Principal
Complex variables
Complex variables
Stephen D. Fisher
Hundreds of solved examples, exercises, and applications help students gain a firm understanding of the most important topics in the theory and applications of complex variables. Topics include the complex plane, basic properties of analytic functions, analytic functions as mappings, analytic and harmonic functions in applications, and transform methods. Perfect for undergrads/grad students in science, mathematics, engineering. A threesemester course in calculus is sole prerequisite. 1990 ed. Appendices.
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Año:
1999
Edición:
2nd ed
Editorial:
Dover
Idioma:
english
Páginas:
445
ISBN 10:
0486406792
ISBN 13:
9780486406794
Series:
Dover books on mathematics
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Complex Variables Second Edition Stephen D. Fisher DOVER BOOKS ON MATHEMATICS An Introduction to Linear Algebra and Tensors, M. A. Akivis and V. V. Goldberg. (635457) Vectors, Tensors and the Basic Equations of Fluid Mechanics, Rutherford Aris. (661105) Asymptotic Expansions of Integrals, Norman Bleistein and Richard A. Handelsman. (650820) NonEuclidean Geometry, Roberto Bonola. (6002Z0) Introduction to Partial Differential Equations, Arne Broman. (66158X) An Introduction to Ordinary Differential Equations, Earl A. Coddington. (659429) Matrices and Linear Transformations, Charles G. Cullen. (663280) Introduction to Nonlinear Differential and Integral Equations, H. T. Davis. (609715) Some Theory of Sampling, W. Edwards Deming. (64684X) Statistical Adjustment of Data, W. Edwards Deming. (646858) Introduction to Linear Algebra and Differential Equations, John W. Dettman. (651916) Linear Programming and Economic Analysis, Robert Dorfman, Paul A. Samuelson and Robert M. Solow (654915) Theory of Hp Spaces, Peter L. Duren. (411842) The Thirteen Books of Euclid's Elements, translated with an introduction and commentary by Sir Thomas L. Heath. (600882, 600890, 60090^4) Threevolume set Calculus of Variations with Applications, George M. Ewing. (648567) Differential Forms with Applications to the Physical Sciences, Harley Flanders. (661695) An Introduction to the Calculus of Variations, Charles Fox. (654990) Foundations of Modern Analysis, Avner Friedman. (640620) Technical Calculus with Analytic Geometry, Judith L. Gersting. (67343X) Introduction to Difference Equations, Samuel Goldberg. (650847) Probability: An Introduction, Samuel Goldberg. (652521) Differential Geometry, Heinrich W. Guggenheimer. (634337) Numerical Methods for Scientists and Engineers, Richard Hamming. (652416) Probability: Elements of the Mathematical Theory, C. R. Heathcote. (41149^4) Ordinary Differential Equations, E. L. Ince. (603490) Lie Algebras, Nathan Jacobson. (63832^4) Greek Mathematical Thought and the Origi; n of Algebra, Jacob Klein. (272893) Theory and Application of Infinite Series, Konrad Knopp. (661652) Applied Analysis, Cornelius Lanczos. (65656X) (continued on back flap) COMPLEX VARIABLES Second Edition m i ^ i m Stephen D. Fisher Northwestern University DOVER PUBLICATIONS, INC. Mineola, New York Copyright Copyright © 1986, 1990 by Stephen D. Fisher All rights reserved under Pan American and International Copyright Conventions. Bibliographical Note This Dover edition, first published in 1999, is a slightly corrected, unabridged republication of the work originally published in 1990 by Wadsworth & Brooks, Pacific Grove, California. Library of Congress CataloginginPublication Data Fisher, Stephen D., 1941 Complex variables / Stephen D. Fisher. — 2nd ed. p. cm. "This Dover edition, first published in 1999, is a slightly corrected, unabridged republication of the work originally published in 1990 by Wadsworth & Brooks, Pacific Grove, California"—T.p. verso. Includes index. ISBN 0486406792 (pbk.) 1. Functions of complex variables. I. Title. QA331.7.F57 1999 515\9^dc21 9913026 CIP Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y 11501 For my patents with love and appreciation Preface to the Second Edition I was gratified by the very positive responses and reviews the first edition of this text received. Nonetheless, I wished to modify the text or exercises at a number of points; this second edition is the result. The fundamental purpose of the text has not changed—it still provides a direct route, with a minimum of prerequisites, to the most important topics in the basic theory and applications of complex variables. The greatest changes from the first edition occur in Chapter 2, principally in the presentation of the allimportant Cauchy's Theorem. In Section 3,1 give only the Green's Theorem proof of Cauchy's Theorem (which assumes the continuity of the derivative); the Goursat version of Cauchy's Theorem is now in the special starred subsection 2.3.1. In the remainder of Section 3,1 discuss simple connectivity (but not starshaped domains) and derive some of the consequences of Cauchy's Theorem. These parallel developments, one assuming continuity of the derivative and the other not, are brought together in the first theorem of Section 4, where I derive the powerseries representation of an analytic function. In Sections 5 and 6 of Chapter 2,1 have made other rearrangements and clarifications in the presentation of residues, Laurent series, and the use of the Residue Theorem to evaluate definite integrals. In Chapter 4,1 have made a number of alterations in the discussion of flows. Moreover, I have modified the text and many of the exercises, throughout—adding here and deleting or rearranging there—all for added clarity. Otherwise, I have done my best to correct typographical errors that were brought to my attention by readers and teachers of the first edition. I would like to thank several people who heeded my invitation to send me suggestions for improved exposition on one topic or another; their comments and discoveries were instrumental in the development of this edition. My colleagues at Northwestern, John Franks and Sandy Zabell, were especially generous with their time in this regard. Other important comments came from Ireena Erteza, Albert G. Fadell, Thomas W. Gage, Jeffrey Nunemacher, James Okon, and Richard Troxel. Additional comments from Martin Billik, R. Carmichael, F. W. Carroll, Fred Goodman, Abel Klein, Steven Krantz, Michel Lapidus, A. Matheson, Philip McCartney, Justin Peters, Walter Rudin, J. C. Taylor, John Wermer, Alvin White, Robert Whitley, and Joan Wyzkoski were also helpful. I would like to thank the manuscript reviewers: Frank Forelli, VII viii Preface to the Second Edition University of Wisconsin; Tom Garrety, Williams College; B. Frank Jones, Jr., Rice University; YueKuen Kwok, Hong Kong University of Science and Technology; David Minda, University of Cincinnati; Burt Rodin, University of California, San Diego; and Mitchell Tableson, Washington University. Once again it is a pleasure to thank my editor at Wadsworth & Brooks/Cole, John Kimmel, for his sustained interest and thoroughly professional approach. As before, I will appreciate it if readers call my attention to any typographical and mathematical errors. Stephen D. Fisher Preface to the First Edition This textbook is intended for undergraduate or graduate students in science, mathematics, and engineering who are taking their first course in complex variables. Its only prerequisite is a threesemester course in calculus; no prior knowledge of Green's Theorem or line integrals is needed. A previous course in differential equations would be useful but is not necessary. The presentation and level of rigor in this book fit this background in several different ways. First, I have not given definitions and theorems in their greatest possible generality, and I have presented only results that are of central importance to elementary complex variables. (However, many important secondary topics are to be found in the exercises that appear at the end of each section.) Second, although virtually all of the theorems are proved in full there are a few places where I refer the reader to other sources for a fact or a complete proof (for instance, the theorem that a continuous realvalued function on a closed and bounded set attains its maximum and minimum). Third, the presentation in Chapters 1 and 2 emphasizes areas of complex variables that have much in common with concepts that the student has studied before; for instance, limits of sequences, continuity of functions, and convergence of series. The exponential function is developed from its representation in terms of ex, siny, and cosy so that the known continuity and differentiability of these functions can immediately be brought to bear. Furthermore, once analytic functions are defined and their basic properties are developed, I devote a section to complex power series; this includes a derivation of the fact that a convergent power series is an analytic function within its disc of convergence, whose derivative is given by the expected power series. This fact can be obtained later by a different and shorter route, but I have found it more sound pedagogically to begin in areas in which the student has had some previous experience and later move into newer areas. Fourth, there is a plethora of solved examples (more than 220 altogether) in the text, which illustrate exactly how each concept or theorem is to be applied. In addition to the examples, there are numerous exercises at the end of each section, a total of 730 throughout the book. The easiest reinforce basic concepts, while the more challenging extend or expand upon themes in the section or relate ideas from that section to earlier sections. Sections and topics within the exercises marked with IX x Preface to the First Edition a star can be omitted with no loss of generality; they are not needed for future developments. There is more material in this book than can be included in a course lasting a single quarter or semester. In teaching this material at Northwestern University I have found that in a tenweek quarter I can cover all of Chapters 1,2, and 3, and many topics from the final two chapters. The variety and extent of the applications and techniques presented in Chapters 4 and 5 allow the instructor to pick his or her favorites and permit ambitious students to read others on their own. Complex variables is, simultaneously, a practical tool of great utility in the hands of a skilled practitioner and a mathematical structure of enormous beauty and elegance. I hope that readers of this book will retain a residue of each of these facets of the subject. This preface would not be complete without acknowledging those people who assisted me during the writing of this book. Professor Robert B. Burckel, Kansas State University, provided me