E S*, then the tangent vector of f[c(O; r)] turns monotonically and its argument increases by 2" as z descril}es C(O; r). Hencef[C(O; r)] is a convex curve whose interior domain is convex. A converse can be ,proved in an analogous manner. 10.2.11. !' = F/z with FE S* (cf. Ex. 10.2.8).
10. UNIVALENT FUNCTIONS
265
10.2.12. We have reF(z) > 0 for z EK(O; 1) and also for z
o<
1
~
o
~
E
[Zl' Z2], hence
~ reF[zl + t(Z2-Z1)]dt = re {(Z2-Z1)-1 ~ (1+z) (l-zr1dz}
= re{2(z2-Z1)-lLog[(I-z1) (I-Z 2)-1]-I}. 10.2.13.p(z)
= 2y!'(z)-1
E
& by Exercise 10.2.11. Hence by Exercise 10.1.3: 2" ....
p(z) = ~ (e'fJ+z)/(eifJ-z)d",«()) o
and this implies 2" 2"
f'(z)
= ~ ~ ei"'eitp(el'P-z)-l(e'tp-z)-ld",(p)d",(Ip) o
0
(where", is increasing, ",(0) = 0, ",(27t) = 1). Integrating w.r.t. z we obtain:
> t.
and hence by Exercise 10.2.12: re[z-lf(z)]
10.2.14. From Exercise 10.2.13 it follows that IZif(z)--:-11 Schwarz's lemma.
<
1. Now apply
10.2.1S. g ESc, hence reg!'(C)IJ(C)} = re{(l-1C1 2)-lC/g(C)} ~ since reg/g(C)}
~
(l-1C1 2 )-1(1-ICD = (l+ICD-1 > t
l-ICI (cf. Ex. 10.2.14).
10.2.16. By Exercise 10.2.15 z!' if -< (l-z)-t, hence z!' if takes inside K(O; r) the values contained in the disk with diameter [1-r, l+r]. 10.2.17. By Exercise 10.2.13 z-y -< (l-z)-t, hence the estimates are the same as in Exercise 10.2.16. 10.3.1. Consider any partition of [0, 27t] containing all the end points of monotoneity intervals of e])«() and consider a corresponding integral sum for the Stieltjes 2"
integral ~ h(R«(j»)de])«(). Any ray e]) = Pk situated inside an angle obtained o
r
by the partition intersects at an 9dd number of points: R~k) > R~k) > ... > R~r:.>k+l and if LIe]) is the measure of the corresponding angle, then the integral sum is equal to L Lle])k[h(R~k»-h(R~k»+ .,. +h(R~~+l)] > 0 because the k
SOLUTIONS
266
orientation is positive. Take now the limit for a normal sequence of partitions. 10.3.2. We have:
r-;'
2"
2"
o
0
~ g(R)dO = r ~
2"
:r
g(R)~O = r ~ g'(R)R
2"
=
~
2"
g'(R)R
~~ dO = ~
o
10.3.3. If g(R)
=
~
Rg'(R)d6W(r, 0).
0
R2, then by Exercise 10.3.2 we obtain
2"
r ;,-
:r 10gRdO
0
2"
If(re"6)1 2d8
o
=
r ;,
~
2"
R 2dO
=
0
which is ~ 0 in view of Exercise 10.3.1 (h(R)
~
2R2(r, O)d6W(r, 0)
0
=
2R2 ).
o FIG. 7
10.3.4. The values of F do not cover'C completely (Liouville's theorem); if F(z) i= Wo for any z E K( 00; 1), then by the argument principle the images of C(O; r) by F(z)-wo contain w = 0 inside, hence 2"
I(r)
=
~ IF(reiB)-woI2dO = 27t[r2+lbo-woI2+lblI2r-2+lb212r-4+ ... ] o
is an increasing function of r and consequently I' (r) ?- O. N
10.3.5.
2..: n IbnI2r-2n-l :S;; r; .=1
then make N
-+
+ 00 .
N
make first r -+ 1 which gives
2..: n Ibnl2 :S;; 1 .=1
and
10. UNIVALENT FUNCTIONS
267
10.3.6. If IE S and Izl > I, then F(z) = IIf(z-1) belongs to 1:0 • If FE Eo and ICI < I, then/(C) = I/F(C- 1) belongs to S. The function F(z) = z(1+z-3)Z/3 corresponds in this way to a function IE S*. 10.3.7. G(z) = [F(zZWIZ = z+tboZ-1+ ... E Eo, hence by Exercise 10.3.5 we have: -} Ibol ~ I. The case of equality occurs only for G(z) = z+e i "z- 1, or F(z) = z+2e'''+e Zi ..z-1, with oc real. 10.3.8. F-Wk
E
Eo, hence IWk-bol ~ 2 (k = 1,2) and
IW1-wzl ~ IW1-bol+lbo-wzl ~ 4. 10.3.9. If IE S, then F(C) = 1/I(C-1) = C-a2+ ... E Eo so that lazi ~ 2 (cf. Ex. 10.3.7). Equality holds only for F(C) = C+2e i "+e Z:"C- 1, i.e. for I(z) = I .. (z) = z(1+ei"z)-2. 10.3.10. cp is univalent and analytic in K(O; I) being a superposition of a linear transformation and IE S; moreover, cp(z) = z+(a z +h-1)zZ+ ... , hence laz+h- 11 ~ 2, i.e. Ih- 11 ~ 2+la21 ~ 4. Equality holds only, if la21 = 2 which means that I = I ...
