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Trang 2 Kluwer Text in the Mathematical Sciences VOLUME21 A Graduate-Level Book Series Trang 3 through Examples and Exercises by Endre Pap Institute of Mathematics, University of Novi

Kluwer Text in the Mathematical Sciences VOLUME21

A Graduate-Level Book Series

The titfes published in this series are listed at the end 0/ this vofume

through Examples and Exercises

by

Endre Pap

Institute of Mathematics,

University of Novi Sad,

Novi Sad, Yugoslavia

SPRINGER-SCIENCE+BUSINESS MEDIA, B.V

A c.I.P Catalogue record for this book is available from the Library of Congress

ISBN 978-90-481-5253-7 ISBN 978-94-017-1106-7 (eBook)

DOI 10.1007/978-94-017-1106-7

Printed an acid-free paper

AII Rights Reserved

© 1999 Springer Science+Business Media Dordrecht

Originally published by Kluwer Academic Publishers in 1999

N o part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means, electronic or mechanical,

inc1uding photocopying, recording or by any information storage and

retrieval system, without written permission from the copyright owner

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Preface

The book Complex Analysis through Examples and Exercises has come out from the

lectures and exercises that the author held mostly for mathematician and physists The book is an attempt to present the rat her involved subject of complex analysis through an active approach by the reader Thus this book is a complex combination

of theory and examples

Complex analysis is involved in all branches of mathematics It often happens that the complex analysis is the shortest path for solving a problem in real circum-stances We are using the (Cauchy) integral approach and the (Weierstrass) power

se ries approach

In the theory of complex analysis, on the hand one has an interplay of several mathematical disciplines, while on the other various methods, tools, and approaches

In view of that, the exposition of new notions and methods in our book is taken step

by step A minimal amount of expository theory is included at the beinning of each section, the Preliminaries, with maximum effort placed on weil selected examples

and exercises capturing the essence of the material Actually, I have divided the problems into two classes called Examples and Exercises (some of them often also

contain proofs of the statements from the Preliminaries) The examples contain complete solutions and serve as a model for solving similar problems given in the exercises The readers are left to find the solution in the exercisesj the answers, and, occasionally, some hints, are still given Special sections contain so called Composite Examples which consist of combinations of different types of examples explaining,

altogether, some problems completely and giving to the reader an opportunity to check his entire previously accepted knowledge

The necessary prerequisites are a standard undergraduate course on real tions of real variables I have tried to make the book self-contained as much as possible For that reason, I have also included in the Preliminaries and Examples

func-some of the mathematical tools mentioned

The book is prepared for undergraduate and graduate students in matheniatics, physics, technology, economics, and everybody with an interest in complex analysis

We have used for some calculations and drawings the mathematical software

ix

and advice about the text, and to Ivana Stajner for reading some part of the text

I would like to express my thanks to MarCicev Merima for typing the majority of the manuscript It is my pleasure to thank the Institute of Mathematics in Novi Sad for working conditions and financial support I would like to thank Kluwer Academic Publishers, especially Dr Paul Roos and Ms Angelique Hempel for their encouragement and patience

Chapter 1

The Complex N umbers

1.1 Algebraic Properties

1.1.1 Preliminaries

The field of complex numbers Cis the set of all ordered pairs (a, b) where a and b

are real numbers and where addition and multiplication are defined by:

(a, b) + (c, d) = (a + c, b + d) (a, b)(c, d) = (ac - bd, bc + ad)

We will write a for the complex number (a,O) In fact, the mapping a 1-+ (a,O)

defines a field isomorphism of IR into C, hence we may consider IR as a subset of C If

we put z = (0,1), then (a, b) = a + bz For z = a + zb we put Re z = a and Imz = b

Real numbers are associated with points on the x-axis and called the real axis Purely imaginary numbers are associated with points on the y-axis and called the imaginary axis

Note that z2 = -1, so the equation z2+ 1 = 0 has a root in C If z = x+zy (x, y E

IR), then we define

Izl = Jx 2 + y2

to be the absolute value of z and z = x - zy is the conjugate of z We have Izl2 = zz

and the triangle inequality

Iz + wl s:; Izl + Iwl (z,w E C)

By the definition of complex numbers, each z in C can be identified with a unique

point (Rez,Imz) in the plane IR2 •

1

E Pap, Complex Analysis through Examples and Exercises

© Springer Science+Business Media Dordrecht 1999

cosO = Tz! and sm = Tz!