with an infinitude of improvements in the grammar and style of the first draft. Various other helpful stylistic and pedagogical suggestions were made by Professors M. D. Arthur, Michael Beals, Steven Bell, Carlos Berenstein, Bruce Berndt, James Brennan, David Colton, Michael Cullen, Abel Klein, Steven Krantz, Michael O'Flynn, James Osborn, Kent Pearce, Thomas Porsching, Glenn Schober, Daniel Shea, and Howard Swann. The computer graphics were produced by Benjamin Slivka using the Control Data CYBER 170 model 730 at Northwestern University's Vogelback Computing Center. Randall Kamien provided detailed solutions to the exercises in Chapters 1,2 and 3; his contribution is very much appreciated. I also want to express my thanks to Professor Ezra Zeheb of the Department of Electrical Engineering, Technion, Haifa, Israel, for two very informative conversations on the Ztransform and related matters. I also had helpful talks with Professor Danny Weiss of the Department of Aeronautical Engineering, Technion, and Professor Nadav Liron of the Department of Mathematics, Technion. They all helped to increase my knowledge of the applications of complex variables. Special thanks go to my editor at Wadsworth & Brooks/Cole, John Kimmel, who not only got me started on this book but stuck with me in what at times seemed to be an endless task. Michael Michaud of Unicorn Productions saw the book through its production; his technical expertise is in evidence throughout. Typographical errors and mathematical slips are the bane of any author. I would appreciate any such being called to my attention. Stephen D. Fisher A Note to the Student This textbook presents an introduction to the theory and applications of complex variables. The presentation has been molded by my belief that what you have already studied in calculus can be successfully applied to learning complex variables, which at its basic level is just the calculus of complexvalued functions. Where there are strong, and even obvious, analogies between the new material and calculus, I have pointed them out and arranged the presentation to emphasize these analogies. However, there are critical points at which the study of complex variables differs intrinsically from the calculus that you know, and at these points I have provided more details to explain the new material. At all times, your comprehension of the subject will be aided greatly by reading with a pencil and paper close at hand. Write things down, fill in computations that may be omitted or only partially worked out, and work through the examples by yourself. By all means, do as many exercises (assigned or not) as you can. Mathematical knowledge is not gained passively—you must be an active participant in the learning process. xt This page is intentionally left blank Contents 1 The Complex Plane 1.1 The Complex Numbers and the Complex Plane 1 1.1.1* A Formal View of the Complex Numbers 10 1.2 Some Geometry 12 1.3 Subsets of the Plane 22 1.4 Functions and Limits 30 1.5 The Exponential, Logarithm, and Trigonometric Functions 43 1.6 Line Integrals and Green's Theorem 56 2 Basic Properties of Analytic Functions 77 2.1 Analytic and Harmonic Functions; The CauchyRiemann Equations 77 2.1.1* Flows, Fields, and Analytic Functions 86 2.2 Power Series 93 2.3 Cauchy's Theorem and Cauchy's Formula 106 2.3.1 * The CauchyGoursat Theorem 119 2.4 Consequences of Cauchy's Formula 123 2.5 Isolated Singularities 135 2.6 The Residue Theorem and Its Application to the Evaluation of Definite Integrals 153 3 Analytic Functions as Mappings 171 3.1 The Zeros of an Analytic Function 171 3.1.1* The Stability of Solutions of a System of Linear Differential Equations 183 3.2 Maximum Modulus and Mean Value 191 XIII xiv Contents 3.3 Linear Fractional Transformations 196 3.4 Conformal Mapping 208 3.4.1* Conformal Mapping and Flows 219 3.5 The Riemann Mapping Theorem and SchwarzChristoffel Transformations 224 4 Analytic and Harmonic Functions in Applications 245 4.1 Harmonic Functions 245 4.2 Harmonic Functions as Solutions to Physical Problems 254 4.3 Integral Representations of Harmonic Functions 284 4.4 BoundaryValue Problems 298 4.5 Impulse Functions and the Green's Function of a Domain 309 5 Transform Methods 318 5.1 The Fourier Transform: Basic Properties 318 5.2 Formulas Relating u and d 335 5.3 The Laplace Transform 346 5.4 Application: of the Laplace Transform to Differential Equations 356 5.5 The ZTransform 365 5.5.1* The Stability of a Discrete Linear System 374 Appendix 1 Locating the Zeros of a Polynomial 381 Appendix 2 A Table of Conformal Mappings 389 Appendix 3 A Table of Laplace Transforms 395 Solutions to OddNumbered Exercises 397 Index 425 COMPLEX VARIABLES Second Edition 1 The Complex Plane 1.1 The Complex Numbers and the Complex Plane The theory and utility of functions of a complex variable ultimately depend in large measure on viewing the usual x and ^coordinates in the plane as separate components of a single new variable, the complex variable z. This new variable z can then be manipulated in the same way as conventional numbers are. The familiar numbers, such as —1,3, y/2, and 7r, which are represented by points on a line, will be referred to as real numbers. A complex number is an expression of the form z = x + iy, where x and y are real numbers and i satisfies the rule (0a = (0(0 — i. The number x is called the real part of z and is written x = Re z. The number y, despite the fact that it is also a real number, is called the imaginary part of z and is written y = Im z. Thus, for instance, we have 1 = Re(l + 3f) and 3 = Im(l + 3J). The modulus, or absolute value, of z is defined by \z\ = y/x2 + y2, z = x + iy. Each complex number z = x + iy corresponds to the point P(x, y) in the xyplane (Fig. 1.1). The modulus of z, then, is just the distance from the point P(x, y) 1 2 Chapter 1 The Complex Plane z = x + ly Figure 1.1 to the origin, which in complexnumber notation is 0. In this way, we see that we have three inequalities relating x, y, and z, namely x ^ z, \y\ ^ z, and \z\ ^ \x\ + \y\. The first two of these are obvious; the third is obtained by noting that \z\2 = x2+y2^x2 + 2\x\ \y\ + y2 = (\x\ + y)2. The complex conjugate of z = x + iy is given by z = x — iy. Occasionally in engineering books, one encounters the notation z* for z, as well as the use of j instead of /; we shall not use either of these. For the specific complex number z = 1 + 3/, z = 1  3/ and \z\ = y/l6 (see Fig. 1.2). 1 + 3/ h  1  3/ ■ • 1  3/ = z Figure 1.2 1.1 The Complex Numbers and the Complex Plane 3 Addition, subtraction, multiplication, and division of complex numbers follow the ordinary rules of arithmetic. (Keep in mind that i2 = — 1, and, as usual, division by zero is not allowed.) Specifically, if z = x + iy and w = s + it, then z + w = (x + s) + i(y + t) z — w = {x — s) + i(y — t) zw = (xs — yt) + i(xt + ys) z wz (xs + yt) + i{ys — xt) — = — = 2 2 > w/0. Here, to obtain the formula for the quotient of z and w we used the device of multiplying both numerator and denominator by w = s — it. Example 1 To illustrate these rules in a particular case, let us take z = 1 + 3/ and w = — 2 + 5i. Then z + w = 1 + 8/ zw = [(l)(2)  (5)(3)] + i[(l)(5) + (3)(2)] = 17  i z _ wz _ 13 Hi _ 13 11. w~tixv~ 29 ~ 29 ~ 29*' Two facts are of particular importance. The first, which is used repeatedly, is that zz = (x + iy)(x  iy) = x2 + y2 = \z\2. The second is that z and z have the same absolute value: Z = v/?Trr = in Another useful relation is derived with the following computation. \zw\2 = (xs  yt)2 + (xt + ys)2 = x2s2 + y2t2 + x2r2 + y2s2 = (x2 + y2)(s2 + t2) = z2w2, 4 Chapter 1 The Complex Plane so after taking square roots, we obtain \zw\ = M kiln a similar vein, we compute zvv: zw = {xs — yt) — i(xt + ys) = (x — iy)(s — it) = zvv. Polar Representation The identification of z = x + iy with the point P(x, y) in the xyplane has further interest and significance if we make use of the usual polar coordinates in the xyplane. The polar coordinate system gives x = r cos 9 and y = r sin 0, where r = yjx2 + y2 and 9 is the angle measured from the positive xaxis to the line segment from the origin to P(x, y). We immediately see that r = z, so z = z(cos 9 + / sin 9). This is the polar representation of z (see Fig. 1.3a). Example 2 Find the polar representation z = — 1 + /. Solution \z\ = yjl and 9 = 3tt/4. Thus, ■ 1 + i = y/l\ cos — + i sin — , which you can easily verify as correct, since cos %n = —\^/l and sin \n = i>/2 (see Fig. 1.3b). n By now you have probably noticed that 9 could equally well be replaced in the formulas by 9 + 2rc, by 9 — An, or, indeed, by 9 4 2nn, where n is any integer. This ambiguity about the appropriate angle to use in the polar representation of a complex number is not just a question of semantics. Later we shall see that this causes some fundamental problems. Putting this aside for the moment, let's proceed with other properties of the polar representation. Suppose that z = z(cos 9 + / sin 9) and w = w(cos \j/ + / sin ij/) 1.1 The Complex Numbers and the Complex Plane 5 z =  z  (cos 6 + i sin 0) Figure 1.3 are two complex numbers. Then zw = z w{(cos 0 cos ^ — sin 0 sin ij/) 4 /(cos 0 sin ^ 4 cos ^ sin 0)} = zw{cos(0 + ^) + * sin(0 4 ^)}. Moreover, z \z\(cos 9 + i sin 9) w w(cos ^ + /sin ^) = f — ) {cos 9 cos ^ 4 sin 9 sin ^ 4 /(cos ^ sin 0 — cos 9 sin ^)} = ^{cos(0iA) + /sin(0iA)}. Here we have made use of the trigonometric identities for the sine and cosine of the sum and difference, respectively, of two angles. Hence, the polar representation of the product (or quotient) of two complex numbers is found by multiplying (or dividing) their respective moduli and adding (or subtracting) their respective polar angles (Fig. 1.4). In other words, multiplying w by z = z(cos 9 + i sin 9) produces a rotation of w in the counterclockwise direction of 9 radians and stretches (or shrinks)  w by a factor of z. Example 3 Find the polar representation of zw and z/w if z = — 1 4 i and w = Solution From Example 2, the polar representation of z is = ^2^08 — + * sin —J. 6 Chapter 1 The Complex Plane zw Figure 1.4 The polar representation of w is w = ^3 + i = 2 cos  + i sin  , since w = 2 and \jj = n/6. Therefore, (v/3  1) + i(y/3  1) = zw = 2^2^08^ + J sin i^l ^V + ^V = w=TLCOST2 + 'smT2} The foregoing is now used to derive De Moivre's Theorem*: (cos 9 + i sin 9)n = cos n9 + i sin n0, for any positive integer n and any angle 0. The formula is clearly true when n = 1; we shall use mathematical induction to prove it true for all n. Suppose there is a positive integer m for which Then (cos 6 + i sin 9)m = cos m9 + i sin m9. (cos 9 + i sin 0)m+1 = (cos 9 + i sin 0)m(cos 9 + i sin 9) = (cos m0 + / sin m0)(cos 0 + / sin 0) = cos(m + 1)0 + i sin(m + 1)0, * Abraham De Moivre, 16671754. 