10.3.11. rp is univalent in K(O; I), moreover, rp(O) = 0, cp'(O) = I. Hence cp E S. Now, z(t) =f:. b and this implies that rp(t) =f:. rJ(b)-/(a)]t'(a)If'(a) = h for any t E K(O; I); t(z) denotes here the inverse of z(t). The result follows immediately from Exercise 2.9.21 and Exercise 10.3.10. 10.3.12. (i) Take b = 0 which gives laf'(a)lf(a) I ~ (I + lal)(I-lal)-1 ; (ii) on integrating both sides in (i) we obtain: I/(a) I ~ lal (I-Ial)-z; taking a = 0 in Exercise 10.3.11 we obtain Ibl(I+lbl)-z ~ I/(b)l; (iii) "P E S, hence 1"P(t)1 ~ Itl(I-ltl)-z by (ii) and putting t = -z we obtain Izf'(z)lf(z) I ~ (I-Izl) (1 + Izl)-1 .
Equality can be attained in all cases for I = I ... 10.3.13. Multiply both estimates for I/(z) I and Iif' (z)/I(z) I side by side. 10.3.14. We have I'(b) = rp'(O) (1-lbI Z)-l,
f'(a) = rp'(C1)(I-lbl z ) (l-a6)-2
with C1 = (a-b) (l-ab)-1. Hence f'(a)If'(b) = (1-lbI Z)Z(I-ab)-zcp'(C1)/rp'(0).
Now, [rp(C)-rp(O)]/rp'(O) = "P(C)
E
S, thus
1"P'(C1)1 ~ (1+IC1 1) (1-IC1 0- 3
268
SOLUTIONS
and finally
I ~ (1-lbI ) (ll-abl+ la-b l )2 If'(a) f'(b) (1-laI 2) 11-abl-la-bl 2
(cf. Ex. 1.1.8 (iii»). The lower estimate is obtained by interchanging a, b. 10.3.16. (i) (Jzl-l)2Izl- 1 ~ IF(z) I ~ (Izl + 1)2Izl-1; (ii) (Izl-l) (lzl+1)-1 ~ IzF'(z)/F(z) I ~ (Jzl+l) (lzl-IJ)-l, which is easily obtained by using Exercise 10.3.12 and Exercise 10.3.6. 10.3.17. (i) Take a sequence {zn} such that IZnl -.. 1, F(zn) is convergent and
limIF(zn)-bol = lim IF(z)-bol. JzJ-l+
Then Wo = limF(zn) is not a value of F, hence F-w o E Eo. Now, by Exercise 10.3.7, Ibo-wol ~ 2. • (ii) lim !z(F(z)-z-b o)!2 ~ lim JF(z)-b ol+l ~ 3, moreover, !z(F(z)-z-b o)! IzJ-+l+ JzJ-l+ -.. Ib11 ~ 1 as z -.. +00, hence !z(F(z)-z-b o)! ~ 3 by the maximum principle.
10.3.18. (i) IF'(z)-11 ~ Iznl·lb11+V2Izl-1y'2Ib21+ ... ) ~ Izl- 2 y(l-lzl- 2)-2 Vlb112+2Ib212+ '" ~ (lzI2-1)-1
by Schwarz's inequality and the area theorem. (ii) IF' (z)1 ~ IF'(z)-11+1 ~ IzI2(lzI2-11-1. 10.3.19. F(z) is a superposition of a(C +C- 1)+b and C = i(h2-1)-1/2(hz-l) hence it must be univalent. The latter transformation carries C(O; 1) into a circle through =F1 and by Exercise 2.5.2: C'"F[K( 00; 1)] is a circular arc. 10.3.20. (ii) g
E
S hence in view of Exercise lO.3.12 (ii) we have
If(z) I 11 +ei":{(z)/MI ~ Izl (1-lzJ)-2 for an arbitrary real ce. Thus
If(z)I(I-lf(z)IIMt2 ~ Izl(I-lzJ)-2 = Iwl(I-lwl/M)-2, where w = fM(Z). Now, u(l-uIM)-2 is an increasing function of u e:'(0, M). hence If(z) I ~ Iwl = IfM(Z)I. A similar proof for the lower estimate. (i) M(lzl,f)
= Izl+la2i1zI2+O(JZJ)3 =
~
M(jzl,fM)
Izl+2(I-M- 1)lzI 2+O(JzI3), hence la21 ~ 2(1-M-1).
10.3.21. g(z) = z+C-i-(I-ICI 2)f"(C)lf'(C)-i-(I-ICI 2)2{f. C}Z-1+ ... Now, by the area theorem (Ex. 10.3.5) the coefficient of Z-1 is at most 1 in absolute
10. UNIVALENT FUNCTIONS
269
value, hence Itf, 01 ~ 6(1-ICI2)-2. If h is analytic and univalent in K(O; 1), then f = ah+b E S for suitably chosen constants a, band {h, z} = {J, z} by Exercise 7.2.14. If h is meromorphic and univalent in K(O; 1) and h(z) ¥= Wo for any z E K(O; 1), then H = (h--wo)-l is analytic and univalent and again
{H, z}
{h, z}.