The point z = x+zy =I- 0 has polar coordinates (r, 0) : x = r cos 0, y = r sin O Clearly

r = Izl and 0 is the angle between the positive real axis and the line segment from

° to z Notice that 0 plus any multiple of 271" can be substituted for 0 in the above

equations The angle 0 is called the argument of z and is denoted by 0 = argz Let Zl = rl ( cos Ol + z sin Ol and Z2 = r2 ( cos O 2 + i sin ( 2 ) then

In particular, if z = r( cos 0 + z sin 0), then

zn = rn(cos(nO) + zsin(nO» (1.1 )

As a special case of (1.1) we obtain DeMoivre's formula:

The n-th root of z = r( cos 0 + z sin 0) are

For more explanations see the chapter on power series

1.1.2 Examples and Exercises

Example 1.1 Find the real numbers p and q such that the complex numbers

1

z = p + zq, W = P + z- be equal

q

1.1 ALGEBRAIC PROPERTIES 3 Solution We have that z = w is equivalent with Rez = Rew and Imz =

Immw Therefore p = p and q = !, pEllt, q2 = 1, i.e., ql = 1, q2 = -1 and pEllt

Finally we have w = x(l - z) for x E llt and x i= 0

d) From Re (~) = ° it follows that for every x E llt :

w x + zy 1 - z x + y + z(x - y)

- = - - - - =

z l+z 1-z 2 Hence W = x(l - z) for every x E llt

Example 1.3 Prove that

Hint It is easy to prove the case n = 2 and then use mathematical induction

to prove the general case

Example 1.4 Find for z = 1 + 2z the following numbers

a) zn j b) l/zj c) l/zn j d) Z2 + 2z + 5 + z

E (~)(2z)k

IJ-l)k(n)22k+Z 't(-l)k( n )22k+1,

k=O 2k k=O 2k + 1 where [xl is the greatest integer part of x

Example 1.5 Find the positions of the following points in the complex plane:

a + za, a - za, -a + za, -a - za fOT a E lR?

Solution Using the trigonometrie representation (p,O) wc obtain

1.1 ALGEBRAIC PROPERTIES 5 Hence the four given points are the corners of the square in the circle with the center

at origin and the radius

J21a1-Example 1.6 Which subsets of the complex plain correspond to the complex bers with the following properties:

d) Imz 2 0; e) Izl :::; 2; f) 1 < Izl < 3;

g) Izl > 2; h) -7r < argz < 7r; i) ~ < argz < f?

Solution

a) Rez = Imz {::::=:} x = y, where z = x + iy

The desired sub set consists of the points of the straight line y = x (Figure 1.1)

x

Figure 1.1 Rez = Imz

b) Rez < 1 {::::=:} x < 1 and y is an arbitrary real number, where z = x + iy The desired subset is the half plane left from the straight li ne x = 1 (without the points

x

Figure 1.2 Rez < 1

c) -1 :::; Re z :::; 1 means -1 :::; x :::; 1 and y is an arbitrary real number The

desired subset is the strip between straight lines x = -1 and x = 1, Figure 1.3

Figure 1.3 -1 :::; Rez :::; 1

d) Im x ~ 0 means y ~ 0 and x is an arbitrary real number The desired subset

Figure 1.5 JzJ ~ 2

f) The ease 1 < JzJ < 3 reduces in a sirnilar way as in e) on 1 < JzJ = P < 3, where () is an arbitrary angle from the interval [0,27l'J The desired sub set is the

Figure 1.6 1 < Izl < 3

g) For the case Izl > 2 we have JzJ = p > 2, () E [0, 27r] Therefore the desired subset is the whole complex plane without the disc Jzl ~ 2, Figure 1.7

Figure 1.7 Izl > 2 h) The condition -7r < arg z < 7r implies that p is arbitrary and -7r < () < 7r

Therefore the desired subset is the whole complex plane without the negative part

Together with the condition 0 ::::; Re z ::::; 1 we have completely described the points

in the given triangle with vertices Zt, Z2 and Z3

Example 1.8 Prove that tor every z E C

Izl ::::; IRezl + IImzl ::::; -12 ·14

Solution The left part of the inequality follows from

Izl = v'Re 2z + Im2z::::; y'(IRezl + IImzl)2 = IRezl + IImzl

11

Example 1.9 Find the complex numbers which are the corners 01 the triangle with

equal sides with vertices on unit circle and whose one vertex is on the negative part the real axis