1.1 The Complex Numbers and the Complex Plane 7 by invoking the formula derived above for the polar representation of the product of two complex numbers. Thus, if the equality holds for m, then it holds for m + 1. Since we know it is true for n = 1, it is true for all positive integers n. Example 4 Let 0 = n/4; then cos 0 = sin 0 = ^2/2. Thus, A/5 V2Y ( n (2 +I2j =(COS4 + i sin — ) = cos n + i sin n = — 1. Example 5 De Moivre's Theorem can be used to derive trigonometric identities for cos n9 and sin n6. For instance, by cubing, (cos 0 + i sin 0)3 = cos3 9 + 3i cos2 0 sin 9 — 3 cos 9 sin2 0 — 1 sin3 9 = (cos3 0  3 cos 9 sin2 9) + i(3 cos2 0 sin 9  sin3 0). However, De Moivre's Theorem gives (cos 9 + / sin 0)3 = cos 30 + i sin 30. After equating the real and imaginary parts of these expressions, we find that cos 30 = cos3 0 — 3 cos 0 sin2 0 = 4 cos3 0 — 3 cos 0 sin 30 = 3 cos2 0 sin 0 — sin3 0 = 3 sin 0 — 4 sin3 0. Similar formulas can be derived for cos 40, sin 40, cos 50, and so on. (See Exercise 19.) □ We define an argument of the nonzero complex number z to be any angle 0 for which z = z(cos 0 + i sin 0), whether or not it lies in the range [0, 2n); we write 0 = arg z. To repeat, arg z = 0 is equivalent to z = z(cos 0 + i sin 0). A concrete choice of arg z is made by defining Arg z to be that number 0O in the interval [ — ft, n) such that z = z(cos 0O + i sin 0O). We may then write Arg(zw) = Arg z + Arg w (mod 2n% where the expression (mod 2n) means that the two sides of this last formula differ by some integer multiple of 2n. For example, if z = — 1 + i and w = f, then 8 Chapter 1 The Complex Plane zw = — 1 — i, so Arg(zw) = — 3rc/4. However, Arg z = 3rc/4 and Arg w = n/2, so Arg z + Arg vv = (5/4)7r = — 3n/4 + 27i. Complex Numbers as Vectors If z = x + ry and w = s + it are two nonzero complex numbers, then \z  w = V(x  s)2 + (y  I)2 is nothing but the distance in the xyplane from the point P(x, y) to the point g(s, 0 Moreover, the sum z + w = (x + s) + i(y + t) and the difference z — w = (x — s) + i(y — t) correspond exactly to the addition and subtraction of the vectors OP and OQ (Fig. 1.5). Z  vv w P(*. y) O Figure 1.5 Note also that the angle a between the vector OP and the vector OQ is found by using the usual dot product of two vectors: cos a = (OP OQ)/OPOQ _ xs + yt J*2 + y 2y/s2 + t2 _ Re(zw) ~ \z\\w\' In particular, OP and OQ are perpendicular if and only if Re(zw) = 0. The relations among the lengths of the sides of the triangle formed by z, vv, and z — vv, which is just the law of cosines, is formulated here as \z  w\2 = (x  s)2 + (y  t)2 = x2 + s2 + y2 + t2  2(xs + yt) = z2 + w22Re(zw). 1.1 The Complex Numbers and the Complex Plane 9 In summary, the usual xyplane has a natural interpretation as the location of the complex variable z = x + iy, and all the rules for the geometry of the vectors P(x, y) can be recast in terms of z. Henceforth, then, we refer to the xyplane as the complex plane, or simply, the plane. The xaxis will be called the real axis, and the >'axis will be called the imaginary axis. EXERCISES FOR SECTION 1.1 1. Let z = 1 + 2/, w = 2  i, and £ = 4 4 3i. Compute (a) z + 3w; (b) 2w + C; (c) z2; (d) w3 + w; (e) Retf"1); (f) w/z; (g) C2 + 2C + 3. 2. Use the quadratic formula to solve these equations; express the answers as complex numbers, (a) z2 + 36 = 0; (b) 2z2 + 2z + 5 = 0; (c) 5z2 + 4z + 1 = 0; (d)z2z= l;(e)z2 = 2z. 3. Sketch the locus of those points w with (a)  w\ = 3;(b)w  2 = l;(c)w + 22 = 4; (d) w + 2 = w2; (e) w2  2w  1 = 0; (f) Re[(l  i)z] = 0; (g) Re[z/(1 + i)] = 0. 4. Find Re(l/z) and Im(l/z) if z = x + iy, z ^ 0. Show that Re(iz) = — Im z and Im(iz) = Re z. 5. Give the polar representation for (a) — 1 + i; (b) 1 + iy/3; (c) —/; (d) (2 — i)2; (e) 4 + 3/; (f) ^5  i; (g) 2  2i; (h) ^2/(1 + i* (i) [(1 + O/x/2]4. 6. Give the complex number whose polar coordinates (r, 9) are (a) (y/3, n/4); (b) (1/^2, n); (c) (4, 7t/2); (d) (2, */4); (e) (1,4*); (f) (^2, 9ji/4). 7. Let a, fe, and c be real numbers with a ^ 0 and b2 < 4ac. Show that the two roots of ax2 + bx f c = 0 are complex conjugates of each other. 8. Suppose that A is a real number and £ is a complex number. Show that z2 + >l2 = z + ,422Re(>lz) and z2 + 2 Rq(Bz) = z + £2  B2. 9. Show that z = 1 if and only if 1/z = z. 10. Let z and w be complex numbers with zw = 0. Show that either z or w is zero. 11. Show that z + w2 — z — w2 = 4 Re(zvv) for any complex numbers z, w. 12. Let zl5 z2, ..., zn be complex numbers. Establish the following formulas by mathematical induction: (a) z1z2...zj = z1z2...zw (b) Re(zx + z2 + • • • + z„) = Re(z!) + Re(z2) + • • • + Re(z„) (c) Im^i +z2 +  + zH) = ImCzJ + Im(z2) f • • • + Im(zw) (d) z1z2...zn = zlz2...zn. 13. Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i, 2 + 2/, 5 + i; (b) 2 + i, 3 + 5i, 4 + i; (c) 6 + 4/, 7 + 5i, 8 + Ai. 14. Show that cos 8 = cos \// and sin 6 = sin i/< if and only if 0 — i/< is an integer multiple of 2n. 15. Show that the triangle with vertices at 0, z, and w is equilateral if and only if z2 = w2 = 2Re(zw). Chapter 1 The Complex Plane 16. Let z0 be a nonzero complex number. Show that the locus of points tz0, — oo < t < oo, is the straight line through z0 and 0. 17. Show that if w ^ 0, then \z/w\ = \z\/\w\. 18. Prove the identity 1 + z + z2 + ••• + zn = (1  zn+1)/(l  z) valid for all z, 19. Show that cos n0 can be expressed as a combination of powers of cos 9 with integer coefficients. (Hint: Use De Moivre's Theorem and the fact that sin2 9 = 1  cos2 0.) The Schwarz* Inequality 20. Let B and C be nonnegative real numbers and A a complex number. Suppose that 0 ^ B — 2 Re(XA) + A2C for all complex numbers X. Conclude that A\2 ^ BC. (Hint: If C = 0, show that A = 0. If C # 0, then choose X = 4/C.) 21. Let fll9 ..., a„ and bl9 ..., fc„ be complex numbers. Establish the Schwarz inequality: "MIM (Hint: For all complex numbers X, we have 0 ^ £"=1 \a} — Afyl2. Expand this and apply Exercise 20 with A = Yj=i afy, B = £j=1 a,.2, C = £j=1 Ifyl2 22. Verify the Schwarz inequality directly for the case n = 2. 23. When does equality hold in the Schwarz inequality? 24. Use the Schwarz inequality to establish that (Hint: Expand £"=1 ay + bj\2 and apply the Schwarz inequality.) A Formal View of the Complex Numbers* The complex numbers can be developed in a formal way from the real numbers. A complex number z is defined to be an ordered pair (x, y) of real numbers; we write z = (x, y). Two complex numbers zx = (xl9 yx) and z2 = (x2, y2) are equal when xx = x2and>>1 = y2. The basic arithmetic operations of addition and multiplication are defined, respectively, by, addition: zx + z2 = (xx + x2, ^t + y2) (1) multiplication: zxz2 = (xxx2 — yxy2^ X\y2 + xiy\) (2) * Hermann Amandus Schwarz, 18431921. 1.1.1 A Formal View of the Complex Numbers 11 The additive identity is 0 = (0, 0), since (0, 0) + (x, y) = (x, y) + (0, 0) = (x, y) for all (x, y). The multiplicative identity is 1 = (1, 0), since (1, 0)(x, y) = (x, y)(l, 0) = (x, y) for all (x, y). Further, it is elementary but somewhat tedious to show that the arithmetic operations of addition and multiplication are commutative: Zi r Z2 == ^2 ' ^l» ^1^2 = ^2^1 and associative: (zx + z2) + z3 = zx + (z2 + z3); (ZiZ2)*3 = z1(z2z3). Each z = (x, y) has a unique additive inverse — z = ( —x, — y), since z + ( — z) = 0. A nonzero z = (x, y) necessarily satisfies the condition x2 + y2 > 0, and its unique multiplicative inverse is since (z)(z"1) = (l,0) = l. The mathematical system of the complex numbers so constructed is one example of a field. There are many other examples of fields besides the complex numbers; for instance, the real numbers themselves form a field, as do the rational numbers. The complex number (0, 1) has the interesting property that its square is 1: (0, 1)(0, 1) = (1, 0). Further, (0, 1) and (0, 1) are the only two complex numbers with this property (see the exercises that follow). We denote (0, 1) by the symbol /. Each complex number then can be written z = (x, y) = (x, 0) + (0, y) = (x,0) + (y,0)(0,l) = (x, 0) + (y9 0)/. Complex numbers of the form (a, 0) are just the real numbers with their usual rules of arithmetic: (<i,0) + (&,0) = (fl + &,0) (a,0)(b,0) = (ab,0), and it is entirely natural to identify (a, 0) with a. In this way we may write 2 = (*, y) = (*, 0) + i(y9 0) = x + iy. This brings us back to the point where Section 1 began. 12 Chapter 1 The Complex Plane EXERCISES FOR SECTION 1.1.1 Throughout, the usual rules of arithmetic are assumed for the real numbers; in particular, a2 > 0 for any nonzero real number a. 1. Show directly from rule (2) for multiplication that z2 = ( — z)2. 2. Suppose that z = (x, y) and z2 = (— 1, 0). Show that z = i or z = — i. 3. Solve the equation z2 = (0, 1). 4. Suppose that z2 is real and negative; that is, z2 = (a, 0), a < 0. Show that z = (0, b) and find b in terms of a. 5. Show by computation that addition of complex numbers is associative: (zl + z2) + z3 = zx 4 (z2 + z3); and commutative: zx + z2 = z2 + zx. 6. Show by computation that multiplication of complex number is associative: (z1z2)z3 = z1(z2z3); and commutative: zxz2 = z2zx. 7. Define the absolute value, z, of z = (x, y) by z = y/x2 + y2. Show directly thatz1z2 = z1z2. 8. Define the complex conjugate, z, of z = (x, y) by z = (x, — y). Show that zz = (z2,0). 9. Show that zxz2 = 0 implies that either zx or z2 is zero. 10. Let z = (x, y). Show that (a) x ^ z; (b) \y\ < z; (c) \z\ ^ x + \y\. 1.2 Some Geometry The Triangle Inequality Let us begin with an important inequality that has a simple geometric interpretation. Suppose z = x + iy and w = s + it are two complex numbers. Then \z + w\2 = (x + s)2 + (y 4 02 = x2 + s2 + y2 + t2 + 2(xs + jtf) = z2 + w2 + 2Re(zvv) ^z2 + w2 + 2zw = z2 + w2 + 2zW = (M + w)2. Taking the square root of both sides yields the inequality z + w ^z + w. This is the triangle inequality, since it simply expresses the fact that any one side of a triangle is not longer than the sum of the lengths of the other two sides (see Fig. 1.6). 