Classical Complex Analysis: a geometric approach (2 vols) [Illustrated] 981426122X, 9814271292

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Classical Complex Analysis A Geometric Approach — Vol.1

Classical Complex Analysis

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Classical Complex Analysis A Geometric Approach — Vol.l

I-Hsiung Lin National Taiwan Normal University, Taiwan

World Scientific NEW JERSEY

•

LONDON

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SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TAIPEI

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CHENNAI

Published by World Scientific Publishing Co. Pte. Ltd.

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CLASSICAL COMPLEX ANALYSIS — Volume 1 A Geometric Approach Copyright © 201 1 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

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To my wife Hsiou-0 and my children Zing, Ting, Ying and Fei

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Preface

Complex analysis, or roughly equivalently the theory of analytic functions of one complex variable, budded in the early ages of Gauss, d’Alembert and Euler as a main branch of mathematical analysis. In the 19th century, Cauchy, Riemann, and Weierstrass laid a rigorous mathematical founda¬ tion for it (see Ref. 47). Nourished by joint effort of generations of brilliant mathematicians, it grows up into one of the remarkable branches of exact science, and serves as a prototype or model of other theories concerned with generalizations of analytic functions such as Riemann surfaces, ana¬ lytic functions of several complex variables, quasiconformal and quasireg¬ ular mappings and complex dynamics, etc. Its methods and theory are widely used in many branches of mathematics, ranging from analytic num¬ ber theory to fluid mechanics, elasticity theory, electrodynamics and string theory, etc. Elementary complex analysis stands as a discipline to the whole math¬ ematical training. This book is designed for beginners in this direction, especially for upper level undergraduate and graduate mathematics majors, and to those physics (or engineering) students who are interested in more theoretically oriented introduction to the subject rather than only in com¬ putational skills. The content is thus selective and its level of difficulty should be then adequately arranged. Besides its strong intuitive flavor, it is the geometric (mapping) proper¬ ties, derived from or characterized the analytic properties, that makes the theory of analytic functions differ so vehemently from that of real analysis and so special yet restrictive in applications. This is the reason why I favor a geometric approach to the basics. The degree of difficulty, as a whole, is not higher than that of Ahlfors’s classic Complex Analysis (see Ref. 1). But I try my best to give as detailed and clear explanations to the theory as possible. I hope that the presentation will be less arduous in order to be more available to not-so-well-prepared or not-so-gifted students and be easier for self-study. Neither the greatest possible generality nor the most vii

viii

Preface

up-to-date terminologies is our purpose. Please refer to Ref. 9 for those purposes. I would consider my purpose fulfilled if the readers are able to acquire elementary yet solid fundamental classical results and techniques concerned. Knowledge of elementary analysis, such as a standard calculus course including some linear algebra, is assumed. In many situations, mathemat¬ ical maturity seems more urgent than purely mathematical prerequisites. Apart from these, the work is self-contained except for some difficult theo¬ rems to which references have been indicated. Yet for clearer and thorough understanding where one stands for the present in the whole mathematical realm and for the ability to compare with real analysis, I suggest readers get familiar with the theory of functions of two real variables.

Sketch of the Contents If one takes a quick look at the Contents or read over Sketch of the Content at the beginning of each chapter, then s/he will have an overall idea about the book. A complex number is not only a plane vector, but also carries by itself the composite motion of a one-way stretch and rotation, and hence, is a two-dimensional “number” . They constitute a field, but cannot be ordered. Mathematics based on them is the one about similarity in global geomet¬ ric sense and is the one about conformality in local infinitesimal sense. Chapter 1 lays the algebraic, geometric, and point-set foundations barely needed in later chapters. Just as one experienced in calculus, we need to know some standard ele¬ mentary complex-valued functions of a complex variable before complex dif¬ ferentiation and integration are formally introduced. It is the isolated-zero principle (see (3) and (4) in (3. 4.2.9)) that makes many of their algebraic properties or algebraic identities similar to their real counterparts. Owing to the complex plane C having the same topological structure as the Euclidean plane R2, their point-set properties (such as continuity and convergence) are the same as the real ones, too. It is the geometric mapping properties owned by these elementary functions that distinguishes them from the real ones and make one feel that complex analysis is not just a copy of the latter. In particular, the local and global single-valued continuous branches of argz are deliberately studied, and then, prototypes of Riemann surfaces are intro¬ duced. Chapter 2 tries to figure out, though loosely and vaguely organized,

Preface

ix

the common analytic and geometric properties owned by these individual elementary functions and then, to foresee what properties a general analytic function might have. A complex-valued function f(z), defined in a domain (or an open set) Q, is called analytic if any one of the following equivalent conditions is satisfied:

(1) f(z) is differentiable everywhere in Q (Chap. 3). f(z)dz = 0 for any triangle contained (2) f(z) is continuous in Q and Q and in (Chaps. 3 4). Q, f(z) can be expressed as a convergent (3) For each fixed point zq power series Y^=oan(z — zo)n in a neighborhood of zq (Chap. 5). An analytic function f(z) infinitesimally, via the Cauchy-Riemann equations, appears as a conformal mapping in case f'(z) 0 and f(z) can be interpreted as the velocity field of a solenoidal, irrotational flow (see (3.2.4.3)). Chapter 3 develops the most fundamental and important ana¬ lytic and geometric properties, both locally and globally, which an analytic function might possess. The most subtle one, among all, is that a function f(z) analytic at zq can always be written as f(z) = f^z^) + (z — zo)ip(z) where ^p{z) is another function analytic at zq. From this, many properties, such as the isolated-zero principle, maximum-minimum modulus principle and the open mapping property, inverse and implicit function theorems, can be either directly or indirectly deduced. This chapter also studies some global theorems such as Schwarz’s lemma, symmetry principle, argument principle and Rouche’s theorem and their illustrative examples. After proving homotopic and homologous forms of Cauchy integral the¬ orem, most part of Chap. 4 is devoted to the residue theorem and its various applications in evaluating integrals, summation of series, and the Fourier and Laplace transforms. Chapter 5 starts with various local power series representations of an analytic function and analytic continuation of power series which even¬ tually lead to the monodromy theorem. Besides power series represen¬ tation, an entire function can also be factorized as an infinite prod¬ uct of its zeros such as polynomials do, and meromorphic function can be expanded into partial fractions via its poles as rational functions. The most remarkable example is Euler’s gamma function T(z) and its colorful properties. Riemann zeta function is only sketched. The essen¬ tial limit process in the whole chapter is the method of local uniform

Preface

X

convergence. Weierstrass’s theorem and Montel’s normality criterion are two of the most fundamental results in this direction. Both are used to prove Picard’s theorems via the elliptic modular function. These classical theorems can also be obtained by Schwarz-Ahlfors’s lemma, a geometric theorem. Riemann mapping theorem initiated the study of global geometric mapping properties of univalent analytic functions, simply called univa¬ lent mappings. Schwarz-Christoffel formulas provide fruitful examples for the theorem. As a consequence, Chap. 6 solves the Dirichlet’s problem for an open disk, a Jordan domain and hence, a class of general domains via Perron’s method. This, in turn, is adopted to determine the canoni¬ cal mappings and canonical domains for finitely connected domains. Based on our intuitive and descriptive knowledge about Riemann surfaces of particularly chosen multiple-valued functions, scattered from Chaps. 2 to 6, Chap. 7 tries to give a formal, rigorous yet concise introduc¬ tion to abstract Riemann surfaces. We will cover the fundamental group, covering surfaces and covering transformations, and finally highlight the proof of the uniformization theorem of Riemann surfaces via available classical methods, even though most recently it admits a purely differential geometric one-page proof (see Ref. 17). Almost all sections end up with an Exercise A for getting familiar with the basic techniques and applications; most of them also have an Exercise B for practice of combining techniques and deeper thinking or applications; few of them have an Exercise C for extra readings of a paper or Appen¬ dices A, B, and C. As far as starred sections are concerned, see “How to Use the Book” below.

Features of the Book

(1) Style of writing. As a textbook for beginners, I try to introduce the concepts clearly and the whole theory gradually, by giving definite explanations and accompanying their geometric interpretations when¬ ever possible. Geometric points of view are emphasized. There are about 546 figures and many of them are particularly valid or meaningful only for complex variable, but not for reals. Most definitions come out nat¬ urally in the middle of discussions, while main results obtained after a discussion are summarized and are numbered along with important formulas.

Preface

xi

(2) Balance between theory and examples. As an introductory text or reference book to beginners, how to grasp and consolidate the basic theory and techniques seems more important and practical than to go immediately to deeper theories concerned. Therefore, there are suffi¬ cient amount of examples to practice main ideas or results. Exercises are usually divided into Part A and Part B; the former is designed to familiarize the readers with the established results, while the latter contains challenging exercises for mature and minded readers. Both examples and exercises are classic and are benefited very much from Refs. 31,52,58,60, and 80. What should be mentioned is that many exercise problems in Ref. 1 have been adopted as illustrative examples in this text. (3) Careful treatment of multiple-valued functions. Owing to historical and pedagogical reasons, complex analysis is conventionally carried out in the (one-layer) classical complex plane. Later development shows that the most natural place to do so is multiple-layer planes, the so-called Riemann surfaces or one-dimensional complex manifolds in its modern terminology. Multiple-valuedness is a difficult subject to most begin¬ ners and most introductory books just avoid or sketch it by focus¬ ing on y/z and log z only. To provide intuitive feeling toward abstract Riemann surfaces in Chap. 7, Chaps. 2-6 take no hesitation to treat multiple-valued analytic functions whenever possible in the theory and in the illustrative examples. Once the troublemaker arg z, the origin of multiple-valuedness, is tamed (see Sec. 2.7.1), what is left is much easier to handle with. Also we construct many (purely descriptive and nonrig¬ orous) Riemann surfaces or their line complexes of specified multiple¬ valued functions, merely for purposes of clearer illustration, wherever we feel worthy to do so. (4) Emphasis on the difference between real and complex analyses. The origin of all these differences comes from the very character of what a compfex number is (see the second paragraph inside the title Sketch of the Contents). This fact reflects, upon differentiating process, in the aspects of algebra, analysis as well as in geometry (see (3.2.2) for short; Secs. 3.2.1, 3.2.2, and 3.2.3 for details). (5) Paving the way to advanced study. The contents chosen are so arranged that they will provide solid background knowledge to further study in fields mentioned in the first paragraph of this Preface. Besides, the book contains more materials than what is required in a Ph.D. qualifying examination for complex analysis.

xii

Preface

How to Use the Book (A Suggestion to the Readers)? The book is rich in contents, examples, and exercises when comparing to other books on complex analysis of the same level. It is designed for a variety of usages and motivations for advanced studies concerned. The whole content is divided into two volumes: Vol. 1 contains Chaps. 1Appendix A, while Vol. 2 contains Chaps. 5-7 plus Appendices B plus 4 and C. I may have the following suggestions for different proposes: Chapters 1 and 2 are preparatory. Except those basic concepts such as limits and functions needed, topics in these two chapters could be selective, up to one’s taste.

(1) As an introductory text for undergraduates. Sections 2.5.2, 2.5.4, 2.6, 2.7.2, and 2.7.3 (sketch only), Sec. 2.9 (sketch only); Secs. 3.2.2, 3.3.1 (only basic examples and y/z, logz), Secs. 3.3.2, 3.4.1-3.4.4, 3.5.1-3.5.5, 3.5.7 (sketch only); Secs. 4.8, 4.9, and 4.10 (sketch only), Sec. 4.11 (sketch only), Secs. 4.12.1-4.12.3; Secs. 5.3.1, 5.4.1, and 5.5.2 (optional and sketch). As a whole, Examples and Exercise A should be selective. Minded readers should try more, both Examples and Exercises, and pay atten¬ tion to more elementary multiple-valued functions and their Riemann surfaces, if possible. In a class, the role played by a lecture to select topics is crucial. (2) As a beginning graduate text. With a solid understanding of materials in (1), the following top¬ ics are added: Secs. 2.8; 3.4.5, 3.4.6; 4.1-4.7, 4.12.3A-4.15 (selective); Secs. 5.1.3, 5.2, 5.3.2, 5.5-5.6 (selective), Secs. 5.8.1-5.8.3; Chap. 6 except Sec. 6.6.4. Examples and Exercise A (even B) should be emphasized. Of course, the adding or deleting of some topics are still possible. (3) To readers who are interested in Riemann surfaces. Pay more attention to multiple-valued functions and their descriptive Riemann surfaces such as Secs. 2.7, 3.3.3, 3.4.7, 3.5.6; 5.1, 5.2, and end up with the whole Chap. 7. (4) Several complex variables. Sections 3.4.7, 3.5.6; Chap. 7.

Preface

xiii

(5) Quasiconformal mappings and complex dynamical systems. Section 3.2.3, Example 2 in Secs. 3.5.5; 5.3.4, 5.8; 6.6.4; Chap. 7 and Appendix C. (6) As a general reference book supplement to other books on complex analysis.

Acknowledgments The following students in Mathematics Department helped type my hand¬ written manuscript: Jing-ya Shui; Ya-ling Zhan; Yu-hua Weng; Ming-yang Kao; Wei-ming Su; Wen-jie Li; Shuen-hua Liang; Shi-wei Lin; D. C. Peter Hong; Hsuan-ya Yu; Yi-hsuan Lin; Ming-you Chin; Che-wei Wu; Cheng-han Yang; Kuo-han Tseng; Yi-ting Tsai; Yi-chai Li; Po-tsu Lin; Hsin-han Huang.