=
10.4.1. The proofs for Steine~ and P6lya symmetrization are similar. For sake of simplicity we give here the proof for Steiner symmetrization Suppose that a(xl), a(x z ) meet G* and also G. Then a(x) meets G for any x E [Xl' x 2] since in the opposite case G would be disconnected. This implies that [Xl, X 2 ] C G* and obviously two arbitrary points Zl' Z2 E G* with rezl = Xl' rez 2 = X2 can be joined by the polygonal line [Zl' Xl' X2' z 2] in G*. Hence G* is arc-wise connected, if G is connected Evidently C",-G* is also simply connected in this case. Suppose now G is an open set and Zo = xo+iyo E G*. It is easily verified that the linear measure lex) of a(x) n G is a lower semicontinuous function of X for open G (note that a(x) n G is an at most countable system of open segments: replace the countable system by a finite system of closed sub-segments with a slightly less total measure and use Heine-Borel theorem for the latter system). Now, in our case l(x o) > 21Yol. From lower semicontinuity it follows that for any Yl > 0 such that l(xo) > 2Yl > 21Yol we can choose lJ such that flex) > Yl for any X E (xo-lJ, xo+lJ). Then the open rectangle {x+iy: Ix-xol < lJ, lyl < yd c G* which proves that G* is open. In case of Steiner symmetrization G* is always simply connected if G is a domain. In case of P6lya symmetrization we should make an additional hypothesis that G is a simply connected domain. 10.4.2. If G is a domain starshaped w.r.t. the origin 0 and WO EC",-G, then the whole ray emanating from Wo whose prolongation contains 0 belongs to C",,- G. This implies that the function l(r) introduced in the definition is decreasing which again implies that G* is starshaped w.r.t. the origin. 10.4.3. Suppose G is a convex domain. It is sufficient to consider bounded, convex domains since each convex domain is a sum of an increasing sequence of bounded convex domains e.g. G n K(O; n). Now the boundary of a bounded convex domain consists of the graphs of 2 functions rpl' rp2' a ~ X ~ b, and possibly two segments rez = a, rez = b, the functions rpl' -rp2 being convex. This implies that t(rpl -rp2) = lex) is convex and non-positive, hence G* being bounded by two convex arcs and possibly two segments must necessarily be convex. D = [S n K(O; 2)] u [S n K(O; 4) n {z: rez > O}], where S
=
{z: IJim zl < I}, is convex, however n* is not convex.
270
SOLUTIONS
00
~~
1f'(rei6 )1 2 rdrd() =
7t
K(O; 1)
Suppose that
< +00.
L nlanl2 n=1
8> 0 is arbitrary and choose m ~ 2 such
00
that
2: nlanl
2
<
n=m
+82.
Thus 00
00
Lnlanlrn-1 n=m
= Lyn1anl ' ynr n- 1 n=m·
and therefore 00
(1-r) Lnlanlrn-1 ~ +8(I+r)-1
<
+8
n=m
for any r
E
(0,1), if m = m(e) is large enough. Choose now ro, m being fixed
as before, such that (1-r)
<
+8 and also
m
m
n=1
n=1
2..: nlanl < +8 for all r E(r o, 1). Then (1-r) L nlanlr n- 1 00
(l-r)M(r,J') ~ (1-r) Lnlanlrn-1
<
28/2 = 8
n= 1
for all r E (ro, 1) which proves our result. 10.4.5. By Exercise 8.1.9: r(w, G) = (1-lzI 2 )1f'(z)1 where f is univalent in K(O; 1), w =f(z), andf[K(O; 1)] = G. Hence r(w; G) is continuous. Note that
the area IGI is finite and use Exercise 10.4.4. 10.4.6. (i) In view of the maximum principle we need only to estimate the difference h(z, zo)-h(z, Z1) on the boundary of G. We have for!; E frG: . h(z, zo)-h(z, Z1)
<
]og[I!;-zolll!;-z11]
=
logll+(z1-zo) (!;-Z1)-11
~I 1(Z1- Z0) (!;-Z1)-11 ~ IZ1-Zol/b;
after interchanging Zo, Z1 we obtain Ih(z, zo)-h(z, z1)1 < IZ1-zollb which gives (i). The inequality logll +zl ~ Izl = loge izi is equivalent to another, quite obvious inequality:
10. UNIVALENT FUNCTIONS
271
(ii) the inner radius r(zo; G) satisfies the equality: logr(zo; G) = h(zo, zo), hence Ilogr(zo; G)-logr(zl; G)I
=
Ih(zo, zo)-h(zo, zl)+h(Zl' zO)-h(Zl' zt)1
in view of (i). We used the symmetry of h which is equivalent to the symmetry of the Green function. <Xl
10.4.7. Suppose that G is the union
U
G. of an increasing sequence of bounded .=1 domains each being regular with respect to the Dirichlet problem. If for some m: dist(z'; frG m) then the same is true for all n
~
>
r5,
dist(z"; frG m)
~
r5,
m and also
Ilogr(z'; Gn)-logr(z"; Gn)1 Z 2Iz'-z"l/r5
for all n ~ m. Hence either both sequences r(z'; G.) tend to to a finite limit. In the limiting case we have:
+ 00, or both tend
Ilogr(z'; G)-logr(z"; G)I Z 2Iz'-z"I/r5
which proves continuity of r(z; G). We have proved even more: r(z; G) is Lipschitzian on compact subsets of G. 10.4.8. r(z; H) attains a finite maximum at some point CE H in view of Exercise 10.4.5. Consider now Steiner symmetrization w.r.t. a line through Cparallel to the major axis. This gives r(C l ; H) ~ r(C; H), Cl being the projection of C on the major axis. Another Steiner symmetrization w.r.t. a line through Cl parallel to the minor axis shows that reO; H) ~ r(C; H), 0 being the center of H. 10.4.9. If G E <§ and the positive, real axis is the -half-line of P61ya symmetrization, then the symmetrized domain G* E <§. Moreover, if , Go = K(O; R)",(-R, -r], then G* c Go. Hence reO; G)
Z
reo; G*)
Z
r(O; Go)
=
4rR2(R+r)-2
(cf. Ex. 2.9.19, 8.1.9). The l.u.b. is actually attained because GoE<§. 10.4.10. r5 is the root of the equation 4R 2r5(R+ r5)-2
=
1, i.e. r5
=
R[2R-1-
-2VR(R-1)1. For suppose, contrary to this, that there exists Go E<§(R) such that dist(O; r) = r5 0 < r5, r being the component of C", G containing 00. If we symmetrize Go w.r.t. the positive real axis and replace G~ by G1 = K(R)~ (- R, - r5 01 which contains G~, then reO; Go)
= 1Z
which is a contradiction because bo
reo; G~) Z reo; G1 )
< b implies
reO; G1)
< 1.