1.2 Some Geometry 13 An illustration of the triangle inequality: Z + w  <  z  +  w  Figure 1.6 If C and £ are two (other) complex numbers, then by putting z = £ — £, and w = £wegetC<C£l + mor C{XCfl. Likewise, iaici<ica which together yield a variation of the triangle inequality, ICI{<IC{. Straight Lines The equation of a (nonvertical) straight line, y = mx + b, m and b real, can be formulated as 0 = Re((m + i)z + b). More generally, if a = ^4 + iB is a nonzero complex number and b is any complex number (not just a real number), then 0 = Re(az + fc) is exactly the straight line Ax — By + Re(fe) = 0; this formulation also includes the vertical lines, x = Re z = constant. (See Fig. 1.7.) Roots of Complex Numbers The computation of the fractional powers of a nonzero complex number is possible with the techniques developed in Section 1 of this chapter. It was in an attempt to 14 Chapter 1 The Complex Plane Re (1 Figure 1.7 find the roots of such equations as x2 + 1 = 0 that the whole subject of complex numbers first arose; here, a certain "completeness" will be evidenced by the complex numbers but not by the real numbers. Suppose w is a nonzero complex number and n is a positive integer. A complex number z satisfying the equation zn = w is called an nth root of w. We shall determine all the distinct nth roots of w. Let w =  w(cos ij/ f i sin i//) be the polar representation of w, where we specify that \// lies in the range [ — n, n). Let z = z(cos 0 + i sin 0); the relation zn = w and De Moivre's Theorem from Section 1 then yield three equations: \z\n = w, cos(n0) = cos ^, and sin(n0) = sin ij/. Thus, we must have \z\ =  w1/n; 0 is not so well determined. Of course, one possibility for 0 is 0 = ^/n; however, there are others. We define % **'(t> fc = 0, !,...,« 1. Then ndk — if/ + Ink and so cos nOk = cos \j/ and sin n9k = sin ^. Complex numbers z0,..., zn_x are defined by the rule zk =  w1/w(cos 0k + i sin 0k), k = 0, 1,..., n  1. Then each of z0,..., zn_x is distinct (see Exercise 14, Section 1), and each satisfies zjj = w, k = 0,1,..., n — 1. Moreover, these complex numbers z0,..., zn_x are the only possible roots of the equation zn = w. For if cos nd = cos ij/, sin n0 = sin ty, 1.2 Some Geometry 15 then (again by Exercise 14, Section 1) we have n0 = \j/ + 2nj for some integer j. The values j = 0,..., n — 1 yield distinct numbers cos 0} + i sin 0p whereas other values of j just give a repetition of numbers already obtained. The geometric picture of the nth roots of w is very simple: the n roots lie on the circle centered at the origin of radius p =  vv1/w; the roots are equally spaced on this circle, with one of the roots having polar angle 90 = Arg w/n; for instance, see Figure 1.8. Example 1 Find the 12th roots of 1. Solution Since w = 1 = cos 0 + i sin 0, the modulus of all the 12th roots is 1. The roots are equally spaced on the circle of radius 1 centered at the origin. One root is z0 = 1; the others have polar angles of 27r/12,47r/12,67r/12,..., 227r/12, respectively. Example 2 Find the 5th roots of i + 1. Solution The polar representation of 1 + i is 1 + i = yfli cos + isinJ, so the modulus of all the 5th roots of i + 1 is 21/10 = 1.0717, the real, positive 10th root of 2. One of the roots is located with polar angle 7r/20, and the others have polar angles of 7i/20 + 2rc/5, rc/20 + 4rc/5, 7r/20 + 6n/5, and rc/20 + 8rc/5, respectively. (See Fig. 1.8.) □ The 5th roots of w = 1 + / Figure 1.8 Example 3 Solve the equation 4z2 + 4  2i = 0. Solution The equation may be rearranged as 4z2 + 4 = 2/ 16 Chapter 1 The Complex Plane or This has solutions (z2  2)2 = 2\ = (1 41)2. ^2'1 + ' 1  i. Equivalently, z2 These may be solved to give the four solutions of the original equation, zt = <yiO(cos 0O + i sin 0O), z2 = — ^10 (cos 0O f / sin 0O), 0O = arctan and z3=^/2(cos9l + ISU10!), 1 7T z4 = ~y2(cos 9X + i sin 0J, 0X = arctan(1) = . Circles A circle is the set of all points equidistant from a given point, the center. If z0 is the center and r the radius, then the circle of radius r and center z0 is described by the equation \z — z0\ = r. There are, however, other ways to use complex numbers to describe circles. If p and q are distinct complex numbers, then those complex numbers z with \zp\ = \zq\ are equidistant from p and q. The locus of these points is precisely the straight line that is the perpendicular bisector of the line segment joining p to q. However, if p is a positive real number not equal to 1, those z with \zp\ = p\zq\ form a circle. To see this, suppose that 0 < p < 1 (otherwise, divide both sides of the equation by p). Let z = w + q and c = p — q; then the equation becomes 1.2 Some Geometry 17 \wc\ = pw. Upon squaring and transposing terms, this can be written as w2(l p2)2Rewe + c2 = 0. We complete the square of the left side and find that \c\ (1  p2)\w\  2 Re wc + cV 1  p2 \p2 Equivalently, 1P2 = c \p 2' Thus, w lies on the circle of radius R = cp/(l — p2) centered at the point c/(l — p2), and so z lies on the circle of the same radius R centered at the point Zn = pp q p l  p2 \p2r \p2 Example 4 To confirm in one special case what was just done, let us look at the locus of points z with zi=ir_l. After multiplying both sides by 2 and squaring, 4{z2  2 Re(z7) + j2} = z2  2 Re z + 12, or after simplifying, 3z2  8>> + 2x = 3. More algebra yields 3x2 + 2x +  + 3y2  $y + — = 3 + — = . Thus, the locus is HN'a" 18 Chapter 1 The Complex Plane This is a circle of radius 2(^/2/3) centered at —1/3 + 4i/3. Now, in the notation that preceded the example, p = i, q = 1, and p = £. The radius should be and the center at p \P~q\p J2&) 2^2 z0 = p p 1 lp2 4 1 4 "3 + 3'' which, of course, is just what was previously found. Note that the center of the circle is on the line through 1 and i (Fig. 1.9). Q The circle \z  i\ = ^ \ z  1 Figure 1.9 We now apply the information just derived to produce a beautiful geometric pattern: two families of mutually perpendicular circles. Let Cx be the family of circles of the form \zp\ =p\zq\, 0<p< oo, where we include the case p = 1 (which yields a straight line) for completeness. Let L be the perpendicular bisector of the line segment from p to q. Take C2 to be the family of circles through p and q and centered on the line L. We shall show that each circle in the family Cx is perpendicular to each circle in the family C2 at their two points of intersection. The computation is considerably simplified by locating the origin at the point of intersection of the line L and the line L', which passes through p and q. L' can then be taken to be the real axis and L to be the imaginary axis; in this way, we may assume that 0 < p = — q. A circle from the family Cx is then centered at a point on the real axis, and because there is no loss in assuming 0 < p < 1, the center of that circle is at the point s = p(l + p2)/(l  p2). The center 1.2 Some Geometry 19 of the circle from the family C2 is at the point t = ioc (a real), and this circle must pass through p and —p. Let z = x + iy be on both circles (see Fig. 1.10). x is center of C, v is center of C2 0<p< 1 Figure 1.10 Since z is on the circle Cx, zp = pz + p, and consequently, x2(l  p2)  2px(l + p2) + p2(l  p2) + y2(l  p2) = 0. For notational convenience, set v = (1 + p2)/(l — p2); the above equation then becomes x2  2pvx + p2 + y2 = 0. On the other hand, since z = x + iy is also on the circle C2, \z  i(x\ = p  ia = p  ia = >/p2 + a2, and so x2 + y2 — lay = p2. In order to prove that the segment from ice to z is perpendicular to the segment s to z, it must be shown that the Re[(z — ia)(z — s)] = 0; this follows from the discussion , 1 ia* P k —^^f2 ►——""""' yr \ P ^Z s IC yc{ 20 Chapter 1 The Complex Plane of perpendicularity near the end of Section 1. However, using s = pv, 2 Re[(z  ioc)(z  s)] = 2x(x  s) + 2y(y  a) = 2(x2 + y2 — pvx — ay) = x2 + y2  2pvx + x2 + y2  2ay = p2 + P2 = 0. This is the desired conclusion. An illustration (with p = — q = 1) of several circles from each of these families is shown in Figure 1.11. This pattern is often called the Circles of Appolonius. These will come to your attention again later in the book. Figure 1.11 EXERCISES FOR SECTION 1.2 In Exercises 1 to 10, describe the locus of points z satisfying the given equation. 1. z + l = zl 2. z4 = 4z 3. Re [(4 + i)z + 6] = 0 4. Im(2iz) = 7 5. \z + 2 + \z  2 = 5 6. \z  t = Re z 7. Re(z2) = 4 8. z  12 = z + 12 + 6 9. z2  1 = 0 10. \z + 12 + 2z2 = z  12 1.2 Some Geometry 21 In Exercises 11 to 17, write the equation of the given circle or straight line in complex number notation. For example, the circle of radius 4 centered at the point 3 — 2i is given by the equation \z — (3 — 2i)\ = 4. 11. The circle of radius 2 centered at 4 + i. 12. The straight line through 1 and — 1 — i. 13. The vertical line containing — 3 — i. 14. The circle through 0, 2 + 2i, and 2  21 15. The circle through 1, i, and 0. 16. The perpendicular bisector of the line segment joining — 1 + 2i and 1 — 2i. 17. The straight line of slope —2 through 1 — i. 18. Show that the two lines Re(az + b) = 0 and Re(cz + d) = 0 are perpendicular if and only if Re(ac) = 0. 19. Let p be a positive real number and let T be the locus of points z satisfying \z — p\ = ex, z = x + iy. Show that T is (a) an ellipse if 0 < c < 1; (b) a parabola if c = 1; (c) a hyperbola if 1 < c < oo. 20. Let zx and z2 be distinct complex numbers. Show that the locus of points tzx + (1 — t)z2, — oo < t < oo, describes the line through zx and z2. The values 0 < t < 1 give the line segment joining zl and z2. 21. Let a be a complex number with 0 < a < 1. Show that the set of all z with (a) z — a < 11 — az is the disc {z: \z\ < 1} (b) \z — a = 11 — az is the circle {z: z = 1} (c) \z — a > 11 — az is the set {z: z > 1}. (Hint: Square both sides and simplify.) 22. Let z and w be nonzero complex numbers. Show that z + w = z + w if and only if z = sw for some positive real number s. In Exercises 23 to 26, follow the technique outlined in the text to find all solutions of the given equation. 23. z5 = i 24. (z + l)4 = 1  i 25. z8 = 1 26. z3 = 8 27. Suppose that n is an odd integer and w is a negative real number. Show that one solution of the equation zn = w is a negative real number. (For instance, — 2 is a root of z3 = — 8.) 28. Let a, b, and c be complex numbers with a ^ 0. Show that the solutions of az2 + bz + c = 0are zl5z2 = (2V)( — b ± >/b2 — 4ac), just as they are in the case when a, fc, and c are real numbers. 29. Let b and c be complex numbers. Show that the roots of the quadratic equation z2 + bz + c = 0 are complex conjugates of each other if and only if the quantity b2 — Ac is real and negative, b is real, and c is positive. 30. Let A be a complex number and B a real number. Show that the equation z2 + Re(Xz) + B = 0 has a solution if and only if \A\2 ^ 4B. If this is so, show that the solution set is a circle or a single point. 31. Let C be a circle and let A and B be any two distinct points on C. Show that if P is selected on the smaller arc of C joining A to B, then the angle from the segment AP to the segment BP is independent of P. This angle is n/2 HA and B are on opposite ends of a diameter. The result remains true if "smaller" is replaced by "larger." 22 Chapter 1 The Complex Plane 32. Let zl9 ..., zn be complex numbers. Show by mathematical induction that ki+" + zJ<z1 +  + zJ. Translation and Scaling* 33. Let C be a circle or a straight line. Show that the same is true of the locus of points z + (i9ze C, and /? a fixed complex number. 