Yan-yu Chen graphed all the Figures appearing in this book. Yan-yu Chen, Aileen Lin, Wen-jie Li, and Ming-yang Kao helped edit the final manuscript for printing. Here may I pay my sincerest thanks to all of them. Without their unselfish dedication, this book definitely cannot be published so soon. Also, teaching assistant Jia-ming Ying helped improve and correct par¬ tially my English writing. My colleagues Prof. Tian-yu Tsai proof-read the entire manuscript, and Prof. Yu-lin Chang proof-read Chaps. 5-7 and adopted parts of the content in his graduate course on complex analysis. Both of them pointed out many mistakes and gave me valuable sugges¬ tions. It would be my pleasure to express my gratitude toward them for their kindest help. As usual, teaching assistant Ching-yu Yang did all the computer work for the several editions of the manuscript. And Ms. Tan Rok Ting, an editor in World Scientific, copyedited the whole book with carefulness and expertise. Thank them so much. I-hsiung Lin 21 January 2009 Taipei, Taiwan

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Contents

Preface

v

Volume 1 Chapter 1. Complex Numbers Sketch of the Content 1.1. 1.2. 1.3. 1.4.

1 1

How to Visualize Geometrically the Existence of the So-Called Complex Numbers in Our Daily Life Complex Number and Its Geometric Representations Complex Number System (Field) C Algebraic Operations and Their Geometric Interpretations (Applications) Conjugate complex numbers 1.4.1. Inequalities Applications in (planar) Euclidean geometry .... Steiner circles and symmetric points with respect to a circle (or line) De Moivre Formula and nth Roots of Complex Numbers ... Spherical Representations of Complex Numbers: Riemann Sphere and Extended Complex Plane Complex Sequences Elementary Point Sets Completeness of the Complex Field C

1.4.2. 1.4.3. 1.4.4. 1.5. 1.6. 1.7. 1.8. 1.9.

Chapter 2. Complex- Valued Functions of a Complex Variable Introduction Sketch of the Content 2.1. 2.2. 2.3.

2 8 15

25 25 30 35 52 57 66 76 86 97

104 104 106

Limits of Functions 107 Continuous Functions 110 Uniform Convergence of a Sequence or Series of Functions . . 123 XV

Contents

xvi

2.4. 2.5.

Curves Elementary Rational Functions Polynomials 2.5.1. The power function w = zn (n 2) 2.5.2. Rational functions 2.5.3. Linear fractional (or bilinear or Mobius) 2.5.4. az + b „ . transformations w =

2.6.

2.7.

2.8. 2.9.

cz + a Joukowski function w = (2 + j) 2.5.5. Elementary Transcendental Functions The exponential function ez 2.6.1. Trigonometric functions cos z, sin z, and tan z .... 2.6.2. Elementary Multiple-Valued Functions The origin of multiple-valuedness: arg z 2.7.1. 2.7.2. w = y/z (n 2) and its Riemann surface (etc.) . . . 2.7.3. w = logz (the natural logarithm function with base e) and its Riemann surface 2.7.4. w = cos-1 z and w = tan-1 z and their Riemann surfaces Differentiation in Complex Notation Integration in Complex Notation

129 142 142 148 152 157

175 185 186 191 200 204 214 241

258 272 281

Chapter 3. Fundamental Theory: Differentiation, Integration, and Analytic Functions

314

Introduction Sketch of the Content

314 314

3.1. 3.2.

3.3.

(Complex) Differentiation Differentiability: Cauchy-Riemann Equations, their Equivalents and Meanings 3.2.1. (Linearly) Algebraic viewpoint 3.2.2. Analytic viewpoint 3.2.3. Geometric viewpoint *3.2.4. Physical viewpoint Analytic Functions 3.3.1. Basic examples 3.3.2. The analyticity of functions defined by power series *3.3.3. Analyticity of multiple-valued functions and the Riemann surfaces (revisited)

317

325 328 329 334 338 345 347

362 372

Contents

3.4.

3.5.

Analytic Properties of Analytic Functions 3.4.1. Elementary properties derived from definition .... 3.4.2. Cauchy integral theorem and formula (simple forms) : The continuity (analyticity) of the derivative of an analytic function and its Taylor series representation 3.4.3. The real and imaginary parts of an analytic function: Harmonic functions 3.4.4. The maximum-minimum principle and the open mapping property 3.4.5. Schwarz’s lemma 3.4.6. The symmetry (or reflection) principle *3.4.7. The inverse and implicit function theorems Geometric Properties of Analytic Functions 3.5.1. Local behavior of an analytic function at a point: Conformality, etc 3.5.2. The winding number: Its integral representation and geometric meaning 3.5.3. The argument principle 3.5.4. The Rouche’s theorem 3.5.5. Some sufficient conditions for analytic functions to be univalent *3.5.6. The inverse and implicit function theorems

3.5.7.

xvii

381 381

387

435 450 464 479 493 498 498

524 531 550

558

(revisited) 579 Examples of (univalently) conformal mappings . . . 609

Chapter 4. Fundamental Theory: Integration

(Advanced) Introduction Sketch of the Content

4.1. 4.2.

Complex Integration Independent of Paths: Primitive Functions The General Form of Cauchy Integral Theorem: Homotopy . The line integral of an analytic function 4.2.1. along a continuous curve Homotopy of curves 4.2.2. Homotopic invariance of the winding numbers . . . 4.2.3. Homotopic form of Cauchy integral theorem .... 4.2.4.

639 639 639

641 645 646 651 655 657

xviii

Contents

The General Form of Cauchy Integral Theorem: Homology . Cycles and homology of two cycles 4.3.1. Simply and finitely connected domains: 4.3.2. Homology basis Homologous form of Cauchy integral theorem .... 4.3.3. Artin’s proof 4.3.4. 4.4. Characteristic Properties of Simply Connected Domains (a Review): The Single-Valuedness of a Primitive Function 4.5. The Branches of a Multiple-Valued Primitive Function on a Multiple-Connected Domain 4.6. The General Form of Cauchy Integral Formula 4.7. Integrals of Cauchy Type and Cauchy Principal Value .... 4.8. Taylor Series (Complicated Examples) 4.9. Laurent Series The Laurent series expansion of an analytic function 4.9.1. in a (circular) ring domain Examples 4.9.2. 4.10. Classification and Characteristic Properties of Isolated Singularities of an Analytic Function 4.10.1. Removable singularity 4.10.2. Pole 4.10.3. Essential singularity 4.11. Residues and Residue Theorem 4.11.1. Definition of residues 4.11.2. The computation of residues and examples 4.11.3. The residue theorem 4.12. The Applications of the Residue Theorem in Evaluating the Integrals 4.12.1. /(cos#, sm9)d0, where f(x,y) is a function in x and y. etc xa~ 2(1 — x)~a f (x)dx (0 < a < 1), etc 4.12.2. 4.12.3. Improper integrals over (—oo, oo) 4.12.4. Improper integrals over (0, oo) 4.12.4.1 f^ffxjdx 4.12.4.2 f(x)eimxdx (m G R) f(x)dx (with periodic /(a:)) 4.12.4.3

4.3.

661 662 666 669 673

678

683 697 701 718 733 736 743

760 761 767 785 797 798 806 826 845

846 858 869 910 910 921 941

Contents

xix

4.13. The Integral 951 /(z^dz along a Line Re 2 = xq 4.13.1. Fourier transforms 957 4.13.2. Laplace transforms 969 4.14. Asymptotic Function and Expansion of Functions Defined by Integrals with a Parameter 994 4.15. The Summation of Series by Residues 1007 4-15.1. E^n) 1010 4.15.2. EEc(-l)n/N 1027 Appendix 1036 Appendix A. The Real Number System R 1036 References 1039

Index of Notations

1043

Index

1049

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CHAPTER 1

Complex Numbers

This opening chapter is to introduce the basic knowledge about the complex numbers in three aspects and the interactions among them.

Sketch of the Content Algebraic: The imaginary solution i = of the equation t2 + 1 = 0 creates the complex numbers z = x + iy which obey the same basic laws of arithmetics as the reals do (Sec. 1.3). The main distinction between them lies on the fact that the complexes cannot be ordered, but compensated by the conjugate operation z —> z (Sec. 1.4.1). This enables us to interrelate both by the operation |z|2 = zz and to introduce the inequalities among various |z| (Secs. 1.4.1 and 1.4.2). An instant consequence is that every polynomial with complex coefficients is always solvable; in particular, zn — 1=0 has exactly n-distinct roots (Sec. 1.5).

Geometric: z = x + iy can be understood not only as a point or a vector in the Euclidean plane R2, but also as a planar motion composed of one¬ way stretch and rotation (Secs. 1.1, 1.2, and 1.4). It comes naturally that z = x+iy can be represented as the point (x, y) in R2 which is thus renamed as the complex plane C, or in the polar form reie (r = \z\,0 = argz) with the origin 0 as pole and the positive a;-axis as the polar axis. And then, they can be used effectively to describe planar geometric objects or to solve geometric problems (Secs. 1.4.3 and 1.4.4). Therefore, almost every algebraic operation about complex numbers has its illuminative geometric meaning. Topological (point-set properties): Owing to |rr|, |y \z\ = \/x2 + y2 (z concepts sequences the of real and series and x limit iy), = + + 1^1 M their properties can be carried verbatim over to the complex ones except the one appeared in (3) of (1.7.3), where arg zn —> arg zq needs to be treated carefully. So do the concepts of open, closed, compact, and connected sets for both R2 and C (Sec. 1.8). The addition of the point at infinity oo to C 1

Chap. 1. Complex Numbers

2

to obtain the extended complex plane C*, realized as the Riemann sphere S in R3, seems more naturally for the need of the limit concept than the algebraic operations (Secs. 1.6 and 1.9). The most importance of all is that C is a complete metric space in which every Cauchy sequence is convergent (Sec. 1.9). 1.1.

How to Visualize Geometrically the Existence of the So-Called Complex Numbers in Our Daily Life

As already well known, man creates

(1) the natural number system N for counting; (2) the integer number system Z for the negative of a quantity; (3) the rational number system Q for fractions of a whole quantity or for measurement such as length, and

(4) the real number system R for measurement (see Appendix A). These numbers are vivid in our daily life. Though we cannot see what irrational numbers such as e and tt would exactly look like, nowadays we are able to approximate them as accurately as we want by rational numbers. T, where a, b R, the so-called complex Numbers of the form a + numbers, appeared as early as 16th century when mathematicians tried to solve quadratic equations. Man imagines that there exists a “number” i satisfying x2 + 1 = 0. This i denotes 1. As a consequence, a quadratic equation ax2 + bx + c = 0 with a, b, c e R and b2 — 4ac < 0 would have two roots They are not real but only imaginary. They seem so freaky because man does not even know how to approximate them by known numbers from the well-established number systems. It was C. Gauss and I. Argand who interpreted the numbers a + bi (also denoted as a + ib) geometrically as the point (a, b) on the plane and hence, laid a firm basis for the development of the complex function theory. For historical account, see Ref. [47]. In what follows, we assume the readers are familiar with basic knowledge about the Euclidean plane R2. Fix the rectangular rry-coordinate system on the plane R2. The point (x, y) can be considered as the position vector from (0, 0) pointed to (x, y) itself. See Fig. 1.1. Under this circumstance, vector operations are applied to (x, y) as follows to form the two-dimensional real space R2:

(1) Addition Cci,3/i) + (x2,y2) = fa +x2,yi + y2)

Sec. 1.1. How to Visualize Complex Numbers in Our Daily Life

3

satisfying the following properties:

(i) (commutative) ($1,3/1) + (x2, y2) = ($2,3/2) + ($1,3/1); (ii) (associative) (($1, + (x2, 3/2)) + ($3, Vs) = ($1, Z/i) + (($2, Z/2) + (^3,2/3)); (iii) (zero vector) (x, y) + (0,0) = (x,y); (iv) (inverse vector) For each (x,y),

(x,y) + (-x, —y) = (0,0). (2) Scalar multiplication

a(x,y) = (ax,ay), a

R

satisfying:

(i) Kx,y) = (x,y\ (ii) (a/3)($, y) = aWx, y)\ a,(3

R.

(3) Distributive law

(a + (3)(x,y) = a(x, y) + /3(x,y), a,/?eR;

(1-1-1)

a((zi,yi) + ($2,2/2)) = a(zi,yi) + o‘(x2,y2)Owing to lack of product operation among vectors, conceptually it is not enough to identify fully the imaginary number x + iy (also denoted

Chap. 1. Complex Numbers

4

alternatively as x + yi) with the point (x, y) or the vector it induces. The point is that we have to define what the product

(xi,yi)(x2,y2') means properly so that it still represents a number a + bi and satisfies nice operational properties such as commutative and associative laws, etc. Hence, we designate the notation z = x + yi, where x, y R and are not both equal to zero. O

(1) z is the point (x,y) or the vector (x,y) in R2, and (2) z represents the one-way stretch along the z-axis with scale factor r = y/ x2 +y2 and then followed by a rotation with center at (0, 0) through an angle in the counterclockwise direction so that the rr-axis will coincide with the line generated by the

vector (z,y).

(1-1-2)

Note that in condition 2, we can perform rotation first and one-way stretch second. Both are commutative. See Fig. 1.2. In particular, i = (0,1) = 0 + H,

(1.1.3)

represents the counterclockwise rotation of 90° of the point (1,0) or any nonzero vector. Let i2 = i i denote two such consecutive motions, etc.