SOLUTIONS
272
10.4.11. We may suppose that G = f[K(O; I)] and fE S. We have:
r(wo; G) where Wo
=
=
(1-lzoI2)1f'(zo)l,
f(zo); hence
1f'(zo)lf(zo) I = r(wo; G) [lwol(I-l zoI2)]-1
< (1+lzol) [lzol(I-lzol)]-1
by Exercise 10.3.12 (i). Hence
r(w o ; G)
< IWol(I+lzoI)2Izol-1 =
4Iwol+lwo!(I-lzoD2Izol-1
< 4Iwol+l,
(cf. Ex. 10.3.2 (ii)). 10.4.12. If 0 E G and Wo =1= 00 is such that IWol = dist(O; C"""G) , then -lwol] c: G* and from the definition of circular symmetrization it follows that C""" G* is arcwise connected. Hence G* is of hyperbolic type. If G = {w: im w > and the real axis is the line of Steiner symmetrization then G* = C which is not of hyperbolic type.
c"""
( - 00,
+}
10.4.13. From the conformal invariance of Idwl/r(w; G) (cf. Ex. 8.1.11) it follows that Idwl/r(w; G) = Idzl/r(z; K(O; 1)) = (l-lzI2)-lldzl, hence
)[r(w; G)]-lldwl
P(WI' W2; G),
=
r
rbeing the h-segmentjoining WI to w2 • We now associate with r a curve y c G* joining WI to W2 in the following manner. Suppose that w = Rei> E Then the point CE Y associated with w is the point C= R. From Idwl 2 = dR2+R 2d¢2 it follows that IdCI < Idwl. We consider a ray Ow, w E r and the symmetrized domain G! of G w.r.t. ray Ow. We have: r(w; G!) ~ r(w; G) due to P6lyaSzego theorem. However, G!, G* are congruent and a rotation round the origin by the angle q;, carries G* into G!. Since the inner radius is invariant under motion, r(w; G!) = r(C; G*). Thus IdCl/r(C; G*) < Idwl/r(w; G) and after integration we obtain
r.
~ [r(C; G*)tlldCI 1
<
~ [r(w; G)r 1 ldwl
=
P(Wl, W2; G).
r
Observe that y joins WI to W2 in G* and
P(WI' W2; G*)
=
inf ~ [r(C; G*)tlldCI, C
c
the g.l.b. being taken w.r.t. all regular curves joining
WI
to
W2
in G*.
10.4.14. Observe that g(Wl' w2; G) = -logtanhp(wl' W2; G), (cf. Ex. 8.5.11) and use the result of Exercise 1O.4.l3.
10. UNIVALENT FUNCTIONS
273
10.4.15. f takes real values on the real axis which follows from the uniqueness and symmetry considerations. Suppose that w = f(zo) , Zo =F 0, and IWol = M(lzol,n. If we symmetrize G w.r.t. the ray OWo, we obtain th.e domain Go which arises from G after rotation by the angle arg Wo. In view of Exercise 10.4.14:
g(O, wo; G)
-loglzol :::;; g(O, Wo; Go)
=
=
g(O, Iwol; G)
-logr,
=
wheref(r) = IWol (I(x) takes on (0,1) all values between 0 and sup/f(z)/, hence Izl
it takes the value /woD. This means that /f(zo)/ = fer) = /wo/, while /zo/ Now, f strictly increases on the real axis, hence f(lzo/)
~.f(r)
~
r.
= /f(zo)/ = M(lzol,j).
10.4.16. Suppose that Zo =F 0, If(zo)1 = M/(zo/, nand Wo = f(zo). The symmetrized domain G~ obtained from G by symmetrizing w.r.t. the ray OWo becomes G* after a rotation hence Wo = ei '1*(r*). Now, g(O, wo; G~) = -logr*, g(O, wo; G) = -logr, hence by the result of Exercise 10.4.14: r* :::;; r. Thus M(r,J*)
~
M(r*,J*) =f(r)
=
IWol = /f(zo) I = M(r,f).