34. Let C be a circle or a straight line. Show that the same is true of the locus of points <xz, ze C, and a a fixed nonzero complex number. Inversion* 35. Let L be the line y = a, a > 0. Show that the locus of points 1/z, z e L, is the circle of radius \/2a centered at — i/2a. 36. Let L be a line through the origin. Show that the locus of points 1/z, z e L, is a line through the origin. What is the relationship of the slopes of the two lines? 37. Let C be the circle \z — c\ — r90 < r < c. Show that the locus of points 1/z, z e C, is the circle centered at c/(c2 — r2), of radius r/(c2 — r2). 38. Let C be the circle \z — r\ = r, r > 0. Show that the locus of points 1/z, z e C, is the vertical line through l/2r. 1.3 Subsets of the Plane To understand the fundamentals of complex variables, it is necessary to single out several special types of subsets of the complex plane that will be used in our discussion of analytic and harmonic functions in subsequent chapters. This section gives the definitions and basic properties of these sets. Open Sets The set consisting of all points z satisfying \z — z0\ < R is called the open disc of radius R centered at z0. A point vv0 in a set D in the complex plane is called an interior point of D if there is some open disc centered at w0 that lies entirely within D (see Fig. 1.12). A set D is called open if all of its points are interior points. An open disc of radius R centered atZo Figure 1.12 A set D and an interior point c, ofD 1.3 Subsets of the Plane 23 Example 1 Each open disc D = {z: \z — z0\ < R} is an open set. For if r = lwo — z0\ < R, choose e — (R — r)/3. Then, for any z with z — w0 < e, we have z  z0 ^ \z  w0\ + w0  z0 < e + r = (K  r)/3 4 r < K, by the triangle inequality. Thus, the open disc of radius e centered at vv0 lies within the set D. Hence, each point of D is an interior point, so D is open. □ Example 2 The set R = {z: Re z > 0} is an open set. To see this, let vv0 e R; then a0 = Re vv0 > 0. Let e = \a0 and suppose that \z — w0\ < e. Then — e < Re(z — vv0) < e, so Re z = Re(z — vv0) + Re vv0 > e + (j0= a0 > 0. Consequently, z also lies in R. Hence, each point w0 of R is an interior point, so R is open. □ Example 3 The set {z: Im z > 1} is an open set. This can be shown in the manner of Example 2. □ Example 4 The set of all points z = x 4 iy9 with x2 < y9 is also an open set. Let Zo = xo + *yo be *n this set; then there is a positive 5 with xj 4 3 < y0. We may also assume that S is so small that 2x0S + £3 < 1 + S. (The latter inequality will be needed only at one rather technical point.) Suppose now that z = x + iy satisfies \z — z0\ < S2. Then S2 > \x — x0, and S2 > \y — y0\. Therefore, x2 < (x0 + S2)2 = x2 + 2x0^2 + S* <y0S + 2x0S2 + S* <yS2S + 2x0S2 + 5*. However, the choice of S gives — S2 — 8 + 2x0<52 + <54 < 0, so x2 < y whenever z = x + iy satisfies \z — z0\ < 82. D Example 5 The set of all z with Re z < 6 is not an open set. □ The Boundary of a Set The set just described in Example 5 is not open because, for instance, the point w0 = 6 is in it, but every open disc centered at vv0, no matter how small the radius, must contain a point z with Re z > 6. The point vv0 = 6 is called a boundary point because it is located on the edge of the set, at the place where the set "almost" meets its complement. The precise definition follows. 24 Chapter 1 The Complex Plane A point p is a boundary point of a set S if every open disc centered at p contains both points of S and points not in S. The set of all boundary points of a set S is called the boundary of S (Fig. 1.13). Let us find the boundary of each of the sets in the preceding examples (see Fig. 1.13). boundary \z  Zo\ = R boundary Re z = 0 A\\\i\\\\, boundary boundary ■—if iWHteal ^ ^s\\W\^^ pltO X boundary ;\$ v Imz =  1N # *; Figure 1.13 Ik K$ Re z ^ 6 boundary K\\ Rez = 6 Example 6 The boundary of the open disc of radius R centered at z0 is the circle of radius R centered at z0. □ Example 7 The boundary of the set of those z with Re z > 0 is the imaginary axis. □ 1.3 Subsets of the Plane 25 Example 8 The boundary of the set of those z with Im z\ > 1 is the set of those z with Im z\ = 1. □ Example 9 The boundary of the set of those z = x + iy with x2 < y is the parabola y = x2. □ Example 10 The boundary of the set of those z with Re z ^ 6 is the vertical line Re z = 6. □ Closed Sets You have observed by now that the open sets just discussed contain none of their boundary. Indeed, this is an elementary theorem, which is set out below. However, what about exactly the reverse situation? A set C is called closed if it contains its boundary. These definitions prepare us for the following result. THEOREM A set D is open if and only if it contains no point of its boundary. A set C is closed if and only if its complement D = {z: z$C} is open. Proof To establish the validity of this theorem, we start by supposing that D is an open set. Let p be a boundary point of D. If it happens that p is in D, then (because D is open) there is an open disc centered at p that lies within D. Thus, p is not in the boundary of D. This contradiction shows that p is not in D. Conversely, suppose D is a set that contains none of its boundary points; we must show that D is open. If z0 6 D, then z0 cannot be a boundary point of D, so there is some disc centered at z0 that is either a subset of D or a subset of the complement of D. The latter is impossible, since z0 itself is in D. Hence, the disc lies in D, and we have shown that each point of D is an interior point; therefore, D is open. The second assertion of the theorem follows immediately from the fact that the boundary of a set coincides exactly with the boundary of the complement of that set; this in turn is a direct consequence of the definition of boundary point. ■ Be wary here—there are many sets that are neither open nor closed, since they contain part, but not all, of their boundary. For example, the set D of those z with Re z ^ 6 and Im z > 2 is neither open nor closed, since its boundary consists of those w with either Re w = 6 and Im w ^ 2 or Im w = 2 and Re w ^ 6 (Fig. 1.14). There is also one technical point here: The complex plane itself has no boundary and so by definition is both open and closed. It happens to be the only nonempty subset of the plane that is both open and closed. (The truth of this statement depends on a fundamental property of the real numbers.) Since what follows makes no use of this fact about the plane, we will not prove it here. Connected Sets A polygonal curve is the union of a finite number of directed line segments P^, P2P3,..., P„iP„, where the terminal point of one is the initial point of the next (except for the last) (see Fig. 1.15). An open set D is connected if each pair p, q of 26 Chapter 1 The Complex Plane D = \z: Re z < 6, ] A boundary point of D that is not in D Figure 1.14 A boundary point of D that is in D A polygonal curve (with n = 5) joining — 2 — / to I + / Figure 1.15 points in D may be joined by a polygonal curve lying entirely with D. That is, there are points P2,..., P„_i in D such that all the line segments pP2, P2P3, • •, Pni<I lie in£. Example 11 An open disc is connected. Example 12 The set of those z with Re z > 0 is connected. Example 13 The set of those z = x + iy with x2 < y is connected. Example 14 The set of those z with Im z\ > 1 is not connected. Example 15 The set of those z with Re z < 6 is connected. Example 16 The set of those z with Re z / 0 is not connected. □ □ D □ □ D For instance, in Example 16 the points p = 1 and q = 1 lie in the given set, but any polygonal curve joining p to q must necessarily cross the imaginary axis (where Re z = 0), so this curve cannot lie entirely within the set Re z # 0. 1.3 Subsets of the Plane 27 An open connected set is called a domain. Domains are the natural setting for the study of analytic and harmonic functions. A set S is convex if the line segment pq joining each pair of points p, q in S also lies in S. In particular, it is immediately clear that any convex open set is connected (Fig. 1.16). A convex set A set S that is not convex Figure 1.16 Example 17 Each open disc is convex. □ Example 18 The set of those z = x + iy with x2 > y is not convex. □ Example 19 Both of the sets Re z > 0 and Re z ^ 6 are convex. □ An open halfplane is defined to be those points strictly to one side of a straight line—that is, those points z for which Re(az + b) > 0, say. Each open halfplane is a convex set and an open set as well. A closed halfplane is the open halfplane plus the defining line—that is, those z with Re(az + b) ^ 0. Each closed halfplane is convex as well as closed (Fig. 1.17). Figure 1.17 It is instructive to see why each halfplane (open or closed) is convex. If p and q are two points, then 28 Chapter 1 The Complex Plane tq + {l t)p, 0 ^ t ^ 1 describes the line segment from p to q; see Exercise 20, Section 2. If p and q are in the open halfplane given by Re{az + b) > 0, then Re(a(r<? + (1  t)p) + b) = t Re(aq + b) + (1  £) Re(ap + fc) > 0, so the line segment from pioq lies in the same open halfplane. The case of a closed halfplane is almost identical. The Point at Infinity A helpful and frequently used convention is to add the point at infinity to the complex plane. This is understood in the following way. A set D contains the point at infinity in its interior if there is a large number M such that D contains all the points z with \z\ > M. For instance, the open halfplane Re z > 0 does not contain the point at infinity, but the open set D = {z: \z + 1 + \z — 1 > 1} does. One "reaches" the point at infinity by letting \z\ increase without bound, with no restriction at all on arg z. One way to visualize all this is to let w = 1/z and think about  vv being very small; an open set containing the point at infinity will become an open set containing w0 = 0. Further, the statement "z approaches infinity" is identical to the statement "w converges to zero." The point at infinity is denoted by the usual symbol for infinity: oo. EXERCISES FOR SECTION 1.3 For each of the sets in Exercises 1 to 8, (a) describe the interior and the boundary, (b) state whether the set is open or closed or neither open nor closed, (c) state whether the interior of the set is connected (if it has an interior). 