(0,0)

(1,0)

Fig. 1.2

(r,0)

Sec. 1.1. How to Visualize Complex Numbers in Our Daily Life

5

Then (see Fig. 1.3)

i2 = (-1,0) = -(1,0) = -1, i3 = (0,-1) = -(0,1) = —i, Z4 = (1,0) = 1 + «0 = 1.

(1.1.4)

0 = 0 + i0,

(1.1.5)

Finally, we designate

where the two zeros 0 on the right are the zero number in R. Hence, it is reasonable to define 2

= z + iy = 0 = 0 + i0 4^a: = 0 and y = 0.

Also, we define, for z

(1.1.6)

0,

the modulus or absolute value \z\ =

^/a;2 + y2 = r,

and the principal argument Argz = 0 shown in Fig. 1.2.

(1.1.7)

Let z\ = Xi + iyi and = x^ + iy?. We try to define the product Z1Z2 properly. Knowing what z± means as in (1.1.2), we define, assuming zi 0

Chap. 1. Complex Numbers

6

and ^2

7^ 0,

Z1Z2

= the point obtained by one-way stretch along the vector

z\

with scale factor |^2 1 and then followed by a counterclockwise

(1.1.8)

rotation through the principal argument ArgZ2 of z2.

See Fig. 1.4. To pinpoint the coordinate of

Z1Z2,

let 0j = Argzi and

02 = Argz2 for

simplicity. Then ziz2 has coordinate

|*1| |^2|(cos(0i + 02),sin(0i +6*2)), where

cos(0i + 02) = cos0i cos 02 - sin#!sin02 =

h

7—

N

•

|*2 1

1*11 1*2 1

and

+ |*i| 7^7. 7^7 7^7 |*2| 1*2 | 7—7

sin(0i + 02) = sin 0i cos02 + cos0i sin02 = |*i| Hence *1*2

•

•

= (^1^2 - yiV2, x^2 + 0:23/1) = ziz2 - yiy2 + *(^13/2 + ^23/1). (1.1.9)

Sec. 1.1. How to Visualize Complex Numbers in Our Daily Life

7

Note that, in Fig. 1.4, the triangle with vertices at points 0, 1, and z\ is similar to the triangle with vertices at 0, Z2, and ziz2. Suppose the symbol defined by (1-1-2) enjoys the operational properties such as commutative, associative, and distributive laws which we are used to in the real number system R, then (1.1.9) is nothing but the usual product as ziz2

= (^i + fyi)^ + iy^

= X1X2 + xi(ij/2) + (iyi)x2 + (iyfifiiyz) . = ^1^2 + 1X^2 + 22/1^2 +2 yih2 = X1X2 - yiV2 + ®(^lZ/2 + ^21/1), .9

•

by using (1.1.4). In case either zi

= 0 or

z2

= 0 or

zi

ziz2

= z2 = 0, we just define

= 0.

(1.1.10)

We call the symbol defined in (1.1.2) a complex number. It is not only a plane vector, but also carries with itself the composite motion of one¬ way stretch and rotation, and hence, is a two-dimensional “numbe'd'' which cannot be approximated by any number we knew before. Mathematics based on the complex numbers is, geometrically, the one about similarity in global sense and is the one about conformality in local sense via limit processes. Exercises A

(1) Let z = x + iy

0. Try to use (1-1.2) and Fig. 1.2 to show that the unique w satisfying zw = 1 is

x

— iy

x2 + y2 which is denoted as

z-1 or

(2) Try to locate the following complex numbers: l-2i;

jl-;

(3 + 20(1-0;

(3) Try to use (1.1.2) to show that the product operation satisfies (i) the commutative law ziz2 = z2zi; (ii) the associative law (ziz2)zs = zi(z2zs); and (iii) the distributive law zi(z2 + Z3) = ziz2 + Z1Z3. (4) Let zi, z2, and Z3 be three complex numbers, noncollinear when consid¬ ered as points. Fix any zq and let wi = zqZi, W2 = zqz2, and

= Z0Z3.

Chap. 1. Complex Numbers

8

are noncollinear. (a) Show that wi, W2, and (b) Graph the triangles Azi 22^.3 and AW1W2W3. Compute their corre¬ sponding angles and areas.

Exercises B

(1) Try to design a kind of chess game based on the idea shown in (1.1.2). 1.2.

Complex Number and Its Geometric Representations

Section (1) The imaginary unit i Suppose x is a real number. Then x2 0 and hence, x2 + 1 1 > 0 holds. Hence the equation t2 + 1 = 0 does not have any solution in R. By imagination, suppose there exists a “number” , denoted by z=

x/^l

(1.2.1)

satisfying that the “product” of i with itself is —1, namely,

z2 = -l. Then z2 + 1 = 0 holds and z is a solution of z. We call z the imaginary unit.

—

x2 + 1 = 0. Another solution is

Section (2) The complex number

We do formal “addition” and “multiplication” of two real numbers x, y and i into the symbol, denoted as z = x + iy or

x + yi,

(1-2-2)

and is called a complex number formed with the real part Re 2 = x, and

the imaginary part Imz

= y.

(1.2.3)

For example, 1 + Oz, 0 + 2z, %/2 + |z, etc. are complex numbers. A complex number z = x + iy with its Im z = y = 0 is specifically denoted as x + zO = x,

(1-2-4)

and is considered as real number in many occasions. For example, -l + 0z = -1.

Sec. 1.2. Complex Number and Its Geometric Representations

9

A complex number z = x + iy with its Re z = x = 0 is denoted as (1.2.5)

0 + iy = iy,

and is called a pure imaginary. For example, 0 + 2f number x + iy is said to be imaginary if y / 0. In particular, only the zero complex number 0 = Of

= 2i. Hence, a complex (1.2.6)

= 0 + Of,

is both real and pure imaginary.

Section (3) As a point in the Euclidean plane R2 Fix a rectangular ^-coordinate system in the plane. We designate the point (x, y) as the complex number z = x + iy and vice versa. This sets up a oneto-one and onto correspondence between complex numbers and points in the plane which is thus called a complex plane. In particular, real numbers x are in one-to-one and onto correspondence with points in the x-axis, hence called the real axis; pure imaginaries iy and the points y in the y-axis are in one-to-one and onto correspondence and hence the y-axis is called the imaginary axis. See Fig. 1.5. Henceforth, no distinction will be made between the set of all complex numbers

C = {x + iy | x, y and the complex plane, also denoted as C.

Fig. 1.5

R}

(1-2-7)

Chap. 1. Complex Numbers

10

Section (4) As a plane vector pointed from (0, 0) to (x, y)

z = x + iy can also be considered as a plane vector pointed from (0, 0) toward the point (x,y). In this case, x and y are orthogonal projections of the vector on the real and imaginary axes, respectively. See Fig. 1.5. The length r of the vector, denoted as

\z\ = r = y/x2 + y2,

(1.2.8)

is called the modulus or the absolute value of the complex number z. Note

that |z| 0, and |z| = 0 z = 0. In the case z = x + iy 0, we call the angle, denoted as

(1.2.9)

argz,

between the vector z and the positive direction of the real axis an argument of z. Usually, we define arg z

> 0 if arg z is obtained by counterclockwise rotation

and argz < 0 if obtained by clockwise rotation. Figure 1.6 shows that argz is multiple-valued. In the case z

(1.2.10)

= 0, argz

is not defined. We summarize the above as

The multiple-valuedness of argz (z 0). The value of argz that lies on — 7T < 0 < tv (or 0 < 0 < 2tt) is called the principal values of argz or the principal argument of z and is denoted as

Argz.

Fig. 1.6

Sec. 1.2. Complex Number and Its Geometric Representations

11

Then

arg z = Arg z + 2mr,

n = 0, ±1, ±2, . . . .

(1.2.11)

The argument arg z is the main trouble-maker in generating multiple-valued functions in complex analysis. For a detailed treatment, see Sec. 2.7.1. We illustrate an

Example 1. For each of the following z, compute |z|, Argz and argz.

(1) x E R and x 0; (2) iy, y C R and y 0; (3) (4) +

4

^-

Solution. (1) If x > 0, |rr| = x, Argx = 0 and argx = 2n7r, n x < 0, |x| = —x, Arg a: = ir and argx = it + 2nzr, n Z; (2) If y > 0, |zy| = y, Arg(zy) = and arg(zy) = f + 2mr, n y < 0, Arg(zy) = f, and arg(zy) = — + 2n?r, n G Z; (3) |z| = x/2, Argz = and arg.z = + 2nit, n Z; Argz argz and (4) |z| = 1, = = + 2rm, n G Z.

—

^

Z; if Z; if

—

Section (5) Trigonometric or polar representation Consider the origin (0, 0) as the pole and the positive x-axis as the polar axis in a polar coordinate system. Then, the modulus |z| and the argument arg z of a complex number z are, respectively, the polar radius r and polar angle 0 of the vector z in the polar coordinate system (see Fig. 1.5): x = Rez = r cos#,

y = Imz = r sin#, and then,

z = x + iy = r(cos# + zsin#),

—

where r = y/x2 + y2 and # = tan-1 , x

(1.2.12) which is called the trigonometric or polar form of z. For simplicity, we introduce the Euler’s formula

eie = cos # + z sin #, # G R,

(1.2.13)

which is to be justified in Sec. 2.6.1. Then (1.2.12) can be rewritten as the concise form re18 .

Chap. 1. Complex Numbers

12

Remark (The relation between Arg z and tan-1 ^). Suppose Argz tt and f < Arc tan , the principle value of tan-1 Then

—

a Arc tan

—, — y

z in the first or the fourth quadrant,

X

Arg z =

a Arc tan


0 or i < 0 holds. Both imply that

i2 = -1 > 0, => (Multiply both sides by —1 > 0)1 > 0; while (Add both sides by 1 > 0) 0 > 1, which is a contradiction. When computing with complex numbers, one often needs the circular operational properties of i: z1 = i, i2 = —1, i3 = — i, and z4 = 1. In general,

^n+i =

-4n+2

=

-4n+3

=

^n+4 =

n&z

We give three examples. Example 1. Write each of the following complex numbers z in the form x + iy.

(^)2- (2)Hrfer-

(i) (l-z)n, n€Z.

(3) (i+e + (i-i)”,«ez. (4)(i+z)n

—

Chap. 1. Complex Numbers

18

Solution. (1) 2+? 3 — 2?

1

.. / 3

2

A

;V3 13/

4 13

7 . 13

=ife' (2) (1 - ?)5 = l5 + 5 • I4 • (-?) + 10 • I3 • (-?)2 + 10 • l2 • (-?)3

+ 5-i-H)4 + H)5 = 1 - 5? - 10 + 10? + 5 - ? = -4(1 - ?) and

(1 +?)5 = —4(1 + ?). Or,

(1 - ?)2 = —2?

(1 - ?)4 = —4 +> (1 - ?)5 = -4(1 - ?).

Hence,

(l-?)5-l _ (l+?)5 + l

”

=

— 4(1 — ?) — 1

-4(l+?) + l -5 + 4?

=

(5 — 4?)(3 — 4?)

3274!

1Z1 ~1 = 2S

“

(3) and (4), by (2), for m = 0, 1, 2, . . . (1 _ i)4m+l = ((1 _

_ •) (_4)"»(1 _ i). =

(1 _i^m+2 = (1 _ j)4«+3 = —2(—4)m(l + ?);

(1 _^4m+4 = and

(1+

= (_4)m(1 + 0.

(l+?)4m+2 = 2(-4)m?; (l+j)4m+3 = 2(-4)m(-l + ?); (l+i)4m+4 = (_4)”»+l. The final results follow easily.

32!)'

Sec. 1.3. Complex Number System (Field) C

19

Example 2. Let w

=-

(-1 + a/3z) .

(1) Show that w12 + w + 1 = 0 and w3 = 1; (2) Evaluate (a + bu + cw2)(a + 6w2 + cw) and (a + b) (a + baj^a + biv2). Solution. (1) By direct computation,

a,2 = -

1—

\/3i) =

—ai

—1

=> cu2 + w + 1 = 0. Also,

w3 = w2

w

(—1 —

=

1+

=

Or,

w3 = w2

•

w

= (— w

— l)cu = — w2 — w = 1.

^

= 1.

(2) By using (1), direct computation shows that (a + bw + au2)(a + bw2 + co?)

= a2 A b2 A c2 A ufi(tu2 A oj') + bc^cj^ A cu2) A cu(tu2 A cu) = a2 + b2 A c2 — ab — be — ca, and

(flA&)(aAfitu) (uA^xu2 )

= a3

-\-a2b^uj2 Atu A 1) -{-ab2 (cu2 Atu A 1) = g3A^-

By the way, how can one factorize a2 — ab + b2 over the complex field C? Just treat a2 — ab + b2 = 0 as a quadratic equation in a, and then its solutions are

6 ± Vb2 - 462 li^t, , , = 2 b = —bw or ow 2. 2 ab A b2 = (a A ba^fa A buj2). See Exercise B(2).

a=

Hence a2

—

—

Example 3. Solve the following equations, namely, try to find z = x A iy so that

(1) (square root) z2 = — 6 + a/5z; (2) (cube root) z3 = 1.

Chap. 1. Complex Numbers

20

Solution. (1) z2 = x2

— y2 + 2xyi. Hence

z2 = — 6 + \/5i, o x2 — y2 = —6 and

\/5, (z2 + y2)2 = (a:2 - y2)2 + (2sy)2 = 41, => (take the positive square root) x2 + y2 = x/41.

Solving this last equation with

x2 = x=

x2

2xy =

— y2 = —6, we get

(-6 + vOT) and y2 =|^6 + >/41) (—6 + >/41) and y = + a/41^ .