Solution Let us put Zj = pj(cos8j + zsin8j),j = 1,2,3, for the soughtafter points

e) 1 + 1VJ; 1 1) (3 + 13)(1 + 1VJ); g) C +11VJf; h) 1 + 1

1 - 1 Answers

Example 1.11 Find t and 0 so that the eomplex numbers

z = t + tf) and w = t(cosO + lsinO)

would be equal

Solution The equality z = w implies

Izl = Iwl and tan(argz) = tan(argw)

The first condition Izl = Iwl implies t2 = t2 + 0 2 , i.e., 0 2 = O Putting 0 = 0 in z and

w we obtain z = t and w = t for t an arbitrary real number

Example 1.12 Let C* be the set of alt eomplex numbers different from zero a) Prove that the set T of all eomplex numbers with modulus 1 is a multiplicative subgroup of the group (C*,.)

b) The multiplieative group C* is isomorphie with jR+ X T

rl '" r2 <===} rl - r2 = 2br, k is an integer Let i be the corresponding quotient set Prove that R+ x T isomorphie with R+ xi The group c· is isomorphie with

R+ x i (check) Therefore by b) and transitivity of the isomorphisms of groups it follows c)

Example 1.13 Find the sum of complex numbers which are the vertices of a

n-polygon in circle with radius r with the center in (0,0) for n = 4,6, , 2p, p E F:!,

Example 1.14 Prove that the condition Jor Jour points Zl,Z2,Z3 and Z4 be quent vertices oJ a parallelogram is the Jollowing

conse-Solution The equality Zl - Z2 + Z3 - Z4 = 0 implies

Exercise 1.15 For three given complex numbers Zl, Z2 and Z3, find Z4 such that the corresponding points in the complex plane will be the corners oJ a parallelogram

Rint Complex number Z4 can be obtained by Example 1.14:

Example 1.16 Starting with a complex number Z i- 0, find where the complex

numbers 2z,3z, ,nz are?

Solution For Z = p( cos () + z sin ()) we have

nz = np( cos () + z sin ()), the complex numbers nx, for n = 1,2, are on the half straight line y = tan () x

with modulus np

Example 1.17 Find the length oJ a side oJ a pentagon and the length oJ its diagonal

iJ it is inside the unit circle

Solution We have Zl = 1 Then

2 sin 2;, Figure 1.13 Find the length of

Exercise 1.18 Find where the points Z are for a fixed Zo :

a) Iz-zol=l; b) Iz + zol = 1; c) arg (z zo) = 4' ~

d) arg (z - 1) = -; 7r 4 e) 7r 4 < arg (z + z) ::; 4; 7r f) larg (z - zo)1 < 0;

g) Re (zo z) = 0, for Zo E li; h) III < r; i) Re(zz) = !;

j) - = z z

z

1.1 ALGEBRAIC PROPERTIES 17 Answers a) The closed disc with center at (1,0) and radius r = 2, Figure 1.16

e) The region between the straight half-lines

Y = x-I, y = ffx -1 ancl Y 2 1 inclucling the last straight half-line, Figure 1.20

r

-I

Figure 1.20

1.1 ALGEBRAIC PROPERTIES 19 f) For Zo = a + zb the points z :

-0 < arg (z - zo) < 0 are in the region between straight half-lines

y - b = (x - a) tan 0 and y - b = (x - a) tan( -0), x ~ a

and y ~ b (y ::; b for the second), 0 E [0, i-l, and x ~ a and y ~ b (y ::; b for the

second), Figure 1.21

~Y ,

Figure 1.21

g) The points z are on the imaginary axis fOT Zo #-O FOT Zo = 0 the points z are arbitrary complex numbers (Examine the case Zo E C; Y = kx, k = ~~:O)

h) From I!I < r we obtain Izl > ;, the points z are outside of the cirele Izl = ~

i) The points z are on the straight li ne y = -i

j) Using the Euler representation z = pe'tp we obtain from ~ = z that

and give a geometrie interpretation

Solution Since u u = lul2 we can easily obtain the desired equality Narnely,

(u + v)· (u + v) + (u - v)· (u - v) = 2u· u + 2v· v,

where we have used :u:E""V = u ± v

The geornetrical interpretation: the surn of squares of diagonals of a square is equal to the surn of the squares of its sides

Exercise 1.21 Let t be a real parameter ZI = (PI, 81 ) and Z2 = (P2' 82 ) fixed complex numbers Which curves are given in the complex plain by the following relations?

a) The circle with the center 0 and radius r = 2

b) The half straight line y = V3x, y ~ o

c) The circle with the center Z and radius r = 1

d) The straight half-line y = x, y ~ 0 translated for PIon the half straight line