10.4.17. Suppose that fE Sand D = f[K(O; 1)]. If he'f% E C""-D for some h E (0, +), then C"", D being connected, any circle C(O; r), r ~ h, meets C",,- D and this implies that C""-D* contains the ray (-00, -h]. For Do = C""-(- 00, -h] we have reO; Do) = 4h < 1. On the other hand 1 = reo; D) ~ reo; D*) :::;; reo; Do) = 4h < 1 which is a contradiction. 10.5.1. h is harmonic in K(O; 1) except for the points z such that fez) = 00. If Izl -+ 1, then limh(z) ~ 0; if z -+ Z' (z' =F O,j(z') = ao), then h(z) -+ + 00 and finally if z -+ 0, then
h(z)
=
-loglf(z)-aol+logr(ao ; G)+loglzl+o(l)
= --loglall+logr(ao ; G)+o(l) which implies that z = 0 is a removable singularity. The maximum principle yields now h(z) ~ 0 and consequently -logla 1 1+logr(ao; G)
~
0,
lall = 11'(0)1:::;; r(ao ; G).
i.e.
In case G is a simply connected domain the result follows by subordination. 10.5.2. We may assume that r(a o ; G) is finite. Choose p such that f'(pe i6) =F 0 for all () E [0, 27t] and consider the domain G(p) = f[K(O; p)] whose boundary consists of a finite number of arcs on rep) f[e(o; p)]. We can add at possible singular points of (1G(p) small disks so Ihat Ihe domain G(I') so obtained possesses "C"
SOLUTIONS
274
the classical Green function and G(p) c G(p) c G. The result of Exercise 10.5.1 as applied to f(pz) and the domain G(p) gives plf'(O)1 ~ r(a o ; G(p») ~ r(a o ; G);
make now p -. 1. 10.5.3. 11'(0)1 ~ r(a o ; D,); moreover, P6lya-Szego theorem gives r(a o ; D,) r(a o ; D*). On applying the monotoneity property of the inner radius, the result follows. ~
10.5.4. We may assume that ao > 0 (otherwise consider e-:y(z) , where A = arga o). The domain D* obtained from D, by circular symmetrization w.r.t. the positive real axis is contained in Do = {w: largwl < flX7t}. The function
= 00(l+2IXz+ ... ) maps 1: 1 conformally K(O; 1) onto Do with f(O) = a o • r(a o ; Do) being equal fez)
=
ao[(l+z)/(l-z)]~
to 2lXao. Now apply Exercise 10.5.3. 10.5.5. We may assume that ao > O. We symmetrize D, w.r.t. the positive real axis and obtain a domain D contained in Do"'-(- 00, - R]. Now, fo maps K(O; 1) onto Do so that fo(O) = ao, hence r(a o ; Do)
= f~(O) = 4(a o+R).
Use now Exercise 10.5.3. 10.5.6. By Exercise 10.5.5 we have 1 = la 1 1 ~ 4R,. 10.5.7. Iff E S, then D, is connected, hen~e C(O; p) cD, implies K(O; p) cD,. Use now the result of Exercise 10.5.6.
C",-
10.5.S. g omits the values exp[27th- 1 (u+iv)], hence on each circle C(O; p) there exists a value omitted by g and therefore Rg = 0 (Ex. 10.5.5.). Now g(z)
= exp(27tao/h) (1+27ta 1h- 1 z+ ... )
and the inequality 127ta1h- 1 1 ~ 4 follows from Exercise 10.5.5. Equality holds for fez) = ao+h-;'--l~og[(l+z) (l-Z)-l]. 10.5.9. Write the inequality Irp'(O)1 ~ 4(lrp(0)I+R)
for rp(C) = f[(zo+C) (l+Z OC)-l] and R = O. 10.5.10. We have II'(z)lf(z) I ~ 4(1-lzI2)-1 (Ex. 10.5.9) and after integration we obtain:
10. UNIVALENT FUNCTIONS
275 z
~I
Ilog[/(z)I/(O)] I = IV'(z)[/(z)]-ldzl":(;
4'~
o
(l-r 2)-ldr
0
and this implies the right-hand inequality; the left-hand one is obtained by applying the right inequality to [f(Z)]-1. 10.5.11. We may assume that ao = flO) > O. On symmetrizing Drp = tp[K(O; 1)] w.r.t. the positive real axis we obtain a set D* contained in the right halfplane. Write now the inequality 1/,(0)1 ~ r(a o ; Do) of Exercise 10.5.3 for /(1;) = tp[(zo+1;)(I+zo1;)-1] and note that r(ao ; Do) = 21ao l = 21/(0)1. 10.5.12. (i) Suppose that / is not identically O. Then fez)
= anz"+an+1z"+1+ ...
with
an:l= O.