In appearance, we would get four solutions. Owing to the restrained condi¬ tion that xy > 0, we have only two solutions left, namely,

=> x3 — 3a;y2 = 1 and 3a:2y — y3 = y(3a:2 — y2) = 0. In case y = 0, then x3 = 1 and we choose the only real root x = 1. If 3a;2 — y2 = 0, then x3 — Sxy2 = x3 — 9a:3 = —8a:3 = 1 and we choose x= | — and hence, y2 = 3a:2 =|which, in turn, results in y = Therefore, z3 = 1 has solutions z = 1, w and w2 where w = |(—1 + v^z). Exercises A

(1) Express each of the following complex numbers z in the form z = x+iy, where x, y R, and then compute |z| and argz. (a) (4 + 3«)(4 + 2z)(3 - z)(l - «). (b)

^>/3-i)

.

( (1 + 2z)3 — (1 — z)3 ( ) ’ ( (z-l)(z-2)(i-3)(3 + 2z)3 — (2 + z)3 . . (a + ^)2 _ (a ^)2 a bi , , n R (f) + 77, a, be R. (g) (a — bi)2 (a + bi)2 ’ a — bi

v

’

(d)

~

Sec. 1.3. Complex Number System (Field) C

21

(2) Find the real and the imaginary parts of the following complex numbers z = x + iy. (a)

_1

— Z

z +1

.

(b)

1

z

. (c) z5. (d) z\ n G N.

(3) Let w = |(—1 + -\/3«). Compute: (a) (1 — w)(l — w2)(l — w4)(l w8). (b) (1 — w + w2)(l + w

—

— w2).

(c) (aw2 + &w)(aw + &w2). (d) (a + 6w + cw2)3 + (a + 6w2 + cw)3. (4) Let z = x + iy, x, y G R. Solve the following equations. (a)

z2 = — i. (b) z2 =

(e)

z2 + (6 + 7i) +

— (1

. (c) z3 = +

\

—

(d) z4 = —1.

+ 5i = 0. (f) z2 - (3 + ^z + (1 + 3z) = 0.

Exercises B

(1) Suppose a, b G R. Show that, in the complex field C,

if a

0, b

if a

0, b = 0,

0

if a < 0, b = 0,

—

^

where sgn i> = = 1 if 6 > 0, = 1 if & < 0 and the square root of a positive real number is chosen to be positive. (2) Suppose a, b, c G C, and a 0. Show that

,

/

az 2 +bz + c = alz + \

— b 2a

b+

+

4ac — b2

Vh2 — 4ac 2a

b

— y/b2 — 4ac 2a

where V^2 — 4ac is as in Exercise (1). Note that, as long as b2 — 4ac / 0, y/b2 — 4ac always have two values with positive and negative sign, respectively, and y/b2 — 4ac could be any one of them in the above expression. So the quadratic equation az2 + bz + c = 0 has exactly two

Chap. 1. Complex Numbers

22

solutions

—b ± y/b2 as in the real case. (3) Let w = |(—1 + y/3i). Suppose a, b, c

(a) a3

— Aac C. Show that

— b3 = (a — b)(a2 + ab + b2) = (a — b)(a — bw)(a — ba;2).

(b) a3 + b3 + c3 — 3abc = (a + b + c)(a2 + b2 + c2

— ab — be — ca)

= (a + b + c)(a + bw + cw2)(a + buj2 + co;). (4) For simplicity, let

elQ = cosa + isino for real a (see (1.2.13)). Suppose a 0. Show that the cubic equation z3 = a always has three distinct solutions II—?—

ii—?—

|ape 3 , |a|3e

3(v,

II—?—

|ape

2

3a;

,

where 0 = Arg a and w = |(—1 + y/3i\ Try to find solutions of zn = a. (5) Suppose a, b C. Show that z3 — 3abz + (a3 + b3) = 0 has solutions

— (a + b), — (aw + bw2), — (aw2+bw), where w =

|(—1 + y/3i).

(6) ( Cardano formula for cubic equations) Given a cubic equation

z3 + O2Z2 + a-^z + ao = 0,

— ^a2 into the equation to

with complex coefficients. Substitute z = w obtain

w3 + pw + q = 0. (a) By comparing to the cubic equation in Exercise (5), let p = — 3ab and q = a3 + b3. Show that a3 and b3 are roots of the quadratic equation

t2 - qt -

^p3 = 0.

By Exercise (2), we may suppose that 3

q

q2

+

P3

.

and

,3

q

q2

p3

=2-Vt + 27-

Sec. 1.3. Complex Number System (Field) C

23

(b) Choose a and b as suitable cubic roots of

respectively, subject to the constrained condition ab = —^p- Then, according to Exercise (5), w3+pw+q = 0 has roots —(a+b), (aw+ bw2), (aw2 + bw). Therefore, the original equation has roots — (aw2 + 6w), and (a + b), — (aw + 6w2). (c) Solve z3 + 3z2 3z 14 = 0.

—

— —

— —

—

— —

(7) Suppose z = x + iy is not a negative real number and z there exists a unique w, Rew > 0, so that w2 = z.

0. Show that

Exercises C

(1) In (1.1.1), replace the real scalars a,(3 by complex numbers a,(3 and then, we can view R2 as one- dimensional complex vector space over the field C. We denote this vector space by C itself. (a) Show that any linear transformation T : C —* C is of the form

T(z) = az,

z

C

where a is a complex scalar. Let a = a + bi and z = x + iy. Then T(z) can be rewritten as the vector form Ta(x,y) = (ax by, bx + ay) or as the matrix form, with respect to the natural basis for R2,

—

Ta(x,y) = (x y)

a b —b a

which is a special kind of linear transformations on (b) Conversely, given a linear transformation on R2 as

T(x,y) = (x y)

R2 .

a, b,c,d C R.

Let z = (x y) = x + iy. Suppose this T turns out to be a linear transformation Ta(z) = az on C. Then

T(z) = (x y)

=> (Let z = 1.) (1 0)

b az d =

for all z

C

b = (a b) = a 1 = a or a = a + bi; d

Chap. 1. Complex Numbers

24

6 = (c d) = ai d or a = —i(c + di) = d — ic

(Let

Z

= Z.) (0 1)

a = a + bi = d

— ci

or a = d

and b = —c.

Hence, T should be of the form

T(x,y) = (x y)

= Ta(z),

_b

where a = a + bi.

(a) and (b) suggest that the following peculiar set of matrices

50(2, R) =

a b —b a

a, b

R>,

is worthy being emphasized.

(c) Under the operations of matrix addition and multiplication, show that SO(2, R) is a field (namely, having properties listed in (1.3.2)). (d) Show that the mapping $ : 50(2, R) —> C defined by $

a b —b a

= a + bi,

is a field isomorphism (namely, $ is one-to-one, onto and preserves operations of addition and multiplication).

Therefore, one can treat a b =T —ba

cos #

— sin 0

sin 0 cos 0

where r = \/ a2 + b2 and tan# =

-a

as a complex number, especially in its polar form rel6 (see (1.2.14)). Since the expression on the right side represents, geometrically, a stretch of vectors followed by a rotation through the angle 0 (for details, refer to Ref. [56], Vol. 2), a complex number is a two-dimensional number, taking care of both stretch and rotation at one time by its very nature. See the end of Sec. 1.1. (2) Fix a point (a, /3), with (3 0, in R2. Then (x, y) = (x — ^)(l,0) + ^(a,[3) always holds. Define: On R2, equality: (aq, yr) = (x2,y2) & xx = x2 and yx = y2: real number: (aq,0); addition: (aq,yi) + (x2,y2) = (oq +z2,yi +3/2), and

Sec. I.4. Algebraic Operations and Their Geometric Interpretations

25

multiplication: ,

7i, yj 0(2:2, 3/2)x = x„z

r/^1 L\ . 7 +

-

ayiW -5-

P

^2

ayi\

-

-5-

J\

P

3/13/21 (1,0) nx -55P /,

-

J

J

3/1I «3/1 A 3/2 + V’T-) [y'’~)7 .

(

ay2\

,

flx

Let C(z) denote R2 with these two operations. (a) Show that C(z) is a field. (b) Show that C(z) is field isomorphic to C.

Hence, conceptually we can view C(z) as a complex field with i = (a. (3) acting as the imaginary unit.

(3) Let 1 2 2 -1 Show that A2 + /2 = O where /2 =

and O is the zero matrix.

Can we construct a field F with A as the imaginary unit so that F is field isomorphic to C? If affirmative, try it (see Sec. 2.7.8 in Ref. [56], Vol. 1). 1.4.

Algebraic Operations and Their Geometric Interpretations (Applications)

The section is divided into four subsections. Section 1.4.1 introduces conjugate complex numbers, a unique operation particularly owned by the complex number system C. Section 1.4.2 discusses the relations among real and complex inequalities. Section 1.4.3 uses examples to show how complex numbers can be adopted to solve planar geometrical problems. Finally, the important concept of symmetric points with respect to a circle or a line will be discussed in Sec. 1.4.4. 1.4.1.

Conjugate complex numbers

The symmetric point of a complex number z = x + yi with respect to the real axis, denoted as z

=x

—

iy,

(1.4.1.1)

Chap. 1. Complex Numbers

26

is called the (complex) conjugate of z. See Fig. 1.10. Note that

z = z z is real, i.e., Im z = 0.

(1.4.1. 2)

This indicates that the conjugate operation z —> z is a two-dimensional operation, particularly owned by C but not by R. Also,

zz = \z\2

0,

(1.4.1.3)

shows that how a complex number z and its conjugate z can be so related, via multiplication, to produce a nonnegative real number |z|2, whose posi¬ tive square root |z| is just the absolute value of itself and can be considered as the length of the vector z or the distance of the point z to 0. This com¬ pensates somewhat the fact that C is no more an ordered field. It is easy to prove the following Operational properties.

(1) Rez = |(z + z), Imz = (2) (z) = z (in short, z = zf

^.(z — z).

(^=^(z2^0f

(3) Z1±Z2 = Z1± Z2, ZTZ^=Z!Z2, (4) \z\ = \z\, Argz = -Argz(-7r < Arg z < tt). (5) |z|2 = zz 0, |z| = \fzfi (positive square root as a real number). (6) Zi = z2 O |zi| = |z2|, Argzi = Argz2 zi = Z2(1.4.1.4) Readers are urged to give these relations their geometric interpretations.

Sec. 1.4.1. Conjugate complex numbers

27

In what follows, we discuss two important applications of the relation

zz = |^|2.

It provides an easy way to compute the multiplicative inverse of a nonzero complex number. Suppose z = x + iy / 0. Then

Figure 1.11 also shows that

— - — IHA— I = —

1 Z ) \Z where (i) is called the symmetric motion or reflection with respect to the unit circle |z| = 1, while (ii) the one with respect to the real axis. Therefore,

z

1

> > (i) Z (ii)

w = -, z

(1.4.1.6)

as a mapping from z to is the composite of two such reflections. In computation involving absolute values of complex numbers, the rela¬ tion |^|2 zz plays an essential role. For example,

—

1-21 ± Z2|2 = (-21 ± 22)(^1 ± Z2) = Z1Z1 ± (Z1Z2 + Zl-Zfc) + Z2Z2, => |zi ± z2|2 =

IzJ2 ± 2Re(zif2) +

(1.4.1. 7)

Hence, it follows immediately that

1-21 + 22|2 + 1^1 — Z2I2 = 2(1^1 12 + |Z2|2),

(1.4.1.8)

which reflects the fact that the sum of the square of two diagonals of a parallelogram is equal to the sum of the square of its four sides (see Fig. 1.7).

Fig. 1.11

Chap. 1. Complex Numbers

28

In (1.4.1.7), let zi = a\bi and Z2 = 02^2- Then

|ai bi + «2^2 1 2

= |Oi^i |2 + | |d&i + a2^2|2 = (|d|2 + |d |2)(|#i|2 + |#2|2) — |d&2 — d#i|2(1.4.1.9) This is a special case of the Lagrange identity (see Exercise A(ll)).

Exercises A

(1) Prove (1.4.1.4) in detail and interpret them geometrically. (2) For each of the following complex numbers z, compute Rez, Imz, |z|, Argz and z. (a) (b) (i-l)(i-2)(i-3) • (C) (V^+i) 3- (d) • + Compute Rew, Imw, |w|, and w. (3) Let w = (4) Let z = cos# + isin# or suppose |z| = 1 and z +|= 2cos#. Show that

—

zn + zn = 2 cos nd

and

—

zn zn = 2sinn#,

n

Z.

(5) Let (1 — \/3z)n = xn + iyn, where xn, yn G R for n = 1, 2, 3, . . . . (a) Show that xnyn-i - xn-iyn = 4n-1 • >/3. (b) Compute vn_i + ynyn-i = ? (6) Suppose |zi| = A|z2|,A > 0, show that |zi — A2Z2| = A|zi — Z2I. Con¬ versely, if |zi — A2Z2| = A|zi — Z2I for A > 0 and A 1, then |zi| = A|z2 1 holds.

(7) (a) Show that, for z

0,

|z| = 1 z = t z = z C

for some

£

0.

(b) Suppose |z| = 1 but z — 1. Show that there exists a unique real Try to express t in terms of z. number t so that z = (c) Hence, the set {z C ||z| = 1 butz —1} can be put in one-to-one correspondence onto the real axis. Try to deduce the polar form of a complex number.

Sec.