1.1 ALGEBRAIC PROPERTIES

y = tan ()I X, Y ~ 0, Figure 1.22

o

Figure 1.22

e) The circle with the center ZI and radius r = P2'

f) The line segment [zt, Z2], Figure 1.23

ZI = PI( COS ()I + z sin ()I), Z2 = P2( tos ()2 + z sin ()2)'

In particular, for ZI = Z2 we obtain

x

x = 2PI COS ()I COS t and y = 2PI sin ()I cos t, 0::; t < 211"

21

h) The curve is given by the following parametrie equations

x = t + cos t, Y = sin t, t > 0, Figure 1.24

1.1 ALGEBRAIC PROPERTIES

for k = 0,1,2,3,4

c) We have

lJlO(cos arctan(-1/3) + zsinarctan(-1/3))

417fi10 ( arctan(-1/3)+2k1r arctan(-1/3)+2k1r)

Write down in all examples all cases in the form a + zb, a, b E ~

Example 1.23 Solve the following equations in C :

a) x8 - 16 = 0; b) x 3 + 1 = 0; c) x 6 + z + 1 = O

Solution

a) The zeroes of the equation x 8 - 16 = 0 are the values of m,

m = y'16( cos 0 + z sin 0) = h( cos k: + z sin k47r),

571"

for k = 0,1, ,5

Example 1.24 Using the equality cos t = sin t = y.} find:

a) cos TI;; b) sin f6-

Solution

a) Starting from the equality

we obtain

2 z z d

sm z = sm 2 cos 2 an cosz = cos 2 2 -Z sm • 2 2' Z

Since

cos i + zsin i E {z Ilzl = I},

we have cos2 t + sin2 ~ = 1 Putting this in the second identity we obtain

cos ~ = J cos z + 1

Applying the last formula two times on cos ~ we obtain

b) In this case we start from the equality

Example 1.26 Find the position 0] the vertices 0] the triangle with equal sides i]

the two vertices are -1 and 2 + t, Figure 1.25

Solution First we shall show that for any triangle with equal sides we have

where Zl, Z2, Z3 are the vertices of the triangle with equal sides We have (see Figure 1.25)

Z2 - Zl = e 1r·/ 3 (z3 - zt} and Zl - Z3 = e 1r·/ 3 (z2 - Z3)

Dividing these two equalities we obtain

Since Zl and Z2 are known we can find Z3 by the last equality (we obtain two tions)

solu-Exercise 1.27 Find the following sums:

k=l

sin n tl cos T

c)

cos~

1.1 ALGEBRA1C PROPERT1ES 27 Exercise 1.28 Find the following sums for real eonstants m and n 1= 2k7r

Exercise 1.29 Solve the equation z = zn-l (n is a natural number)

Exercise 1.30 Let m and n be integers Prove for z 1= 0 :

a) that (y'Z)m has ~ different values where (m, n) is the greatest eommon

~n,mJ

divisor of the numbers m and n

b) That the sets of values of (y'Z)m and yrzm are equal, i.e.,

(y'Z)m = ;:;zm if and only if(n,m) = 1, i.e., n and m has no non-trivial eommon divisors

Exercise 1.31 Prove the identity

11 - ZlZ212 - IZI - z212 = (1 - IZlI2) (1 - IZ212)

Exercise 1.32 Prove the inequality

Exercise 1.33 Find the vertices of regular n- polygon if its center is at Z = 0 and one vertex is known

Answer The vertices are

b) 1f Zl + Z2 + Z3 + Z4 = 0 and IZII = IZ21 = IZ31 = IZ41 = 1, then the points

Zl, z2, Z3, Z4 are either vertiees of a triangle with equal sides or they are equal in pairs

By the given conditions: Zl + Z2 + Z3 = 0 and IZil = 1, i = 1,2,3, we have

IZ3 - zll2 = 12z1 + z21 2 = (2z1 + z2)(2Z1 + Z2) = 5 + 2(ZlZ2 + ZlZ2),

and analogously

IZ3 - z212 = 12z2 + zll2 = 5 + 2(ZlZ2 + ZlZ2)'

Therefore IZ3 - zll = IZ3 - z21·

We can prove in a quite analogous way that IZ2 - zll = IZ3 - z21 Therefore the tri angle with vertices Zb Z2, Z3 is with equal sides

Example 1.35 Let the points Zb Z2, , Zn be on the same side with respect 0/ a straight line wh ich cross (0,0) Prove that the points

, , ,

Zl Z2 Zn have the same property and that Zl + Z2 + + Zn # 0, and that

it is obvious that Re Zk > 0 and Re lk > 0 for all k, which implies the desired properties