By the local mapping theorem (cf. [1], p. 131) there exists ~ > 0 such that K(O; b) ~ /[K(O; 1)]. Suppose now that there exists Z1 E K(O; 1) such that /(Z1) = Re i « with R > ~-1. Then for some Z2 E K(O; 1) we have /(Z2) = R-1e- i « E K(O; b), i.e. /(Z1)f(Z2) = 1 which is a contradiction. (ii) From [1+/(z1)][I-/(z1)]"'1 = -[1+/(z2)][I-/(Z2W 1 (Z1 :1= Z2) it follows that 1 = /(Z1)f(Z2) , which is a contradiction. 10.5.13. (i) Write the inequality obtained in Exercise 10.5.11 for cp(z) (ii) Put
= [1+f(z)][I-/(z)]-1.
z = 0 in (il.
10.5.14. The transformation W = W(Z), where W = pz(z+p)(I+pzrl, t = t(Z+Z-l+ p+ p -l) = 2(P_1)2p- 1Z(I+Z)-2, maps 1:1 conformally {Z: IZI < I} onto the W-plane slit along the ray [p, + 00) and the circular arc: IWI = p, larg WI :(; 2arcsinp-l, (cf. Ex. 2.9.22). We have: W = W(Z) = 4p3(1+p)-2Z + ... , hence w = tp-3(1+p)2W(Z) = Z+ ... =/0(Z) is a function which belongs to S and maps K(O; 1) onto the domain G(p) such that C""-G(p) consists of the ray (r, +00) and the circular arc w = r = (1 +p)2 14p2, largwl :(; 2arcsinp-1. We have reO; G(p» = 1. Suppose now that L(r,f) > L(r,/o) for some / E S. We symmetrize/[K(O; 1)] w.r.t. the negative real axis and obtain a domain D* such that /o[K(O; 1)] c D* and D*""-/o[K(O; 1)] :1= 0. Thus we have reO; D*) > r(O'/o[K(O; 1)]) = 1 which contradicts the inequality r(O;/[K(O; 1)]) = 1 :::.;; reO; D*). 10.5.15. Obviously fez) = F[(t-toz)!(I-z)], where y~ = to+iYl-f(P. Similarly as in Exercise 10.5.14 we can show that the extremal domain is the image domain of K(O; I) under the mapping g(z) = /(z)//'(O) and arises
276
SOLUTIONS
from H«()) by a similarity with the ratio r = 1/'(0)1-1 , The set of values not taken by g and situated' on C(O; r) is a circular arc subtending the angle (2-())1t = mp(r), where mp(r) is the maximal value to be evaluated. We have: 11'(0)1 = t[(2+())2+B(2_())2-0]1/2, hence rp(r) = 2-() satisfies the equation:
Bibliography 1. BASIC TEXTBOOKS
1. L. V. Ahlfors, Complex Analysis. McGraw-Hill, New York, 1966. 2. H. Behnke, und F. Sommer, Theorie der analytischen Funktionen I!!iner komplexen Veriinderlichen. Springer, BerIin-Gottingen-Heidelberg, 1955. 3. C. Caratheodory, Funktionentheorie, Vols. I and II. Birkhliuser, Basel, 1950. 4. B. A. Fuchs, V. I. Levin, and B. V. Shabat, Functions of a Complex Variable and Some of Their Applications, Vols. I and II. Pergamon, Oxford, 1961. 5. M. Heins, Complex Function Theory. Academic Press, New York, 1968. 6. E. Hille, Analytic Function Theory, Vols. I and II. Ginn, Boston, 1962. 7. A. Hurwitz, und R. Courant, Vorlesungen iiber allgemeine Funktionentheorie und elliptische Funktionen. Springer, Berlin-Gottingen-Heidelberg, 1964. 8. G. W. Mackey, Lectures on the Theory of Functions of a Complex Variable. Van Nostrand, Princeton, 1967. 9. R. Nevanlinna, und V. Paatero, Einfiihrung in die Funktionentheorie. Birkhliuser, Basel, 1965. 10. S. Saks and A. Zygmund, Analytic Functions. PWN, Warszawa, 1965. 11. E. C. Titchmarsh, The Theory of Functions. Oxford University Press, 1947. 2. COLLATERAL READING
12. S. Bergman, The Kernel Function and Conformal Mapping. Amer. Math. Soc., New York 1950. 13. R. P. Boas, Entire Functions. Academic Press, New York, 1954. 14. R. Courant, Dirichlet's Principle, Conformal Mapping and Minimal Surfaces. Interscience Pub!., New York, 1950. 15. W. H. J. Fuchs, Topics in the Theory of Functions of One Complex Variable. Van Nostrand, Princeton, 1967. 16. G. M. Golusin, Geometrische Funktionentheorie. VEB Deutscher Verlag der Wissenschaften, Berlin, 1957. 17. W. K. Hayman, Multivalent Functions. Cambridge University Press, 1958. 18. W. K. Hayman, Meromorphic Functions. Oxford University Press, 1964. 19. M. Heins, Selected Topics in the Classical Theory of Functions of a Complex Variable. Holt, Rinehart and Winston, New York, 1962. 20. J. A. Jenkins, Univalent Functions and Conformal Mapping. Springer, Berlin~Gottingen Heidelberg, 1958. 21. G. Julia, Exercices d'analyse, Vol. II. Gauthier-Villars, Paris, 1969. 22. J. E. Littlewood, LecflIres Ofl 'he Theory of Functions. Oxford University Press, 1944.