Conjugate complex numbers

29

(8) Prove the following identities. (a) |zi (1 + |z2|2) - ^2(1 + |*112)| = 1^1 -^2|2|1-^1^2|2- (2122-^1^2)2(b) |1 + ziz2|2 + |^i - z2|2 = (1 + |^i I2) (1 + M2)(c) |1 - z^l2 - \Z1 - z2|2 = (1 - kr |2)(1 - N2). (9) Suppose a, b, z C. (a) Show that

—a

z

1 — az

= 1 o |a| = 1 or [z[ = 1.

Discuss what happens if |a| = |z| = 1. (b) Suppose |z| = 1. Show that = 1 where

|

|

|a|2 — |&|2

0.

(10) Use (1.4.1.8) to show that

|a + y/a2 — b2^ + |a — \/a2 — b2^ = |a + b\ + |a — 6|, and hence, deduce that

kil + kl =

2^1 + *2) + V^k2

(11) For aj,bj G C for 1

j

-(zi + Z2) - y/zXZ2

n, prove the Lagrange identity

2

l≤J

|z|

|Rez| + |Imz|.

(1.4.2.1)

These are the most important elementary inequalities involving real and complex numbers. And above all, they connect the real and complex limit processes together (see Sec. 1.7). By using (1.4.1.7) and (1.4.2.1),

|^i + z2|2 = |zi |2 + |^2|2 + 2Re(ziz2) l^i I2 + |^2|2 + 2|ziz2| = (|zi| + |z2|)2 => |^i + z2| |^i| + |z2|, which is called a triangle inequality (see Fig. 1.7). (1.4.2.2) Equality in (1.4.2.2) holds if and only if |zjz2|

1-21 + 22| = 1-211 + |22| o Z1Z2

0 o Argzi

= Re(zjz2), and hence

= Argz2 o

—

0 if z2

± 0.

(1.4.2.2)' Similarly, we have another triangle inequality

I N - N I ki + 22I and the equality “=” holds O zi z2

0O

—

0 if z2

0.

(1.4.2.3)

Sec. 1.4.2. Inequalities

31

This can also be proved by observing, via (1.4.2.2), that |zi| = |(zi + Z2) — z2\ |^i +Z2I + |~ Z2I = |^i +Z2I + |^2|> with equality holds if and only if (zi + z2)(-z2) = -.Z1Z2 - N2 0 or 2:1^2 0; and |z2| l-Zi + Z2I + |zi| with equality if and only if z1Z2 0. We illustrate three examples.

j
0 so that |zi

Wil -I Z2I | holds for any zi, z2 G C. But, we do have

-^

— Z2I

f°r

Chap. 1. Complex Numbers

32

Fig. 1.12

Example 2. Fix 8, 0 Show that

6
1.

—

Try to figure out when equalities hold. (5) Suppose 2i + 22 = 1- Show that 1 |2i| + I22I and, in case 2122 7^ 0, equality holds if and only if both Z\ and 22 are positive real numbers. (6) Suppose Rea > 0 and Re(\/a2 — 1) 0 (see Exercise B(l) of Sec. 1.3).

(a) Show that Re{a\/a2 — 1} 0. (b) Show that |a + Va2 — 1| 1 with equality if and only if a is real and 0 < a

1.

(7) Consider the equation z2 — 2az + 1 = 0. (a) In case a is real and —1 a 1, show that the roots z of the equation satisfy

\z\ = 1.

(b) Otherwise, the equation has one root 21 satisfying |2i| < 1 and another root 22 satisfying (22! > 1(8) Prove the following special case of Minkowski inequality

with equality if and only if

1

n, are equal.

J

Exercises B

(1) Show that the roots z of the equation az2 + bz + c = 0, where ac

0,

satisfy

|c|

+ (2) Suppose 0 < an

an_i

a^z

0

> 0;

— z± < 0; 22 21 Z± left closed half plane : Im — 0 22 21 Zi including the boundary line Im — z

right open half plane : Im

-

Z

Z

22 - Zt

right closed half plane : Im

—

22 - 21

=0

0.

(1.4.3.1)

In particular, the real axis Imz = 0, pointed to right, separates C into

(open) upper half plane : Imz > 0 and (open) lower half plane : Imz < 0;

(1.4.3.2)

Sec. 1.4-3. Applications in (planar) Euclidean geometry

37

while the imaginary axis Rez = 0, pointed upward, separates C into

(open) right half plane : Re z > 0 and (open) left half plane : Re z < 0.

(1.4.3.3)

Also we have The relative positions of two lines (segments'). Let z = zi + t(z^ be two lines. Then, they are z = z( + t(z£

—

(i) coincident

—

z[—zi and z2 z( are real multiples of

z^

— zi) and

— z\ , i.e.,

and Zi 21 are real; is real; (ii) parallel z'2 — z( is a real multiple of z-2, — zi, i.e., having parallel positive and the direction real is O same a (iii) z'2 — z( z' —z multiple of Z2 — zi, i.e., 2?_2* > 0; —z (iv) intersecting at a point o Im z/_2) 0.

The angle from the line z = z(+t(z2~ zj) to the line z = zi+t(z2—zi) is

Arg Z2

~

Zi

4-

’

See Fig. 1.15. Hence, they are

(v) perpendicular Re

— — -= ^2

Z1

^2-^1 Proofs are left to the readers as Exercise A(3).

Fig. 1.15

0.

(1.4.3.4)

Chap. 1. Complex Numbers

38

Suppose zi, Z2, and Z3 are not collinear. Then, the consecutive segments and Z3Z1 form the sides of a triangle [\z\Z2Z3 with zi, Z2, and Z3 as vertices. The ordering z± —> Z2 —> Z3 is said to determine an orientation of the triangle, usually called positive if counterclockwise and negative if clockwise. Z1Z2, Z2Z3.

Some facts about a triangle

(1)

having the same orientation, are similar if and and only if (see Fig. 1.16) Z3

~

-?i

Z2 - zi

If having opposite orientation, then they are similar if and only if (see Fig. 1.17) Z3 Z2

~

zx

~

Zi

21

z{

Z2

z^ z'3

23

Fig. 1.16

Fig. 1.17

Sec. 1.4-3. Applications in (planar) Euclidean geometry

39

(2) AZ1Z2Z3 has area -Im(zi-z3)(z3-z2) = 2

-|z3-z2|2Im^—— = -Im(z1z2+ z2z3+ z3z1) 2

z3

— z2

2

if the triangle is positively oriented; or, •

21

21 1

± ~ 22 22 1 4 23 23 1

where

± is to be chosen so that

the area is nonnegative.

(1.4.3.5)

The details are left as Exercise A(4). In two real variables x and y, a circle has the equation

a(x2 + y2) + /3x + qy + 5 = 0

\ 1 Let x=-(z + z) and y= — -{z — z). Note that zz = |z|2 = x2 + y2. 1

(

^(z + z) — ^{z — z) + 6 = 0, or azz + i^z + ^(/3 + iy)z + ac: a real circle

\z - Zol =r with center zq =

—^

and radius r =

aC

or, in polar form,

z = zq + r(cos 0 + i sin 0) = zq + relB , 0 < 0 r

— zo|

= r itself),

and

(including \z — 2q| = r).

r

(1.4.3.6)

Section (2) Some illustrative examples Example 1. A triangle 2

A2i2223 is an equilateral triangle if and only if 2

2

।

21 + 22 + 23 ।

— ^122 - ^223 - 2321 = 0. n

Proof. The necessity: See Fig. 1.19. Then 23 - 21 22

-

21

_

21 - Z2

Z3

-

Z2

_

22

~

21

-

23 ’ Z3

(*1)

Sec. 1.4-3. Applications in (planar) Euclidean geometry

41

Fig. 1.19

After cross multiplication of the first equality, we have

(z3 - Zl^Zs - Z2) = (z2 - Z^Z! - Z2) = -(21 - Z2)2 Z^ + Z^ + zl Z1Z2 - Z2Z3 - ^3^1 = 0. ~

The sufficiency: Reversing the above process, we will recapture (*1) and hence, Aziz2z3 is equilateral. Or we can do this as follows. The given iden¬ tity can be rewritten as

(zi - z2)2 + (^2 - ^3)2 + (z3 - zi)2 = 0. Now

(^i - Z2} + (z2 - z3) + (z3 - zj = 0

=> (Z! - Z2)2 + (z2 - z3)2 + 2(zi - z2)(z2

~

Z3)

= [~(z3 -21)]2 = (z3 -Z1)2 => (Substitute (zi — z2)2 + (z2

— z3)2

= ~{z3 - zi)2.)(zi - z2)(z2 - z3) = (z3 - zi)2 z3 - Zl _ z2 - z3 z2 - Zl

And so on.

Zl

-

z3

'

Chap. 1. Complex Numbers

42

Example 2. Given four distinct points plane C. Show that the identity

zi, z2, z3,

and Z4 in the complex

(zi - 22)(z3 - 24) + (z>l - 23)(z4 - Z2) + (Zi - Z4)(z2

“

Z3) = 0,

and hence, deduce the inequality

|21 - 22| |23 - 24| + |z2 - 23| 1-21 - 24|

1(^3 - 2i)(z4

~

-22 )|,

with equality if and only if zi, z2, z3, and z^ in this ordering, lie on a circle or on a line. This is the classical Ptolemy theorem. See Fig. 1.20(a) and (c).

Proof. The identity can be rewritten as

(22 - 21)(z4 - 23) _ (z4 - 22)(z3 - 21) (24 2i)(z2 Z3) (z3 - Z2)(z4 - 21) ’ which we simply denoted as — A + 1 = B, where

—

*2

—

(22 2i)(z4 z3) A= (24 - 2i)(z2 - Z3) and

B=

(24-Z2)(23-Zi) (23 - 22)(z4 - Zi)'

The necessity: Suppose these four points lie on a circle as shown in Fig. 1.20(a) or on a line as shown in Fig. 1.20(c). Set

6*i = Arg

.

22

~

24

-

21 Zi ’

a 02 = Arg

24 - Z3

03 = Arg

z4

— z2

23

- 22

22 - Z3

and 23 21 Z4- Zi' ~

^4 = Arg Then,

+ 02 = — 7r Arg B = 03 + 64 = 0 Arg A = #i

=> Arg(— A) = 0.

or

7r,

Sec. 1.4-3. Applications in (planar) Euclidean geometry

43

Hence — A > 0 and B > 0. (*2) means — A + 1 = B which is equivalent to |-A| +1 = \B\, i.e., |^i —^2| \z3-z4\ +\z2-z3\ |zi - z4| = |(z3- zi)| l(z4-z2^ holds. The sufficiency: Reverse the above process. The identity is equivalent to |— A| + 1 = |B|. When comparing to (*2), namely, —A + 1 = B, we have

Chap. 1. Complex Numbers

44

—A > 0 and hence B > 0. These mean that Arg A = + #2 = ArgB = #3 + #4 = 0. Therefore, zi, Z2, 23, and z^ lie on a circle.

—

and

Remark. Suppose we disregard the ordering of appearance of zi, Z2, Z3, and Z4 on a circle or on a line. Then, in addition to Figs. 1.20(a) and 1.20(c), we also have to consider the cases shown in Figs. 1.20(b)-l.20(e). In all cases, the ratio of ratios Z1 - Z3

Z2

-

^3

’

Z2

-

z^

-

Z4

will be a real number-, as a matter of fact, it is positive if z^ and z^ does not separate zi and z?, and it is negative if Z3 and z^ separate zi and Z2. Example 3. Let a and b be two distinct points in C.

(1) The set of points z satisfying

z—a A, z—b =

0

A

00

represents

(i) (ii) (iii) (iv)

the point circle a, if A = 0; the point circle b, if A = 00; the perpendicular bisector of the segment ab, if A = 1, and in case 0 < A < 00 and A 1, the circle

a — X2b 1 - A2

A|a

— b\

= |1 - A2|

'

See Fig. 1.21: If A varies from 0 to 00, the family of circles varies from the point a, via circles with centers lying on the line ab, called Apollonius circles, to the pointe b. (2) The set of the points z satisfying Arg

z—a n = f), z—b

— 7T < u„

7T

represents

(i) the segment ab, if 0 = tt; (ii) the two outward rays eminating from a and b along the line con¬ necting a and b, if 0 = 0, and

Sec. 1.4-3. Applications in (planar) Euclidean geometry

(iii) in case

—

tt

^|CSC0|.

< 0 < tt and 0^0, the circle \z

See Fig. 1.22: If 0 varies from

—

45

— °a2e 0 1 = 2

— tt

to tt, the family of circles are coaxial circles with the segment ab as the coaxis. (3) Any one of the circles in (1) is orthogonal to any one of the circles in (2). They together form the so-called Steiner’s circles. Proofs are left as Exercise A(5). For further discussions concerned, see Sec. 1.4.4.

Example 4. Describe the plane curve

|z2 — a2 1 = A where a > 0 is a constant and 0 Try to find out the point set of z satisfying |z2

Solution. |z2

— a2 1 =

A
a, a Bernoulli lemniscate if A = a, a Cassini oval of two branches if 0 < A < a, and the point set {a, —a} if A = 0. See Fig. 1.23. What is the point set defined by

|z2-a2| 0, = 0 or < 0. (2) Im

w^-0.