Exercise 1.36 Solve the equation (1 + ~) 3 = z

Example 1.37 Prove that the field 0/ complex numbers is the smallest field which contains the field 0/ real numbers and the solution 0/ the equation x2 + 1 = O Solution We will prove the desired result by reductio ad absurdum Suppose the contrary, that there exists a field T which contains IR, the solution of the equation

x 2 + 1 = 0 and is smaller than the field C, C :J T and C # T Then there exists

Zo = a + zb, z E C, such that a, b E IR and Zo (j T On the other side, since T contains

the solution of the equation x 2 + 1 = 0 we have z E T Therefore by T :J IR and the fact that T is a field we have x + yz E T for every x, y E IR Hence Zo ETaIso, which

is a contradiction

1.1 ALGEBRAIC PROPERTIES 29

Example 1.38 Prove that there does not exist a total order in the field of complex

numbers whieh is compatible with the opemtions in this field and which extend the usual order of reals

Solution Suppose that there exists a total order ~ in the field C We shall

compare z = z and z = O Suppose that z ~ O Then Z2 ~ 0, -1 ~ O Contradiction Suppose now that z ~ O Then 0 = z - z ~ i Multiplying both sides by -z (-z ~ 0),

we obtain (-z)(-z) ~ 0, i.e., -1 ~ O Since we have obtained contradiction in both cases, we condude that such a total order ~ can not exist

Example 1.39 Let p be the eommutative ring of all polynomials with real eients endowed with the usual addition and multiplieation

eoeffi-Let J be a ideal ofthe elements oftheform (1+x 2 )Q(x), where Q is a polynomial,

in the ring P We define in P an equivalenee relation'" in the following way:

Prove that

a) the set of all polynomials of first order with respeet to + is isomorphie with the set PI J of all equivalenee classesj

b) PIJ is a fieldj

c) the field PI J is isomorphie with the field of alt eomplex numbers C

Solution a) Each polynomial P from P can be written in the polynomial form

P(x) = (x 2 + 1)Q(x) + ax + b (x E lR)

for some a, b E lR Therefore an equivalence dass from PI J has the form J + ax + b

for some a, b E lR The addition + in PI J is defined by

(J + (ax + b)) + (J + (ex + d)) = J + (a + e)x + b + d (1.3)

This implies the isomorphism between (PI J, +) and (PI, +), where pI is the set of all polynomials of the first order

b) By (1.3) the operation + in PI J is an inner operation The neutral element

is J + 0 and the inverse element of J + ax + b is the element J - ax - b The multiplication in PI J is given by

(J + (ax + b)) (J + (ex + d)) = J + ae(x 2 + 1) + x(be + ad) + bd + ae, i.e.,

(J + (ax + b)) (J + (ex + d)) = J + x(be + ad) + bd - ae (1.4)

b b' - a a' = 1

c) Comparing the usual operations in C :

(az + b) + (cz + d) = (ae)z + b + d (az + b) (cz + d) = (bc + ad)z + bd - ae

with (1.3) and (1.4), respectively, follows the isomorphism between (C, +,.) and

Example 1.40 Prove the isomorphisms:

a) the group (C \ {O},·) with the group of matrices of the following form

[_~ !], a, b E IR and a2 + b 2 =I 0

with the matrix multiplication;

b) The group of quaternions

with matrix multiplications

c) The group of quaternions (K \ {O},·) with the group of matrices of the second order

(u,v E C \ {O})

with matrix multiplication

1.1 ALGEBRAIC PROPERTIES

Rint For c) U se a) and b) in the deeomposition of the following matrix

where u = Wo + iWI and v = W2 + iW3

z-c

Example 1.41 Let Icl < 1 and w = _-

1 - cz What are the following sets in the z-complex plane:

Al = {zllwl < I}, A 2 = {zllwl = I}, and A 3 = {zllwl > I}?

31

Solution For set AI, the eondition Iwl < 1 is equivalent with the inequality

Iz - cl2 < 11 - czl\ i.e., (1 -lcI2)(lzI2 - 1) < O

Henee Al = {zllzl < I} In a quite analogous way we obtain

Rint Use the geometrie interpretation of the complex number

Exercise 1.44 Which curves are given by the following sets

a) {zllz - 111z + 11 = const},

b) {ziliz - 111 = const}?

z+l

Exercise 1.45 Prove the equality

(n - 2) E lakl 2 + I E akr E Elak +ail2

l::;k< i::;n