278
BIBLIOGRAPHY
23. l. P. Natanson, Theorie der Funktionen einer reel/en Veriinderlichen. VEB Deutscher Verlag der Wissenschaften, Berlin, 1954. 24. Z. Nehari, Conformal Mapping. McGraw-Hill, New York, 1952. 25. R. Nevanlinna, Eindeutige analytische Funktionen. Springer, Berlin-Gottingen-Heidelberg, 1953. 26. I. I. Priwalow, Randeigenschaften analytischer Funktionen. VEB Deutscher Verlag der Wissenschaften, Berlin, 1956. 27. M. Tsuji, Potential Theory in Modern Function Theory. Maruzen, Tokyo, 1959. 3. PROBLEMS AND EXERCISES
28. M. A. Evgrafov, Yu. V. Sidorov, M. V. Fedoryuk, M. I. Shabunin :and K. A. Bezhanov, Collection of Problems on the Theory of Analytic Functions (in Russian). Moscow, 1969. 29. K. Knopp, Problem Book in the Theory of Functions. Dover, New York, 1948. 30. G. P61ya und G. Szego, Aufgaben und Lehrsiitze aus der Analysis. Vois. I and II. Springer, Berlin, 1925. 31. M, R. Spiegel, Theory and Problems of Complex Variables. Schaum, New York, 1964. 32. L. I. Volkovyskii, G. L. Lunts, and I. G. Aramanovich, A Collection of Problems on Complex Analysis. Pergamon, Oxford, 1965.
Index Abel's limit theorem, 65 Almost uniformly convergent sequence of functions, 57 Almost uniformly convergent series of functions, 57 Analytic continuation of a function, 95 Analytic element, 95 Analytic function, 19 univalent in a domain, 23 Area of a spherical triangle, 10 Area theorem, 131 Argument principle, 5 4
Bergman kernel function, 119 Bernoulli numbers, 63 Bessel function, 68 Biberbach-Eilenberg function, 137
Cauchy's coefficient formula, 61 Cauchy's integral formula, 38 Cauchy's theorem for a rectangle, 38 for simply connected domains, 38 homological version of, 38 Cauchy-Hadamard formula, 59 Cauchy-Riemann equations, 19 Center of Taylor's series, 60 Chain, 33 Circular symmetrization, 134 Circulation along a contour, 122 Close-to-convex function, 104 Compact family of analytic functions, 75 Complete analytic function, 95 Complete Legendre elliptic integral, 105 Complex function, 19
Complex potential of an electrostatic field, 124 of a flow, 121 Conformal mapping, 23 Conjugate harmonic functions, 21 Continuation of a function analytic, 95 direct analytic, 95, 97 Contour, 34 Convergence of an infinite product, 82 disk of, 59 Convergent infinite product, 82 Convex hull of a set, 6 conv {Zlo z" ... , z.}, 6 Criterion Marty's, for normality, 75 Montel's, 75 Curve Jordan, 34 regular, 17, 18, 33 starshaped image, 94 Cycle 33 homologous to zero, 38 Decomposition of an entire function, 84 .Derivative of a complex domain, 19 Schwarzian, 98 Differentiable complex functIon, 19 Direct analytic continuation of a function, 95, 97 Disk of convergence, 59 Domain Jordan, 110 of hyperbolic type, 135 regular with respect to the Dirichlet problem, 113
INDEX
280 Eilenberg-Biberbach function, 137 Elliptic linear transformation, 15 Elliptic modular function, 99 Electrostatic potential, 124 Entire function, 76 of finite order, 88 Equipotential lines, 121 Essential isolated singularity, 41
Fibonacci sequence, 62 Field of force, 121 stationary, 121 two-dimensional electrostatic, 124 Flow complex potential of, 121 function, 121 stationary two-dimensional, 121 Formula Cauchy's coefficients, 61 Cauchy's integral, 38 Cauchy-Hadamard, 59 for nth derivative, 39 Green's, 125 Jensen's, 80, 81 Legendre's, 107 Poisson'S, 111 Pringsheim's interpolation, 85 Taylor's -42 Formulas, Schwarz.-Christoffel, 100 Fourier series representation, 112 Function analytic, 19 analytic, univalent in a domain, 23 Bergman kernel, 119 Bessel, 68 Biberbach-Eilenberg, 137 cIose-to-convex, 104 en, 105 complete analytic, 95 • complex, 19 differentiable complex, 19 dn, 105 elliptic modular, 99
Function entire, 76 entire, of 'finite order, 88 flow, 121 gamma, 73 Green's, 114 harmonic, 21 holomorphic, 19 integrable complex-valued, of real variable, 33 Koebe's, 131, 166 Pick's 133 regular, 19 subordinate, 92 Weierstrass, t, 106 Weierstrass, f.J 106 Weierstrass, (1 106 Function element, 95 Functions, conjugate harmonic, 21 Gamma function, 73 Gauss and Lucas theorem, 6 Gauss theorem, 125 Genus of a sequence, 84 Goursat lemma, 38 Green's formula, 125 Green's function, 114 Hadamard's three circles theorem, 90 h-area of a regular domain, 18 Harmonic function, 21 Harmonic measure of a system of arcs, 114 h-boundary rotation, 17 h-distance, 18 h-Iength of a regular curve, 18 h-line at infinity, 17 h-parallels, 17 h-rotations, 17 h-segment, 17 h-translation, 17 h-triangle, 18 HeIIy's selection principle, 260 Herglotz theorem, 127 Holomorphic function, 19 Homotopic paths, 98