Solution. Let z = x + iy and w = u + iv. Then

— 3(z + iy) = (x3 — 3xy2 — 3zj + i(3x2y — y3 — 3y) => u = x3 — 3xy2 — 3x = a:(a:2 — 3y2 — 3), v = -y3 + 3x2y - 3y = y(3x2 - y2 - 3).

u + iv = (x + iy)3

Then

x2 — 3y2 = 3 (a hyperbola); u = Re w > 0 x > 0 and a:2 — 3y2 > 3 or x < 0 and x2 — 3y2 < 3; u = Rew < 0 x > 0 and x2 — 3y2 < 3 or x < 0 and x2 — 3y2 > 3. u = Re w = 0 x = 0 or

Similarly,

> 0 O y > 0 and 3a;2 — y2 > 3 or y < 0 and 3z2 — y2 < 3, v = Im w < = 0 o y = 0 or 3a:2 y2 = 3 (a hyperbola), < 0 O y > 0 and 3a:2 — y2 < 3 or y < 0 and 3a:2 — y2 > 3.

—

See Fig. 1.24. Exercises A

(1) Do the following problems. (a) Prove that the direction vector of a line az + az + b = 0 is orthog¬ onal to a. (b) Two points z\ and 22, considered as vectors, are orthogonal if and only if 2122 + Z1Z2

= 0.

(c) Three points 21,22, and z^ are collinear if and only if, there exist real scalars ti, t2, and t^, not all equal to zero, so that

ti2i + t2z2 + (d) Suppose

21

+ 22 + 23 =

= 0 and 212223 holds.

+ t2 + t3 = 0. Show that

cannot lie on the same side of the real axis.

(2) Prove (1.4.3.1) in detail. (3) Prove (1.4.3.4) in detail.

21,22,

and z^

Chap. 1. Complex Numbers

48

(4) Prove (1.4.3.5) in detail. (5) Prove Example 3 in detail. (6) (a) Find the other two vertices of a square with two known vertices ai and a^.

(b) Find the fourth vertex of a parallelogram with three known ver¬ tices 01,02, and 03.

(7) Fix any three noncollinear points ai, 02, and 03 in the plane C. Show that any point z in C can be expressed uniquely as Z

= tlOi + ^2^2 +

tl + ^2 + ^3 = 1-

Usually, we call (ti,t2: fo) the barycentric coordinate of z with respect to the affine basis {01,02,03} for the plane. Show that a point z lies in the interior of the triangle AaiO2O3 if and only if it has barycentric coordinate (ti, 0, tz > 0, ts > 0, and ti + a2,

respectively.

lie on 0. Show that —a, a, z, and a circle. (c) If a circle passes through —a,z(Imz 0) and then it should pass a. (d) Let T be as in (a). Show that z lies inside F o lies outside F.

(b) For any point z with Im z

Sec. 1.4-3. Applications in (planar) Euclidean geometry

51

(18) Describe the following point sets or curves and try to graph them. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (1) (m)

a < Re z < /3, where a, P R. a < Imz < P, where a, P R. a < Arg z < P, where a, P R and a < P < a + 2tt. a < \z — zo| < P, where zg is a fixed point and 0 a
0 and |a| < A.

Re^|0. |z| < 2 and 0 < Argz < j. 0 < Arg(z — 1) < f and 2 < Rez 3. Let ab and a'b' be two line segments. Try to locate these points z so that Azab and /Xza'b' both have the same orientation and are similar.

Exercises B

(1) Suppose

zi, . . . ,

zn

are distinct nonzero points and they all lie on the same side of a line passing 0.

—

(a) Show that , 1 < k < n, all lie on the same side of a certain line passing 0. (b) zi H d 1- zn 0 and T 0 hold. This indicates that, as long as zi + + zn = 0 and zi, . . . , zn do not lie on the same line, then any line passing 0 will separate the points zi,...,zn, i.e., some lie on one side of the line while others on the other side. (2) Let zi,...,z„ be distinct points in C. Denote the convex closure spanned by zi ,..., zn as

±

••■

Con(zi, . . . ,zn)

with zi , . . . , zn as vertices. It is a convex set, namely, the line segment connecting any two of its points lies entirely in the set. In case Xj > 0 |-An = 1, then 52”=1 XjZj is called an interior for 1 j n and Aid

Chap. 1. Complex Numbers

52

point of Con(zi, . . . , zn). If Zi, zn are collinear, then Con(zi, . . . , zn) is the smallest line segment containing zi, .. . , zn.

(a) Suppose zi , . . . , zn are not collinear. Then any line passing an inte¬ rior point of Con(zi, . . . , zn) will separate zi, . . . , zn. (b) If 7=7“ = 0 holds, then z should lie on the set Con(zi, . . . , zn).

52^=1

1.4.4.

Steiner circles and symmetric points with respect

to a circle (or line) We try to give another proof of the fact that an Apollonius circle Ci : = A (0 A 00) and a coaxial circle C2 : Arg = 0 (— 7r < 0 tt) will intersect orthogonally (refer to Example 3 in Sec. 1.4.3). And hence, we introduce how two points are said to be symmetric with respect to a circle (or line). In Fig. 1.26, fix any point Zq on a C2 circle and draw the tangent to the circle at zq so that it intersects the extended line ab at a point p. Now,

A Apzoa

—

, , . (similar) => -*o-a = , P-a = -P-^o . ~ ^pbz0 |z0-h| \p-z0\ |p- b| A

/

-1

1

Note that zq lies on a Ci circle, namely,

Fig. 1.26

Sec. 1.4-3. Applications in (planar) Euclidean geometry

53

Hence,

Ip~«I |p ^ol

—

.

—

—

|p-^o| |p-a| \2 A => = =A , Ip -b\ |p - &|

. . a constant

1^11

This means that, as long as a point Zg lying on a Ci circle: = then will always intersect the tangent to the circle: Arg = 0 = Arg the line ab at a fixed point p. Moreover,

|p-z0|2 = |p-a||p

—

shows that |p — zo| is also a constant. And we prove that such a Ci circle has its center at p and radius |p — zg |, and is orthogonal to any possible C2 circle. We formally summarize the above as

Symmetric or reflection points with respect to a circle (or a line). Fix a circle C with center z0 (including the degenerated case, a line L) and two points z, z* £ C. Then, the following are equivalent:

(1) z and z* are situated on the same half line from zg, namely > 0, and \z — zo| \z* — ^o| = f2 (2) Any circle (or line) passing through z and z* will intersect the fixed circle C (or line L) orthogonally. (3) Any circle (or line) passing z and intersecting C orthogonally will also pass the point z*.

In such a case, we called z and z* symmetric with respect to C (or L), or reflection points of C (or L). In fact:

(a) If C has equation \z — zg\ = r, then z and z* are symmetric w.r.t. C if and only if Z

= Zg +

r2

—.

Z - Zg

In this case, the circle C can be expressed as

z = z0+rietv and z* = then

|7^77| = 77 =

where

z^r^e1^ . If C has equation azH+bz+Sz+c = 0, az*z + bz* + bz + c = 0.

Chap. 1. Complex Numbers

54

(b) If L has equation az + az + b = 0, then z and z* are symmetric w.r.t. L if and only if az + az* if L has equation z = a + 6t(|6|

+ 6 = 0;

= 1, t

G

R), then

z* = a + b2 if L is the line passing two distinct points ai and

z* =

1 02 — al

02,

[(a2 - ai)z + aia2 - aia2] .

then

(1.4.4.1)

See Fig. 1.27 Note that z lies inside C z* lies outside C; z lies on C o z* lies on C and z = z* holds. In case of a line L, z and z* lie on different sides of L. By the way, we obtain partial results of the following Steiner’s circles. Fix two distinct points a and b in the plane C. Then, Apollonius circles

z—a

Ci : z

—b

—

Coaxial circles C2 : Arg - - = 0, z—b together form the so-called Steiner’s circles: a and b are called the limit point of Ci circles, while the line segment ab the coaxis of C2 circles. See Fig. 1.28. They own the following basic properties:

(1) Except the limit points a and b, any point in C lies on exactly one Ci circle and only one C2 circle.

Fig. 1.27

Sec. 1.4.3. Applications in (planar) Euclidean geometry

55

Fig. 1.28

(2) Every Ci circle is orthogonal to every C2 circle. (3) A reflection or symmetric motion (z » z*) with respect to a Ci circle maps every C2 circle onto itself, and maps a Ci circle onto another Ci circle. A reflection or symmetric motion with respect to a C2 circle

—

maps every Ci circle onto itself, and maps a C2 circle onto another C2 circle. See Exercise A(7). (4) The limit points a and b are symmetric with respect to every Ci circle, (1.4.4.2) but not to any other circles.

As a matter of fact, (1.4.4.1) and (1.4.4.2) are easy consequences of the con¬ , respec¬ formality of the bilinear transformations w = *2^ and w = tively. (See Sec. 2.5.4; in particular, Examples 1 and 2 there.) Exercises A

(1) Prove (a) and (b) in (1.4.4.1) in detail. (2) Prove (1.4.4.2). See also Exercise (7) below. (3) Suppose a circle Ci intersects another circle C2 orthogonally and a line passing the center of Ci will intersect C2 at two points a and b. Show that a and b are symmetric with respect to (fl.

Chap. 1. Complex Numbers

56

(4) (a) Show that a point z symmetric with respect to both |z — ai | = ri and |z — a2| = r2 satisfies r2(z - ai)-1 - r^z - a2)-1 = a2 - di(b) Show that a point z symmetric with respect to both a circle |z a| = r and a line passing ai and a2 satisfies

—

a + r2(z - a)-1 = (a2 - ai)-1[(a2 - ajz + dia2 - aid2],

(5) Try to use Exercise (4) to do the following problems.

\z —

(a) Find out all the circles orthogonal to both |z| = 1 and ||= (b) Find out all the circles orthogonal to both |z| = 1 and the line x = 2. (6) Discuss the following families of curves and graph them. (a) Re |= Ai and Im| = A2, are they orthogonal to each other? (b) Rez2 = Ai and Imz2 = A2, are they orthogonal to each other?

(7) (Refer to (3) in (1.4.4.2)) (a) Suppose z and z* are symmetric with respect to a Ci circle: | Ao (0 Aq oo). Then

Arg

——

z -a

z

-

a

- d (-7T < 0 Arg z—o = z —0 =

tt).

(b) Suppose z and z* are symmetric with respect to a z—a z—b Arg

z*

—a

z* —b

z*

|=

circle. Then,

A (0 < A < oo); —b = v - - o

0 O Arg

z

— -b =

—0

(— 7r < 0

7r).

(8) Consider the graph of the circle \z\ = 1 under the translation z Show that, if |z| = 1 and z —1, then 2 Arg(l + z) = Argz.

—> 1+ z.

Sec. 1.5. De Moivre Formula

1.5.

57

De Moivre Formula and nth Roots of Complex Numbers

In (1.2.15), let

= Z2 = z, then z2 = |x|2(cos20 + z sin 20), 0 = arg2; in (1.2.15)', let z? = 1 and z± = z, then = |z|-1(cos(— 0) + i sin(—0)), 0 = argz. zi

z~r

Based on these two identities, we get inductively the following

zn = rn (cos nd + i sin nd),

r = \z\, 0 = arg z

= rnein0, n = 0, ±1, ±2,... . In case |z| = r = 1, we have the

(1.5.1)

De Moivre formula.

(cos 0 T isin0)n = cos nd + isin nd,

n = ±0, ±1, ±2, . . . ,

or,

(ei8)n = ein8.

(1.5.2)

One of the main advantages of this formula is that it provides an easy way to compute the nth roots of a complex number, while the other one is that, via binomial expansion, we can express both cos nd and sinnd as polynomials in cos0 and sin0 (see Exercise B(2)-(4)). Let n be a fixed positive integer and z C. Then, the nth roots of z, denoted as

z^

or

tfz,

(1.5.3)

are defined as any complex numbers w such that If z = 0, designate z^ = 0. Now, suppose z 0. In wn = z, let

z = rel8,

r

= |z|,

and

w = pelv ,

p=|w|,

and

wn = z.

0 = Argz; ip

= argw

=> (By (1.5.1)) pneinv = rei8

pn = r

and

=> p = r f = |z| f y

=

n

nip =

d + 2kir,

k = 0, ±1,±2, ...

(as a positive real number), (Argz + 2fc7r), fc = 0,±l,±2,.... =n

Chap. 1. Complex Numbers

58

Hence, the roots of Wk

wn = z are

= \z\ ie^^T^z+2k7r\

k = 0, ±1, ±2, ....

In case k < 0 or k n, Wk = wm O k = pn + m for some integer p and 0 m n— 1. Hence, wn = z has only n distinct roots Wk for 0 k n— 1. We summarize the above as nth roots (n

2) of a nonzero complex number.

(1) The nth roots of the unit 1. Let co = cos

Then

2tt . . 2tt l-z sin n n

—=

2j

e ".

cun = 1 and nth roots of 1 are 1

, , , ,2

,

,n-l

which form vertices of a regular n-gon inscribed in the unit circle. See Fig. 1.29 (for n = 6). (2) The nth roots of z 0. Let 0 = Argz. Then, the nth roots of z are

& = Wte^0+2k^

—

= (owk, k = 0, 1, 2, . . . , n 1, is usually called the principal value of tfz. Note that Co, Co^, . . . , Co^"-1 form vertices of a regular n-gon inscribed in the circle with center at 0 and radius \z\^. (1.5.4)

Co

A complex number C, such as co, satisfying C" = 1 but Cm 1 for any 1 < m < n — 1, is called a primitive nth root of 1. If C is a primitive nth

Fig. 1.29

Sec. 1.5. De Moivre Formula

59

root of 1, then 1, £2, . . . , £n-1 will be all the nth roots of 1. £ = ivk is a primitive nth root if and only if 1 A: n — 1 and k is relatively prime to n. Equation (1.5.3) can be extended as follows. Let n and m be integers so that n > 0 is relatively prime with |m|. Then, we define

'(zi)m, m > 0, (1.5.5)

z

m < 0 and

0.