1. A = {z = x + iy: x ^ 2 and y ^ 4} 2. B = {z:\z\ < lorz3< 1} 3. C = {z = x + iy: x2 < y} 4. D = {z: Re(z2) = 4} 5. E = {z: zz  2 ^ 0} 6. F = {z: z3  2z2 + 5z  4 = 0} 7. G = {z = x + iy: \z + 11 ^ 1 and x < 0} 8. H = {z = x + iy: — n ^ v < n) 9. Let a and /? be complex numbers with .: ^ 0. Describe the set of points <xz + /? as z varies over (a) the first quadrant, {z = x + iy: x > 0 and y > 0}; (b) the upper halfplane, {z = x + iy: y > 0}; (c) the disc {z: \z\ < R}. Show that in each case the resulting set is open and connected. (Hint: First investigate the set az.) 10. Describe the set of points z2 as z varies over the second quadrant: {z = x f iy: x < 0 and y > 0}. Show that this is an open, connected set. (Hint: Use the polar representation of z.) 11. A set S in the plane is bounded if there is a positive number M such that \z\ ^ M for all z in 5; otherwise, S is unbounded. In Exercises 1 to 8, six of the given sets are unbounded. Find them. 1.3 Subsets of the Plane 29 12. Which, if any, of the sets given in Exercises 1 to 8 contains oo? 13. (a) Show that the union of two nonempty open sets is open. Do the same, replacing "open" with "closed." Do the same replacing "union" with "intersection." (b) Repeat part (a) replacing "two" with "finitely many." 14. Let Dx and D2 be domains with a nonempty intersection. Show that Dt u D2 is a domain. 15. Let Qx = {z: 1 < \z\ < 2 and Re z > £} and Q2 = {z: 1 < \z\ < 2 and Re z < ^}. Show that both Qx and Q2 are domains but Slx n Q2 *s not 16. Let D be a domain and let p and q be points of D. Show that there is a polygonal curve joining p to q whose line segments are either horizontal or vertical (both types can be used). (Hint: Replace a "slanting" segment by (perhaps many) horizontal and vertical segments; see Fig. 1.18.) \ y^ Solid lines are the original polygonal curve X. ^^^^ joiningp to q. The dotted lines are a polygonal curve from p to q, which consists entirely of vertical/horizontal segments. Figure 1.18 17. Fix a nonzero complex number z0. Show that the set D obtained from the plane by deleting the ray {tz0: 0 ^ t < oo} is a domain. 18. An open set D is starshaped if there is some point p in D with the property that the line segment from p to z lies in D for each z in D. (a) Show that the disc {z: \z — z0\ < r} is starshaped, (b) Show that any convex set is star shaped. 19. Determine which of the following sets are starshaped: (a) D = {z:\z l<2orz + 1<2} (b) D = {z = x + iy: x > 0 and \z\ > 1} (c) D = {z: \z\ > 1} (d) D = {z = x 4 iy: x>0 and [x > y + 1 or x > 1  y]} 20. Show that each starshaped set is connected. Chapter 1 The Complex Plane Separation of a Point and Convex Set* Let C be a closed convex set and z0 a point not in C. It is a fact that there is a point p in C with r = \z0 — p\ ^ \z0 — q\ for all q in C. (This last statement requires a bit of proof, but let us assume its validity.) 21. Show that the only point of C in the disc \z0 — z\ ^ r is the point p. 22. Let L be the perpendicular bisector of the line segment from z0 to p. Show that no point of C lies on L or in the halfplane, determined by L, which contains z0. 23. Conclude from Exercise 22 that L separates z0 from C: z0 and C lie in the two open halfplanes determined by L, but not in the same open halfplane. 24. Show that each closed convex set is the intersection of all the closed halfplanes that contain it. Topological Properties* 25. Show that the boundary of any set D is itself a closed set. 26. Show that if p e D, then p is either an interior point of D or a boundary point ofD. 27. Show that a set D coincides with its boundary if and only if D is closed and D has no interior points. 28. Show that if D is a set and £ is a closed set containing D, then E must contain the boundary of D. 29. Show that if D is a set and S is an open set that is a subset of D, then S must be composed entirely of interior points of D. 30. Let C be a bounded closed convex set and let D be the complement of C. Show that D is a domain. Functions and Limits A function of the complex variable z is a rule that assigns a complex number to each z within some specified set D; D is called the domain of definition of the function. The collection of all possible values of the function is called the range of the function. Thus, a function / has as its domain of definition some subset of the complex plane and as its range some other (usually entirely different) subset of the complex plane. We frequently write w = f(z) to distinguish the independent complex variable z from the dependent complex variable w. Example 1 f(z) = 4z2 + 2z + 1 has as its domain of definition the entire complex plane. Its range is also the entire complex plane, for if w is any complex number, then the equation f(z) = w is nothing but the quadratic equation 4z2 + 2z + 1  w = 0. This equation is solved by use of the quadratic formula. Its solutions are 1.4 Functions and Limits 31 where sl9s2 are the two square roots of 4 — 16(1 — w) = —12 4 16w. (It is possible that st = s2; this occurs only when w = f.) □ Example 2 /(z) = l/(z — 1) has as its domain of definition all complex numbers except z = 1, where it is not defined. Its range consists of all complex numbers w except w = 0, since/(z) = w = l/(z — 1) is solved by z = 1 + (1/w); this is a complex number as long as w / 0. □ Example 3 g(z) = z2 has the complex plane as its domain of definition; its range consists of all nonnegative real numbers. □ Example 4 h(z) = i(2 — (Im z)"1) is defined for all z except those on the real axis. For a pure imaginary number w, a solution of w = h(z) = i(2 — (Im z)"1) is any complex number z with Im z = (2 + iw)'1. This defines many complex numbers as long as w / 2i. Hence, the range of h is all purely imaginary numbers except w = 2i. □ Example 5 Show that the range of the function w = T(z) = (1 + z)/(l — z) on the disc \z\ < 1 is the set of those w whose real part is positive. Solution Compute the real part of w: z)(lz) lz2 Re w = Re = Re ^—— \lzj 1 z2 lz2 This last quantity is positive when \z\ < 1. This shows that the range of T is a subset o£ those w with Re w > 0. Now let w' be any point with Re w' > 0; we shall show that z' = (w'  l)/(w' + 1) satisfies \z'\ < 1. Indeed, 1 > (w'  l)/(w' + 1) exactly when \w' + 12 > w'  12. We expand both \w' + 12 and w'  12 and obtain Kl2 + 2 Re w' + 1 > Iw'l2  2 Re w' + 1. This is a correct inequality because Re w' > 0. Thus, z' = (w' — l)/(w' h 1) lies in the disc \z\ < l,and wf1 «, / 1 + z' w'+l 2w' 1 — z w — 1 2 "w' + l (NOTE: T(z) = (1 4 z)/(l — z) is an example of a linear fractional transformation, which will be studied in some detail in Section 3 of Chapter 3.) As with functions of a real variable, the domain of definition of a function is usually easier to determine than the range. In many cases, this text will provide only a general description of certain properties of the range rather than an explicit rule 32 Chapter 1 The Complex Plane for each point in it. For instance, we may be able to state that the range is open or connected or convex. Graphs For realvalued functions of a real variable, like those studied extensively in calculus, the device of displaying the graph of the function is an extremely useful tool in visualizing the behavior of the function. Such a technique is not as readily available for functions of a complex variable. // the function / has only real values, its graph in 3space (x, y, t), t = f(x + iy), can be sketched (see Fig. 1.19). However, if the function has complex values, this type of picture is not possible (at least in our world!). Moreover, almost all the functions dealt with here have complex values. One way to proceed is to graph /, but a more useful way is to use two complex planes: one for the domain variable z and a second for the range variable w = f(z). For instance, we saw in Example 5 that the function T(z) = (1 + z)/(l — z) maps the disc {z: z < 1} onto those w with Re w > 0; this can be "graphed" as shown in Figure 1.20. This type of picture is very helpful in understanding the behavior of the complexvalued function of a complex variable and we will employ it frequently. continues up indefinitely continues up indefinitely Figure 1.19 1.4 Functions and Limits 33 &>ViVi*¥sy*wiWiWi*i*iVi*i > 0 = Re w Figure 1.20 Limits The concepts of the limit of a sequence of complex numbers, or the limit of a function of a complex variable, or even of continuity of a function of a complex variable, are almost identical to those for a real variable. Let us begin with the limit of a sequence. Let {zn}„=l be a sequence of complex numbers. We say that {z„} has the complex number A as a limit, or that {z„} converges to A9 and we write lim zn = A or z„ ► A n*co if, given any positive number e, there is an integer N such that \zn  A\ < e for all n ^ N. A sequence that does not converge, for any reason whatever, is called divergent. If *n = xn + iyn and A = s + if, then zn* A if and only if x„ ► s and y„ ► t. This is due to three inequalities noted in Section 1, namely \xns\^\znAl \ynt\^\znA\9 and \zHA\^\xHs\ + \yHt\. We can equivalently state that the sequence {z„} of complex numbers converges to the complex number A if and only if, whenever D is any open disc centered at A, all but a finite number of the points {z„} lie in D. Example 6 The sequence z„ = 1 + (i/n) converges to 1. Example 7 The sequence z„ = (■£)" + i(\  (l/2n)) converges to i. □ □ 34 Chapter 1 The Complex Plane Example 8 The sequence zn = (l/n)(cos(n7r/4) + i sin (nn/4)) converges to zero. □ Example 9 The sequence zn — n — (1/n) diverges. □ Example 10 The sequence zn = in diverges because its terms are i, — 1, — i, 1, in that order, repeated infinitely often. □ One simple consequence of the definition of convergence is this: Ifz.yl, thenzj>i The converse of this assertion is generally false; for example, (— l)n + i/n\ ► 1, but the sequence {( — l)n + i/n} itself has no limit. You will no doubt recall from calculus results about the sum, product, and quotient of convergent sequences of real numbers. Similar statements hold for sequences of complex numbers; we collect these results in the next theorem. THEOREM 1 Let {z„} and {vv„} be convergent sequences of complex numbers with limits A and B9 respectively. Let X be a complex number. The sequences {zn + Aw„} and {znwn} then also both converge with limits A + KB and AB, respectively. Furthermore, if B / 0, the sequence {zn/wn} converges to A/B. j Proof The proof of each of these results is quite direct. For example, to show that znwn ► AB, we write znwn — AB = (z„ — A)B + (vv„ — B)zn. Let Nt be chosen so big that \z„ — A\ < e' if n ^ Nx and N2 chosen so big that \wn — B\ < s' if n ^ N2. Then \zn\ ^ e' + \A\ < 1 + \A\ if e' < 1; and for n ^ N = Nx + N29 we obtain \znwnAB\^\znA\\B\ + \wnB\\zn\ <£,B + e'(l + \A\) = C(1+^ + B) <e, if e' is chosen initially to satisfy £'(1 + ,4 + 2J)<£. ■ Example 11 The sequence zn = 1 + i[l  (2/n)] converges to 1 + i as n + oo. Hence, the sequence zn2= j + i 2 converges to 2i = (1 + 02 as n > oo. D Example 12 Suppose that a0, al9 a2, and a3 are complex numbers and {z„} is a sequence with z„ ► A. Then the sequence {a0 h «!Zn h a2zl h a3z^}^=1 converges 1.4 Functions and Limits 35 to a0 + axA + a2A2 + a3>43 as n ► oo, as can be seen by several applications of