INDEX
de I'Hospital's rule, 43 Hurwitz theorem, 76 Hyperbolic distance, 17, 18 Hyperbolic length of regular curve, 18 Hyperbolic linear transformation, 15 Hyperbolic metric, 111 Hyperbolic motions, 16 Hyperbolic straight lines, 16
Index of a point, 36 Infinite product, 82 Inner radi us of a domain at a point, 111 of a simply connected domain, 133 Isolated singularity, 41 Intensity, 1~2 Invariant point of a linear transformation, 14 Involution, 14 Integrable complex-valued flmction of a real variable, 33 Integral complete Legendre elliptic, 105 line, of a complex-valued function, 33 Schwarz-Christoffel, 105 StieItjes, 127 unoriented line, 34
Jensen's formula, 80, 81 Jordan curve, 34 Jordan domain, 110 Jordan's theorem for curves starlike w.r.t. an origin, 36 Koebe function, 131, 166 Koebe one quarter theorem, 136
Luarent coefficients, 67 Laurent expansion, 45 Laurent series, 66 principal part of, 66, 67 Legendre formula, 107 Line integral of a complex valued function, 33
2
Lines equipotential, 121 of flow, 121 Liouville's theorem, 40 Lucas and Gauss theorem, 6
Marty's criterion, 75 Mercator projection, 163 Mittag-Leffler representation 77 Montel's compactness condition, 75 Montel's criterion, 75 M -test of Weierstrass, 57
Nevanlinna's characteristic, 81 Normal family of analytic functions, 75 . n-sheeted unit disk, 28
Order of an entire function, 88 ofa pole, 41 Osgood-Taylor-Caratheodory theorem, 110
Parabolic linear transformation, 15 Parallel axiom, 16 Parseval's identity, 59, 119 Pentagram, 103 Phragmen-LindeIOftheorem,115 Pick flllction, 133 Point regular, 61 singular, 61 Points of a hyperbolic plane, 16 Poisson's formula, 111 Poisson's kernel, 112 Pole, 41 Polynomial represented as a product, 84 Potential, 124 electrostatic, 124 complex, of an electrostatic field, 124 velocity, 121 Power series, 58
282 P61ya symmetrization, 134 Principal part of the Laurent series, 66, 67 Pringsheim's interpolation formula, 85 Product convergent infinite, 82 infinite, 82 Weierstrass, 84
Radius inner, 133 of a domain, 111 of convergence of a power series, 59 Reflection of a point with respect to a circle, 12 Regular curve, 17, 18, 33 Regular function, 19 Regular part of the Laurent series, 66 Regular point, 61 Removable singularity, 41 Representation Fourier series, 112 Mittag-Leffler, 77 Residue of an analytic function, 43 Residue theorem, 45 Riemann ~here, 8 Riemann theorem, 11 0 Robin's constant, 126 Rouche's theorem, 54
Schwarzian derivative, 98 Schwarz-Christoffel formulas 100 Schwarz-Christoffel integral 105 Series power 58 Taylor's 60 Sequence Fibonacci 62 of functions almost uniformly convergent 57 Single-valued branch of a function 95 Singular part of the Laurent series 66, 67 Singular point 61
INDEX
Singularity essential isolated 41 isolated 41 removable 41 Sink of intensity 122 Solomon's seal 103 Source of intensity 122 Sphere of Riemann, 8 Spherical derivative of an analytic function, 75 Spherical distance between two points, 9 Spherical image of a complex number, 8 Starshaped image curve, 94 Stationary field, 121 Stationary two-dimensiomil flow, 121 Steiner symmetrization, 134 Stereographic projection, 8 Stieltjes integral, 127 Stieltjes-Osgood theorem, 75 Stolz angle, 65 Subordinate function, 92 Symmetric points w.r.t. a circle, 12 Symmetrization circular, 134 P6lya's, 134 Steiner's, 134 Symmetrization principle, 136 Szego, 134 Taylor coefficients, 61 Taylor's formula, 42 Taylor's series, 60 Theorem . Abel's limit, 65 area, 131 Cauchy's, for a rectangle, 38 Cauchy's, for simply connected domains. 38 Cauchy's, homological version, 38 Hadamard's three circles, 90 Herglotz's, 127 Gauss', 125 Gauss-Lucas, 6 Hurwitz's, 76 Jordan's, 36 Koebe's one quarter, 136
INDEX
Theorem Liouville's, 40 on decomposition of an entire function Weierstrass, 84 Osgoo d-Taylor-Caratheodory, 110 Phragmen-Lindelof, 115 residue, 45 Riemann's, 110 RoucM's, 54 Stieitjes-Osgood, 75 Toeplitz's, 7 two constants, 115 'uniqueness, for the velocity potential, 121 Weierstrass' mean value, 35 Toeplitz's theorem, 7 Toeplitz's transform, 65 Two constants theorem, 115 Two-dimensional electrostatic field, 124
28 Uniqueness theorem for the velocity potential, 121 Unoriented line integral, 34 Velocity potential, 121 Vortex, 122 Weierstrass C-function, 106 Weierstrass SO-function, 106 Weierstrass mean value theorem, 35 Weierstrass M-test, 57 Weierstrass primary factors, 84 Weierstrass product, 84 Weierstrass a-function, 106 Weierstrass theorem on decomposition of an entire function, 84 Winding number, 36