By using (1.5.4), we have

z n = |z| " el

"

\

0 = Argz,

k = 0, 1, 2, . . . , n

— 1.

(1.5.6)

Note that, within this formula, \z\^ = 7klm = ( y/\z])m is chosen to be a positive number. For further discussion, see Exercise A(10). We give four examples. Example 1. Solve Solution,

z6 — 2z3 + 2 = 0.

z6 — 2z3 + 1 = —1 => (z3 — I)2 = —1 = e7". Hence

Z3 - 1 =

A: = 0,1.

In case k = 0 : z3 — 1 = eim = i^z3 = l+ i = y/2eivl. Hence Zl

= 2^ e^i7^21^, I = 0,1,2

zi

1 \ 13 1/1 = 26e4irl = 26 + z—= I ; = \ 72 72/

In case k = 1 : z3

— 1 = — z => z3 = 1 — i = 72e zi+3 =

2h^-iv+2lir\

=> Z3 = zq ,

Z4

Hence

/ = 0,1,2

= Z2 and

z$

= z\.

See Fig. 1.30. Example 2. Compute (1) [(1 + «) a]3, [(1 + z)3] 2 ;

Chap. 1. Complex Numbers

60

Fig. 1.30

(2) [(1 + z)^]2, [(1 + z)2]i, and compare them. Solution. (1) 1 + i =

So

(l + zJ^eiOH,

A: = 0,1,

=> [(l + i)5]3 = namely, zq = 2^63”, zi = — 2Jes1". On the other hand, (1 + z)3 = 2(—1 + z) two values zq and z^. (2) By (1),

fc = 0, 1,

= 2y/2e?’r2 and [(1 + z)3]^

has

[(l+z)5]2 = 2iel(^+2M, fc = 0,1, and we get only one value hence

^/26^’ = 1 + z. While, (1 + z)2 = 2z = 265” and

[(1 + z)2]l = 2ie^^+2^);

k

= 0)

namely, ±(1 + z), which has one more value than [(1 + z)^]2.

Remark. Suppose /i(z) and /2(2) are two multi-valued functions. If, for each z in their common domain of definitions, the set of values of /i (z) is equal to the set of values of f?(z), then we designate /i(z) = /2(2), otherwise fi(z) f2^z). According to this convention, [(1 + z) 2 ]3 = [(1 + z)3] 2 while [(1 + z)s]2 7^ [(1 + z)2]l. For general setting, see Exercises A(9) and (10). Also, refer to (1.2.16).

Sec. 1.5. De Moivre Formula

Example 3. Let n

2 be an integer and

61

wn = e2". Show that

(1) m e Z and k = 0, 1, . . . , n - 1; k = 0, 1, . . . , n - 1; and (2) w£-fc = (3) 1 + wn + w2 4 + = 0 (for geometric meaning, see Fig. 1.29)

Proof. (1) and (2) are obvious. (3) Note that

— 1 = (bln — 1)(1 + =+ 1 + Wn + Or, let p = 1 + w" = 1 + wn + that p = 0.

+

•• •

1

' ‘

wn + • • • + w„_1. Then • + w"-1 = p. Hence

n + w2 • • • + w”-1 + p(wn — 1) = 0 and it follows

Example 4. Factorize the polynomial

x12 + x9 + x6 + x3 + 1 over the integer, the real and the complex number systems, respectively. Analysis: The original polynomial can be rewritten as (a:3)4 + (a:3)3 + (a;3)2 + a:3 + 1. Recall the formula a5 — 1 = (a — l)(a4 + a3 + a2 + a1 + 1). Hence

—=

r15 - 1 x6

a;12 x9 a:6 x3 1 —1 = + + + +

and try to factor the left side. Or, observe that 12 = 5x2 + 2, 9 = 5x1 + 4, 6 = 5 x 1 + 1. If we choose w as a primitive 5th root of 1, then

w12 + w9 + w6 + w3 + 1 = u,uV2 + wV + + w3 + 1 = w2 + w4 + w + w3 + 1 = 0, which means the original polynomial has a factor a;4

+ x3 + x2 + x + 1.

Solution. In C, z15 — 1 = 0 has roots e2^1, 0 k 14; while the roots corresponding to k = 0, 5, 10 are roots of z3 — 1 = 0. Hence, 14

(x - e2^’).

a:12 + a;9 + x6 + x3 + 1 = k k

=o 0,5,10

Chap. 1. Complex Numbers

62

By use of Example 3(2), e2isl(15 for such k,

(x

—

= e 2is‘ for fc = 0, 1, 2, . . . , 7. Hence,

— e 2^l) = x2

l)(x

x + 1.

Therefore, in R,

k

5

In Z g 12 z12 + x9 + x66 + x33 + 1 =

— 1 (a:5)3 — 1 = x6 — 1 —— x6 — 1 a;15

%5 — 1 a:10 + x3 + 1 „ t x — 1 a:2 + a: + 1

= (a;4 + x3 + x2 + x + 1) x (a;8 — x7 + x5 — x4 + x3 — x + 1), where the two factor polynomials in the right cannot be factored any more over Z (why?). Or, according to the Analysis above, divide a?12 + x9 + x6 + x3 + 1 by a?4 + x3 + x2 + x + 1 via long division.

Exercises A

(1) Try to set z = cosO + zsin0 in identities such as 1 + z + 1~1Z_Z > z 7^ 1, to prove the following identities. (a) i + ELicosfc0 =| + (b) (C) (d)

EL1 sinA:0 = ELi cos(2A: - 1)0 = ELi sm(2fc - 1)0 =

^±F,

•

•

+ zn =

02nz

i

’’’

i

aXnz

>

where An is the largest multiple of n, not larger than k. (17) (a) Suppose w is a primitive 5th root of 1. Show that (x + y + z)(x + wy + w4z)(z + w2y +

x (x + w4y + uz) = x5 +y5 + z5

^z^x + w3y + uz)

— 5x3yz + 5a:y2z2.

(b) Use (a) to solve a;5 — 5ax3 + 5a2x + (a5 + 1) = 0. (18) Prove the following factorizations, where a

0.

nZLi.1

(a) x2m — a2m = (x2 — a2) (x2 — 2aa; cos + a2). (b) x2m+1 — a2m+1 = (x — a) TlTL, (z2 — 2a;r cos + a2>) . (c) x2m + a2m =

n^o1 (*2

-

+

2«^ cos

n^o1 ^2 HZTo1

(d) z2m+1 + a2m+1 = (z + a) (e) x2m - ‘2amxm cos 6» + a2m =

-

.

2az cos “

2ax cos

+ . + °2)

•

Exercises B

(1) Factorize the following polynomials over Z, R, and C, respectively: (a) (b) (c) (d)

x5 + x4 + 1. x1 + x6 + x4 + 2x2 + 1. x8 + x6 + x4 + x2 + 1. a + (a + b)x + (a + 2b)x2 + (a + 3b)x3 + 36a:4 + 2bx5 + bx6.

Sec. 1.5. De Moivre Formula

65

(2) By expanding (cosa; + zsina;)m binomially and comparing the real and the imaginary parts of both sides, show that

W

52 (-1)^ cos^"1-2^ zsin2fc x,

cosmx =

fc=0

sinm2;

52

=

k=Q

where [y] denotes the largest integer not larger than y, etc. (3) (a) If m is an even integer and x E R, show that

.

9

sin

x

.2 (2fc+l)7T bm 2m

sin2 x

sinmx cos x

sin2

(b) If m is an odd integer and x

kit m

R, show that

.

cos mx cosz

sin

2

x

2 (2fc-l)7T sin m •

.

sin

sin mx = m sin x

2

x

sin2^

k=l

Note. These identities can be used to derive the infinite product expres¬ sions for cos x and sin#: oo

x2

cosz

(n -I- |)27t2

n—Q

(4) Let z = re10 . Then that:

oo

x2

sin x = x

n27r2

k=l

zn + z n = 2cosn0 and zn — z n = 2zsinn0. Show

(a) If n is an odd integer, c°s” =

°

1

n—1 2

5L ck cos(n

-

2k)0^

k=0

n—1

sin" 0 =

^(-1)^ £ (-l)fcC'fc sin(n fc=0

-

2k)0.

Chap. 1. Complex Numbers

66

(b) If n is an even integer, n-2

1

cos” 0 =

2

cos(n

1

— 2k)0 + — Cn ,

k=0 n-2

sin” 3 =

^-(-1)^ £(-l)fcCfc” cos(n - 2k)3 + k=0

(5) (a) If 0 < 0
R) is usually transformed to the study of the origin 0 and its neighborhood It is from the geometric and the point¬ (|z| < via the reflection z set aspects (Sec. 1.9), but not from the algebraic one, that we view C* as the Riemann sphere S. Section (2) The analytic expression of the stereographic projection $

In Fig. 1.31, let Q = (aq,^,^) G S and z = (x,y, 0) = x + iy. Recall that the north pole N = (0, 0, 1). Then

z, Q, and N are collinear. Xi

x

—0 —0

X2

y

—0 —0

£3

0

—! —1

Sec. 1.6. Spherical Representations of Complex Numbers O xi

=>

= Ax,

A

i+

= 1 — A for some scalar A = A2(z2 + y2) + (1 - A)2 = 1

X2 = Ay

x^ + x% + xl

69

and

x3

K

Summarizing, we have

+ 1X2 1 - X3

$(a;i, ^2,2:3) =

OO,

(xi,X2,x3) e ^{IV}, namely,

^^1

(zi, X2, x3) G S and x3 = 1, namely, (Xl, X2,X3) = (0,0, 1)

and

z+ z

+-10) =


S is the inverse of $ : S —> C*.

Section (3) The spherical distance on C*

“||” cannot be adopted as the distance on C* since each finite complex number z C has the same +00 distance to 00, namely, \z — 00 1 = +00. Now, for any two points zi and Z2 on C*, we define the distance between them as

d(z1,z2) = |$ ^zi)-^ \z2)|

= {fai - yi)2 + (^2 - y2)2 + (^3 - y3)2}5 , where $-1(zi) = {xi, X2, x3) and $-1(z2) = (yi,y2,y3), and it is called the spherical {chord) distance of zi and z2. In fact, it is the length of the chord connecting the points $-1(zi) and $-1(z2).

Chap. 1. Complex Numbers

70

Since

xl + + xl = yl + yl + Us = 1, via (1-6-7), it is easy to see that 2|21 - 22|

'

yoThnorw

Z1, Z2 G

(1.6.8)

2 zi

.7i+W

C,

C and 22 = 00

satisfying the following properties: for

2i,22,23 G

(1) d(zi,22) 0 and =0 21 and (2) d(z1,z2) = d(z2,zi). (3) d(2i,23) d(2i,22) +d(22,23).

22.

C*

Hence, d(, ) defines a metric on C* and, endowed with d(, ), C* becomes a metric space (refer to Exercise B(l) of Sec. 1.8). Note that d(0, 00) = 2. Section (4) Circle-preserving under $ The plane aiXi + a2x2 + 03X3 = ao, where 00,01,02, «3 G R and al + o2 + o3 = 1, in space R3 has a nonempty intersection with the interior xl + Z2 + ^3 < 1 °f 8, if and only if the distance from (0, 0, 0) to the plane is not greater than 1, namely, |ao| 1- Hence, the circle on the sphere S has the equation aiXi

+ a2x2 + 03X3 = ao,


(ao — a3)(ao + a3) = a^ — a2, (2) in (1.4.3.6) guarantees that it is either a point circle or a real circle. A reverse process says that, a circle or line on C* is mapped, via $-1, onto a circle on the sphere S.

Sec. 1.6. Spherical Representations of Complex Numbers

71

We summarize the above as Circle-preserving of the stereographic projection.

(1) The stereographic projection $ maps circles on the Riemann sphere S onto circles or lines on the extended complex plane C*, and conversely. (2) In particular, a circle on S passes through the north pole N O its stereographic image on C* is a line. (1.6.10) Hence, any line in the plane C should pass through the infinite point oo, and every open half-plane (see (1.4.3.1)) does not contain the point oo, which is merely a boundary point (see (1.8.6)). $ also preserves angles (see Exercises B(l)). Exercises A

(1) Prove (1.6.7) in detail. (2) Prove (1.6.8) in detail. (3) Let Qi and Q2 be two points on the sphere S, and Pi = zi and P2 = Z2 are their stereographic images under $, respectively. See Fig. 1.32. (a) Show that, as lengths of segments in R3,

NPj = (l + \zj\2)i and NQj = 2(1 + Iz/)^ for j = 1, 2.

(b) Show that ANQ1Q2 is similar to AAP1F2- Then, give a geometric proof of (1.6.8).

Fig. 1.32

Chap. 1. Complex Numbers

72

(4) Construct the stereographic projection of the sphere x? + x$ + (a?3 — |)2 = with its center at the north pole N = (0,0,1) and the coordinate plane Z3 = 0 as the complex plane. Recapture (1.6.7)(1.6.10) in this case. (5) Introduce the spherical coordinate zi

= cos p cos ip,

X3

= siny?,

x? = cos p simp,

in the sphere S: x^ + a;2 +x% = 1, where p represents the latitude and ip the longitude of a point (a:i, X2, x$) on S.

(a) Show that a point P = (p,ip) on S has its image, under the stereographic projection, on C the point /7T . . . ,. , z = (cos ip + 1 simp) tan

p\

j.