results in Theorem 1. □ Suppose next that / is a function defined on a subset S of the plane. Let z0 be a point either in S or in the boundary of S. We say that / has limit L at the point z0, and we write lim f(z) = L or /(z) * L as z » z0 if, given 6 > 0, there is a 5 > 0 such that \f(z) — L\ < e whenever zeS and \z — z0\ < 5. It is worth stressing here that / has the limit L at the point z0 exactly when the numbers/(z) approach L as z approaches z0 in any manner whatsoever from within S. This is substantially different from the case of a function of a real variable, where the real variable can approach only from the left or right. The complex variable z may approach z0 from infinitely many directions. Once again, it will be informative to look at some examples. Example 13 The function f(z) = z2 has limit 4 at the point z0 = 2L □ Example 14 The function g(z) = 1/(1  z) has limit (1/2)(1 + i) at z0 = i. □ Example 15 The function h(z) = Re(z4 + 4) has limit 0 at z0 = 1 + i. □ Example 16 The function f(z) = (z4 — l)/(z — i) has limit — 4f at z0 = i, since z4  1 = (z  i)(z + i)(z  l)(z + 1), so f(z) simplifies to (z + i)(z + l)(z  1) so long as z # l U Example 17 The function f(z) = zfz, z / 0, has no limit at z0 = 0. For if z is real, then/(z) = 1, while if z is purely imaginary, z = iy, then/(iy) = — 1. Such a function cannot have a limit at z0 = 0. □ We say that the function / has a limit L at oo, and we write lim f(z) = L z*co if, given e > 0, there is a large number M such that \f(z) — L\ < e whenever \z\ ^ M. Note that we only require that \z\ is large; there is no restriction at all on arg z. Example 18 lim^^ l/zm = Oifm = 1,2,.... □ 36 Chapter 1 The Complex Plane Example 19 limz^00 [(z4 + l)/(z4 + 5z2 + 3)] = 1, since the ratio can be written as 5 3' z2 z4 all of whose terms except the l's go to zero as \z\ ► oo. D Example 20 limz^00 [(x + y3)/(x2 + y3)] does not exist; for if we take z = x, then the expression goes to zero, but if we take z = iy, then the expression is identically 1. □ The following theorem on the sum, product, and quotient of functions that have limits at z0 is a counterpart to Theorem 1. THEOREM 2 Suppose that / and g are functions with limits L and M, respectively, at z0. Let k be a complex number. The the functions / + kg and fg have the limits L + kM and LM, respectively, at the point z0; and, if M ^ 0, the function f/g has the limit L/M at the point z0. ■ Continuity Suppose again that / is a function defined on a subset S of the complex plane. If z0 e S, then / is continuous at z0 if lim/(z) = /(z0). z>zo That is, / is continuous at z0 if the values of f(z) get arbitrarily close to the value /(z0), so long as z is in S and z is sufficiently close to z0. If it happens that / is continuous at all points of S, we say/is continuous on S. The function/is continuous at oo if/(oo) is defined and limz_00/(z) = /(oo). Let us review Examples 13 to 17 in the context of "continuity." Example 21 /(z) = z2 is continuous at every point of the complex plane. □ Example 22 g(z) = 1/(1 — z) is continuous at all points of the plane except z = 1. □ Example 23 h(z) = Re(z4 + 4) is continuous at all points of the plane. □ Example 24 /(z) = (z4 — l)/(z — i) is continuous on the whole plane if we define /(0 4i. D Example 25 h(z) = zfz is continuous everywhere except z = 0. Further, there is no way to define h(0) to make h continuous at z0 = 0, since h(z) = 1 for all z that 1.4 Functions and Limits 37 are real and h{z) = — 1 for z that are purely imaginary. Such a function cannot be continuous at z0 = 0, the point where the real axis meets the imaginary axis. □ Example 26 The function g(z) = l/(z + 1) is continuous at all points of the plane and at oo as well, if we set 0(00) = 0. You will no doubt recall that the sum and product of continuous functions is continuous. This follows, in fact, directly from Theorem 2. We state this formally in the next theorem. THEOREM 3 Suppose that/and g are functions, both of which are continuous at the point z0. Let X be a complex number. Then the functions/ + kg and/7 are also continuous at the point z0. Further, if g(z0) / 0, the function f/g is continuous at z0. Finally, if h is a function continuous at each point of some disc centered at the point vv0 = /(z0), then the composition h(f(z)) is continuous at z0. ■ The proofs are left to the exercises at the end of this section. These facts allow us to conclude that each polynomial p(z) = a0 + axz + • • • 4 anzn is continuous on the complex plane; here a0,..., an are complex numbers. Further, if p and q are two polynomials, their quotient r = p/q is also continuous at all points at which q(z) / 0. The ratio of two polynomials is called a rational function. Each complexvalued function / can be written as / = u + iv9 where u and v are realvalued functions: u(z) = Re /(z), v(z) = Im /(z). In this way, the statements about limits, continuity, etc., of / can be recast as statements about u and v. For example, / is continuous at a point z0 if and only if both u and v are continuous at z0. Some exercises on these topics are included at the end of this section. Infinite Series Just as the notion of the convergence of a sequence of complex numbers is almost the same as the convergence of a sequence of real numbers, so the notion of the sum of an infinite series of complex numbers is virtually the same as that of the sum of an infinite series of real numbers. Specifically, if the numbers zx, z2,..., are complex numbers, we define their nth partial sum by n sn = Z ZJ = zi + *'' + z«> n = 1, 2,.... We then examine the behavior of the sequence {s„}. If the sequence {sn} has a limit s, then we say that the infinite series £ j°=1 z3 converges and has sum s; this is written 00 E h = s 38 Chapter 1 The Complex Plane If for any reason the sequence {sn} does not have a limit, we say that the series YjT=i zj diverges. The convergence (or divergence) of the infinite series Yj=i zj °f complex numbers can be formulated in terms of the convergence (or divergence) of two infinite series of real numbers. This follows directly by writing z} = Xj + iyj9 so n n n Sn = Z h = Z XJ + * Z yj = Gn + »V j=l j=l j = l As noted earlier in this section, the sequence {sn} converges if and only if both of the sequences {on} and {t„} of real numbers converge, and, this being the case, s = lim sn = lim an + i lim t„. However, the convergence of the two sequences {on} and {t„} is exactly the statement that the two infinite series Yj=i xjanc^ Zi=i X; ^ot^ converge. We thus arrive at this result: Let Zj = Xj + iy}\ j = 1, 2, 3, The infinite series Y,T=i zj converges, Zj = Xj + iy,, if and only if both 00 00 Z xj and Z yj converge. Furthermore, if Zzj converges, then 00 00 00 Z h = Z xj +l Z *• ;=1 j=l j=l There is more that can be said here. Since \xj\ < \Zj\9 \yj\ < \Zjl the convergence of the series Z^=i If/1 of nonnegative numbers implies, by the use of the usual comparison test for infinite series of real numbers, the convergence of both the series Z x, and f \yj\ Hence, both the series oo oo Z Xj and Z yj i=i y=i converge, so 00 Z (*j + tyj) also converges. Furthermore, 1.4 Functions and Limits 39 kl = 7=1 < I N J=l for all n. We conclude that if Y,f=i \*j\ converges, then Y,f=i zj converges, and j=i < Z z,. j=i This is a useful criterion for convergence, since there are several tests available from calculus (comparison, ratio, root, integral; see the exercises) for the convergence of an infinite series of nonnegative numbers. When £z/ converges, we say that the series Zj is absolutely convergent. Example 27 If a is not equal to 1, the identity 1 ha + ••• + an = 1  aH+: 1 a is easily verified by multiplying both sides of the equation by 1 — a. If a < 1, then an+i _► o as n ► oo, with the result that 00 J l+a + a2 + =YaJ = , j=o 1  a This is the geometric series, with ratio a. Example 28 The series M < 1. converges, since and the series converges, as can be seen by use of, say, the ratio test. Example 29 The series 00 1 I kit k=l * □ 40 Chapter 1 The Complex Plane converges, even though the series diverges. Note that CO 1 T  kkk f i if k = 1,5,9,... 1 iffc = 2,6,10,... i if* = 3,7,11,... 1 iffc = 4,8, 12,... Hence, » i" ( 1 1 1 \ ./ 1 1 1 \ and each series in the parentheses converges by the alternatingseries test. Example 30 The series diverges, since 3(1 + 0" = t id + i)2k k=l *> 2*_)d+'")21* = i\ so the terms of the series do not go to zero. Thus, the series must diverge (see Exercise 30). □ Some Sums* The identity at the beginning of Example 27 has significance beyond the geometric series. Let x be any real number not equal to 1. Then 1 + X + x2 + x3 + • • • + xn = 1  xw+ 1x This identity can be manipulated to obtain other useful identities. For instance, differentiate both sides with respect to x, then multiply both sides by x, and add 1. The result is the identity yfl + l 1 y" + 1 1 + x + 2x2 + 3x3 + 4x4 + • • • + nxn = 1  (n + 1) + x— 1 — X (1 — x) 2 * 1.4 Functions and Limits 41 Repeating this threestep process yields another identity, this time for 1 + x + 4x2 + 9x3 + • • • + n2xn. EXERCISES FOR SECTION 1.4 Limits In Exercises 1 to 8, find the limit of each sequence that converges; if the sequence diverges, explain why. 1 + i\n 1. z „ = 5. zn = n +  n 7. zn = Arg f 1 +  j, a fixed f0\ • f9 8. z„ = n {1 — cos I — I — i sin I — KnJ \n 2. z, 4. zn = Log  cos n6 + i sin n6 6. zn = , 9 fixed 9 fixed In Exercises 9 to 14, find the limit of each function at the given point, or explain why it does not exist. 9. f(z) = 11  z\2 at z0 = i 10. f(z) = Arg z at z0 = 1 11. /(z) = (1  Im z)"1 at z0 = 8 and then at z0 = 8 + i 12. /(z) = (z  2) logz  2 at z0 = 2 13./(z) = ^L, z/0, atzo = 0 z z3  8i 14. f(z) = , z # 2i, at z0 = 2i Continuity In Exercises 15 to 20, find all points of continuity of the given function. z3 + i Z — I z*i 15. f(Z): — 3, z = i 17. /(z) = (I'm z  Re z)_1 19"(Z) = W if W>1 16. f(z) = 1 fz4l z — i 4i, z^i z = I 18. g(z) = (1  lzl2)"3 20. *(z) = z3 In Exercises 21 to 24, find the limit at oo of the given function, or explain why it does not exist. 21. m = i izii 22. fc(z) = —, z#0 z 42 Chapter 1 The Complex Plane 4z6  7z3 21 0(z) = (z2 _ 4)3 24 *(*) = Arg z, z ?& 0 25. Let / and g be continuous at z0. Show that / + g and fg are also continuous at z0. If #(z0) t* 0, show that \/g is continuous at z0. 26. Establish the following result. A function / is continuous at a point z0 of its domain of definition S if and only if, given e > 0, there is a 8 > 0 such that \f(z) — f(z0)\ < e for all z e S with z  z0\ < 5. 27. Let g be defined on a set containing the range of a function /. If/ is continuous at z0 and # is continuous at /(z0), then g(f(z)) is continuous at z0. 