(b) Use (a) to prove (1.6.7). (6) Show that the spherical images, under $-1, of z and |, on the sphere S are symmetric with respect to the cria^-plane. (7) Pinpoint the spherical images, under $-1, of the following points:

z, — z, —z

z

1, z, —z,

—z

z

and compute their coordinates.

(8) Find the coordinates of the vertices of the following figures under (a) A cube inscribed to the sphere S, with sides parallel to the coordinate axis.

(b) A regular tetrahedron inscribed to S. (9) (a) Determine the center and the radius of the circle on S whose projection, under $, has the equation \z zo| = r. (b) Show that the circle on S in (a) is a great circle r2 = 1 + |zo |2(10) Find the spherical image of the ray Arg z = 0 (constant) under 4>-1.

—

Exercises B

(1) Angle-preserving of the stereographic projection (calculus is needed). A continuous mapping 7 : (—1,1) —> S is said to define a curve on S. The point set 7((—1, 1)) is usually called a curve on S and is still denoted

Sec. 1.6. Spherical Representations of Complex Numbers

73

by 7. Let

7W = (xi(t),x2(t),x3(t)), x^t)2 + x2(t>)2 + x3(t)2 = 1,

-l 00 and is denoted as lim

n

—

>00

zn = 2q

or

lim zn = 2q or

zn —> zq,

(1.7.1)

Sec. 1.7. Complex Sequences

77

if for any £ > 0, there exists a positive integer N = N(e) so that \zn — z0| < e as long as n N. A complex sequence zn 6 C is said to diverge to co or converge to oo in C* as n > oo and is denoted as

—

lim

n—>oc

zn = oo

or

limzn = oo or zn —> oo,

(1.7.2)

if for any R > 0, there exists a positive integer N = N(R) so that \zn\ > R, n>N. If a sequence zn C does not converge to any point zq g C or diverge to oo, then zn is called divergent. A convergent sequence zn has a unique limit and is bounded, i.e., there exists M 0 so that |zn| M for n 1. Also, a convergent sequence zn is Cauchy, i.e., for any £ > 0, there exists a positive integer N = N(e) so that for all m,n>N, \zm — zn\ < £ always holds. By use of (1.4.2.1), namely,

|Re zm

Re zn| , |Im zm

Im zn|

|zm

zn |

|Re zm

Re zn|

+ |Imzm - Imz„| and the completeness of the real R (see Appendix A) , it follows easily that every Cauchy sequence does converge (to a point) in C. Hence, the complex field C is complete as a metric space endowed with the metric || (refer to (1) in (1.9.3)). Now, here comes

The necessary and sufficient conditions for convergent sequence. Let and zq C. Then

zn

EC

(1) limn — Zn — Zq, o (2) limn_ooRezn = Rezo and limn_oo Im zn = Imz0; (3) In case z0 0, lim^oo \zn\ = |z0| and lim^oo argz„ = argz0. ,00

Note-. The last means that, for any preassigned value pn of arg zn, n 1, can be chosen so that lim

n—>oc

ipn =

po

ipo

o/argzo, a value

(1.7.3)

holds.

Proof. (1) 4^ (2) is an easy consequence of (1.4.2.1). By using polar forms of zn, (3) => (1) follows immediately. For (1) => (3), lim|zn| = |zq| follows from the inequality | |zn| — |zq|| \zn — zq|. What remains is to prove that arg zn —> argz0.

Chap. 1. Complex Numbers

78

—

Let us start from a concrete example. Set zn = — 1 + U J . Then zn > . = vr— Arctan -A and Argz2n+i = — 7r+Arctan Argzn does not converge and neither does argz„. Choose o = tt + 2mo?r (mo is a fixed integer), a value of arg(—1). Then, we choose

2n1+1

zq = —1. Since Argz2n

Arg z2m + 2m07r, Arg z2m+i

if n = 2m,

+ 2?r + 2m07r, if n = 2m + 1.

See Fig. 1.38. Under this circumstance ipn —> (/So¬ ln the general case, since zq 0, oo, then for all sufficiently small e > 0, the open disk \z z2| < |^o| sins' does not contain 0. See Fig. 1.39.

—

Fig. 1.39

Sec. 1.7. Complex Sequences

79

By assumption zn —» zq, there exists no so that \zn — zq| < |zq | sine, n > no. In case 0 Argzo tt and n — ip0 1 < s if n 2mo7r: Choose

{ Then | oo => zn + z'n -> oo; (ii) zn —» zq G C and zq 0, z'n —> oo => znz'n —> oo; (iii) zn —» z0 G C* and z0 0, z'n —> 0 => > oo; Zo G C,

z„

z^

—

oo

* 0-

(1-7-4)

Proofs are left as Exercise A(l). We illustrate two examples.

Example 1. Prove that, if z = x + iy,

—

lim (1H

n—-oo \

) nJ

= ez . = ex (cos y + i sin y) (def.)

(1.7.5)

Proof. Since lim,,^^ (1 + „) = 1, f°r anY 1 > e > 0, there exists that n no implies that |Arg(l + ^)| < s always holds.

no so

Chap. 1. Complex Numbers

80

For simplicity, let

zn = (1 + ^)n. Then

l^nl — => (by using L’Hospital’s rule on n) 1

lim log |zn| = lim

n—too

n—too

n2

lim |zn| = ex. => n—too

no,

On the other hand, if n Arg zn = nArc tan

IL n

l

+t

—>

—> oo

y as n

/

„ ,

recall that lim

\

e^o

\ Arctand = 1) . 0

Hence, Z\ n

(Id— ) 71 /

-A

lim \zn\elAlsz’' = exeiy = ex(cosy + isiny). = n—>oo

Example 2. Prove that, if z = x + iy lim n( y/z

n—>oo

/ 0,

— 1) = log \z\ + z(Arg2 + 2fc7r)

= log z, k = 0, ±1, ±2, . . ..

(def.)

(1.7.6)

\d

Proof. y[z = fixed k,

= Arg 2 and k = 0, ±1,±2, .... For any

— 1 = |z|" cos 0 +n2fc7T 1 + i|z|" sin 0 +n2fc?r 0 + 2kir \ 1| and => Re qz — 1 > = n z cos n / xfz

1

n(

n

\

Imn(oo

sin

1

0 + 2A:7r n

= log a (a

R and a > 0) or by using

lim / (|z| " — 1) + |z| " — 1) = n—too I

— |^| + = n

= log

0

— |^| (real n

log

0 2kTT cos + n

logarithm of |z|),

Sec. 1.7. Complex Sequences

81

while lim Im n(y/z n—^oo

— 1) = nlim |z|" (0 + 2&tt) — •

>oo

•

sin ®±2krr „ —£_+oo nk = oo. Then the sequence znk, k 1, is «2 < called a subsequence of the sequence zn, n 1. A bounded sequence, such as zn = in, is not necessarily convergent. But we do have

Let rik-k • ••

two basic facts about sequences. (1) A sequence zn converges to a point zq all the subsequence znk ,k >1, of zn converge to the same point zq. (2) (Completeness of C) Any bounded sequence has a convergent subse¬ quence (converging to some point in C). (1-7.7) For (2), refer to Sec. 1.9. Proofs are left as Exercise A(2). A point z g C* is called a limit point of a complex sequence zn, if there exists a subsequence znk of zn which converges to z.

Example 3. Show that every point on the unit circle [z[ = 1 is a limit point of the sequence Zn

= ein, n = 0, ±1,±2, . . ..

Proof. Let a be any irrational number. It is well known from elementary real analysis that the set {n + ma | n, m = 0, ±1, . . .} is dense in R (refer to Section (5) in Sec. 1.8), namely, for each x R, there exists a sequence I) from the set converging to x. xi, if k xk (xk In our case, choose a = 2tt. By Example 1, ez = e^^cosy + zsiny) is a continuous function of x and y. Let Xk = nk + 2mk7r, k 1 and Xk —> x. Then = emk —> etx as k —> oo.

Exercises A

(1) Prove (1-7.4) in detail. (2) Prove (1-7.7) in detail.

Chap. 1. Complex Numbers

82

(3) Try to use complex sequences to explain why we cannot define each of the following expressions as a definite value: 00

+ 00,00 — 00, 0

•

00,

(4) Try to use zn = inn to explain that

77,

0

—, oo°, 00

zn —>

00

I00.

does not necessarily

imply that

|Rezn|

—

and |Imzn|

> +00

— +00. »

00 that limn-^ |zn| = |zo| would imply limn^oo zn = z0. (6) Suppose zn » zg 0. Let zg 0 and zg is not a negative real number. Show that

(5) Show that it is only if zg = 0 or

—

Argz„

Argz0.

(8) For each of the following sequences, find its limit if it converges; oth¬ erwise, find the set of all its limit points. (a)

zn =

^+

inrn(]r\ < 1). (b)

log n! n! n 2n {0 is an irrational number which is not of the form

+ y/Si.

(e) zn = rv

eme

(-l)V1. zn = n+

(f) zn

(g) zn = kir, k = 0, ±1, ±2, . . .), n = 0, ±1, ±2, . . .. (9) Suppose zn —> zg C C and wn (a) Prove that

—> wg

G C.

z± + z^ + •

hm

•

+ zn

n

n^oo

= Zg.

What happens if zg = oc? (b) Prove that lim

n—too

21wn

+ Z2W-1 4

1- ZnWi

n

= ZgWg.

What happens if zg 6 C and wg = 00? (c) Suppose Ai > 0, ... , An > 0 and limn_>O0(Ai Show that

..hm A1Z1 + A222 + Al + A2 +

* * *

+

••

+ Xnzn zg. = + An

•

+ An) =

00.

Sec. 1.7. Complex Sequences

83

(10) Suppose a complex sequence zn satisfying limn_>oo(zn — zn-2) = 0. (a) Show that

zn

hm

Zn—1 fl

n^oo

= 0.

Is the converse true? (b) Show that lim

n—>oe

fl

= 0.

Is the converse true?

(11) In elementary real analysis, we learned the following limit processes: (a) limn^oo nan =0 (a g R and |a| < 1). (b) limn^oo y/a = 1 (a G R and a > 0). (c) limn^oo = 0 (a G R and a > 0). k (d) limn^oo = 0 (a 6 R and a > 1, k a fixed integer). prove Try to these statements by the e—N process. In case a is replaced by a fixed suitable complex number z, where in (d), z is supposed that |z| > 1, do the following questions. (i) Use the e N process to prove that these statements are still valid. Be careful that, in (b), yfz is a multiple-valued function. (ii) Use (1.7.3) to prove them again. Then, try to compare it to the

—

method in 1.

(12) If |zn+i — then

zn\

X\zn — zn-i\,n

1, where 0

< A < 1 is a constant,

zn converges.

Exercises B

Let zn & C, n 1 and Sn = then the complex series

n

57 Zn = ZI + Z2 H

F Zn H

1. If limn-^ Sn = S G C,

Or

n—1

57^

is said to converge to the sum S and is denoted as oo

57^ = n—1

or

52 zn = S.

(1.7.8)

Otherwise, we call 52 zn a divergent series. If 52 I zn I converges (in this case, zn is necessarily convergent) , we call 52 zn absolutely convergent; if 52 I zn| diverges but 52 zn converges, we call 52 zn conditionally convergent.

Chap. 1. Complex Numbers

84

Do the following problems.

(1) Basic criteria for convergence

57 zn converges. O 57 Re zn and 57 Im zn converge. (Cauchy condition) For any e > 0, there exists an integer N = N(e) > 0 so that l^n+i + • • • + ^n+p| < £ for all n N and all p 1. => lim zn = 0.

O

n—>oo

(2) Basic properties (a) For any integer N

1,

together.

53JX1 zn and

zn converge or diverge

52 zn = S => '^(azn) = aS, a & C. 52 zn = S, then for any sequence of integers 0 < ki < k? < < , kn < kn-j-1 bZfc2)d F(^fc„+iH FZfen+1)H — S. If 52 zn = s and 52 wn = S' ^(zn ± Un) = s ± S'. If both = S and 52 wn = S' absolutely, then ^2^=i(ziwn +

(b) If (c) If

• ••

• ••

(d) (e)

Z2Wn-i +

•

+ znwi) = SS'

absolutely.

(3) Some criteria for convergence (a) (comparison test) Suppose zn

then both

0, wn

0, n

1. If

52 \zn\ and 52 |wn| converge or diverge together.

(b) (ratio test) Suppose

lim lim

zn 0,n Zn

Zn

1.

< 1 => 57 Iznl converges; > 1 =>

57 |^n| diverges.

< 1 =>

57 lz«l converges;

(c) (root test) lim

liml/iTJ > 1 => 57 lz"l diverges.

Sec. 1.7. Complex Sequences

(d) (Abel test) Suppose

22 znwn

22 \zn —

Zn+i| and

85

^wn

converges, then

converges.

(e) (Dirichlet test) Let An If the partial sum sequence, namely, verges.

An+i

0 for n > 1 and limn^oo An = 0. zn is a bounded •• • then 22 con¬

+ zn,n > 1, of supn>1 |zi + + zn\ < oo,

Sn = z± +

(f) (Leibniz test) Let An An+i 0 for n > 1 and linin^oo A„ = 0, then 2222i(-1)n-1^ converges.

22^=1

zn = S absolutely O For any per¬ N —> N (a one-to-one and onto mapping), the series = S absolutely. (5) (Conditional convergence) Suppose zn converges conditionally C* is any fixed point, then there exists a permutation and S N so that 22^Li ^(n) = sa:N (6) Test if the following series are convergent or divergent. In case of con¬ vergence, does it converge absolutely or conditionally? (4) (Absolute convergence) mutation