28. Suppose that / = u + iv is continuous at z0. Show that each of the functions u, u, u — iv, and (u2 + v2)1/2 are also continuous at z0. Conversely, show that if u and i? are continuous at z0, then so is /. 29. Let u be a continuous function on a domain D. Suppose that u has this property: For each point peD, there is a disc A centered at p on which u is constant. Conclude that u is constant throughout D. You will have to use the fact that D is a domain. Infinite Series 30. Suppose that Y,n=i an converges. Show that lim,,.^ an = 0. In Exercises 31 to 39, determine whether the given infinite series converges or diverges. IOtt)" *£■(£ 10)" 00 1 00 34 I TT^n 35" Z »(»  W"2' 101 < ! n=l ^ h I n = 2 00 1 °° 1 (\ 4 i 36I^r 37. S1/1 + ' oo oo L2 _j_ .• 38. X (r  z"+>) 39. J —+ n=0 fc = l I* + I) 40. Show that each of the following series converges for all z. <*>.?„* (b).?.'"'<2St (C».?,(2^TT)! 41. Suppose that the series Y,n=ian converges. Let \z\ < 1. Show that the series Y,n=i anzK is convergent—indeed, that Y,n=i \anzn\ converges. (Hint: By Exercise 30, \an\ < M for some M and all n.) The Root and Ratio Tests* In Exercises 42 and 43, {cn} is a sequence of positive numbers. 42. (Root test) Suppose that limn_00(c„)1/n = A exists. Show that the series £c„ converges if 0 ^ A < 1 and diverges if A > 1. (Hint: If A < 1, there is a B with /I < B < 1. Hence, cn<Bnifn^ N. (Why?) If 1 < A, then c„ ^ 1 for all n ^ JV'. (Why?)) 43. (Ratio test) Suppose that 1.5 The Exponential, Logarithm, and Trigonometric Functions 43 lim CJ!±L = C nKx> Cn exists. Show that the series £c„ converges if 0 ^ C < 1 and diverges if C > 1. (Hint: If C < 1, then for £>, C < D < 1, we have cn+1 ^ Dcn for all n^N. (Why?) Therefore, cn+k ^ Dfccn, fc = 0, 1, 2, ..., and £c, converges. If C > 1, then cn+1 ^ c„ for all n ^ N'. (Why?) So {ck} does not converge to zero.) 1.5 The Exponential, Logarithm, and Trigonometric Functions The Exponential Function The exponential function is one of the most important functions in complex analysis. Its definition is simple: ez = ex(cos y + i sin y), z = x 4 iy. The form exp(z) is used sometimes, especially if z itself has some complicated form. The definition of ez allows us immediately to derive one of its most significant properties: For any two complex numbers z and w, ez+w = ezew. To see this, write (as usual) z = x + iy and w = s + it. Then, making use of two basic trigonometric identities for the sine and cosine of the sum of two numbers, ez+w = ex+s[cos(y + t) + i sin(y + 0] = exes[(cos y cos t — sin y sin t) + i(sin y cos t + sin £ cos y)] = ex(cos y + i sin y)es(cos t + i sin t) = ezew. The basic definition allows several more conclusions about the exponential function. First, ez is a continuous function of z. The functions ex, cos y9 and sin y are continuous functions of x and y. Consequently, Re(ez) = ex cos y and Im(ez) = ex sin y are both continuous, so ez is continuous at all points of the plane. Second, \ez\ = eRcz, because \ez\ = ((ex cos y)2 + (ex sin y)2)1/2 = ex = eRcz. 44 Chapter 1 The Complex Plane In particular, \eu\ = 1, treal. Since eif = cos t + i sin t, t real, we see that as t increases, elt moves on the circle of radius 1 centered at the origin in a counterclockwise direction, making one complete circuit when t increases by In (Fig. 1.21). In particular, of course, 1 for m = 0, ± 1,... and eni= 1. >o + 2tt Figure 1.21 Second, the function f(z) = ez never has the value zero, since neither ex nor eiy is ever zero. On the other hand, if w is any nonzero complex number, the equation ez = w has infinitely many solutions. This can be seen, for example, by writing w in polar form: w = r(cos t/f + i sin ^), and then setting x = In r and y = ^ + 27rm, where m is any integer (positive or negative) and In r is the natural logarithm of r; In is log to the base e, studied in calculus. Thus, ex + iy = gin rei(\l/ + 2nm) = r[cos t/f + i sin ^] = w. Furthermore, every solution z of the equation ez = w has the form given above. For 1.5 The Exponential, Logarithm, and Trigonometric Functions 45 if ez = w, then r = \w\ = \ez\ = e\ so x = In r. Consequently, cos y + i sin y = eiy = cos ij/ + i sin ^. Thus, y = ^ 4 2flm for some integer m (see Exercise 14 in Section 1). The mapping of f(z) = ez thus carries the complex zplane onto the complex wplane with the origin deleted; each point w0 has infinitely many preimages z, each of the form z0 + 2nim, m = 0, ±1, + 2,..., where z0 is any solution of ez° = vv0. In particular,/(z) = ez carries each strip y0 ^ Im z < y0 + 2rc, — oo < Re z < oo,onto the wplane with the origin removed. The function f(z) = ez is onetoone on that strip—that is, distinct points have distinct images. For eZl = eZl if and only if ezxz2 = j. ^jg occurs exactly when zx — z2 = 27rifc for some integer fe (Fig. 1.22). Figure 1.22 The function f(z) = ez maps each horizontal line y = c onto a ray from the origin to infinity—specifically, the ray {r cos c + ir sin c: 0 < r < oo}, since exp(x + ic) = ex cos c h iex sin c, and ^x increases from 0 to oo as x increases from — oo to oo; of course, ex is always positive. Furthermore, ez maps each vertical line x = c onto the circle centered at the origin of radius ec. This is because exp(c + iy) = ec(cos y + i sin y), and cos y + i sin y travels along the circle of radius 1 centered at the origin in a counterclockwise direction as y increases. Each point on this circle is the image of infinitely many points on the vertical line, since both cos y and sin y are periodic, with period 2n. The Logarithm Function The inverse of the exponential function is the logarithm function. For a nonzero complex number z, we define log z to be any complex number w with ew = z. 46 Chapter 1 The Complex Plane The preceding discussion of the exponential function leads immediately to the relationship log z = lnz + i arg z, z / 0. However, it is obvious that this is not a single complex number, but rather a set of complex numbers, each two elements of which differ by an integer multiple of 2ni. One way to be definite is to use Logz = lnz + i Argz. Be wary here; it may happen that Log(z1z2) / Log zx + Log z2 exactly because Arg(zj z2) need not equal Arg zx + Arg z2. The function Log z is called the principal branch of the logarithm of z. Other choices of arg z yield other values of log z (more on this later). To investigate the continuity of log z, delete from the plane any ray beginning at the origin (for example, delete all the nonpositive real numbers) and let D be the domain remaining. Let z0 be any point of D and choose and then fix any value for arg z0, say arg z0 = 0O. We then define in D a branch of log z by the rule log z = lnz + i arg z, with the additional specification that log z0 = lnz0 + i00. Then within D the values of arg z lie in a uniquely determined open interval of length In, which contains 0O; furthermore, arg z is continuous in D with this specification. The function lnz is a continuous function on D as well, so the branch of log z determined by the specific choice arg z0 = 90 is continuous on D. For example, if we delete the nonpositive real axis (see Fig. 1.23) and pick z0 = 1 + i and 90 = f rc, then the branch of log z that is determined has arg z in the range (n, 3n). Clearly, there are infinitely many possible branches of log z in D, each continuous. deleted \ • 1 + / Figure 1.23 The definition of log z allows us to complete the discussion of roots begun in Section 2. For a nonzero complex number a, we define az by the rule az = ez\oga 1.5 The Exponential, Logarithm, and Trigonometric Functions 47 This results in many values, but it does agree with the usual definition in the special case when a and z are both positive and real. Example 1 Find the values of (—1)\ Solution Using the relation log(— 1) = (In + l)nU n = 0, ± 1,..., we obtain (iy = eiio*(i) = e(2n+i)n9 n = 0, ±1, ±2,.... □ Example 2 Find the solutions of z1+i = 4. Solution Write this equation as e(l+i)logz = 4 so (1 + 0 log z = In 4 + 2nnU n = 0, ±1, — Hence, log z = (1 — i) [In 2 + nnQ = (In 2 + nn) + i(nn  In 2). Thus, z = 2e7tn(cos(7W — In 2) + i sin(7tn — In 2)) = 2enn{( l)n cos(ln 2) + i( l)n+1 sin(ln 2)} = ( l)"2€""{cos(ln 2)  i sin(ln 2)}, n = 0, +1,.... □ Example 3 Establish the formula lim ( 1 I  I = ez, zdi complex number Solution Look at n Log(l + z/n) for large n. Write n Log ( 1 +  ) = n In n + in Arg[ 1 + The real part satisfies nln n 1 f T 2x x2 + y2l = nln 1 + — + y x, 2 _ n w J as n ► oo by l'Hopital's Rule, for example. Next, if z = r(cos 9 + i sin 0) and \j/n = Arg(l h z/n\ then from Figure 1.24, 48 Chapter 1 The Complex Plane Figure 1.24 tan fa = r . _ sin 0 n 1 +  cos 9 n and so n tan ij/n = r sin 9 1 +  cos 0 n Because ij/n ► 0 as n ► oo, we know that il/n/(tan ^n) ► 1 (say, l'Hopital's Rule again). Hence, n\l/n = (n tan fa) tan fa) ► r sin 0 = y, as n ► oo. Consequently, n Log I 1 +  J ► x + iy = z, so (1 + (z/n))n ► ez, as intended. as n ► oo, Trigonometric Functions The trigonometric functions of z are defined in terms of the exponential function. We begin with the cosine and the sine of z: cos z =(eiz + e~iz) sinz = — (eizeiz). 1.5 The Exponential, Logarithm, and Trigonometric Functions 49 The other four trigonometric functions are now defined in terms of sin z and cos z: sin z cos z 1 1 tanz = , cotz = ——, secz = , cscz = —. cos z sin z cos z sin z Of course, the definitions are only at those points at which the denominator is not zero. If z is real, z = x, then cos z and sin z agree with the usual definitions of cosine and sine. Note that cos(z + 2n) = Uei{z+2n) + ei{z+2n)) = l(eiz + eiz) = cos z. Likewise, sin(z + 2n) = sin z for all z. Thus, cos(z + Ink) = cos z sin(z + Ink) = sin z, for all z and any integer k. Furthermore, Ink is the only number a with cos(z + a) = cos z for all z. For if cos(z + a) = cos z, for all z, then eizeia + e~izeia = eiz + e~iz, so eiz(eia  1) = e~iz(\  e~ia) = e~iaeiz(eia  1). If eia — 1 is not zero, it can be canceled from both sides, leaving eiz = eiae~iz9 for all z, but then setting z = 0 gives 1 = e~ia, contradicting the fact that 1 # eia. Hence, it must be the case that eia = 1, so a = 2nk for some integer k. Consequently, 2n is the basic period of cos z and, likewise, of sin z. 50 Chapter 1 The Complex Plane In the exercises at the end of this section, you will be asked to establish the formulas cos(x + iy) = cos x cosh y — i sin x sinh y, sin(x + iy) = sin x cosh y + i cos x sinh y, where cosh u = (eu + e~u% u real sinh u = (eu — e~u% u real. Let us use the formula given above for sin(x + iy) to establish some basic properties of the function f(z) = sin z. Restrict z = x + iy such that 0 ^ x < n/2 and y ^ 0. In this restricted region, we shall show that the function sin z is onetoone—that is, if sin(x1 + iyx) = sin(x2 + iy2\ where xx and x2 lie in [0, n/2) and yx and y2 are nonnegative, then x1 = x2 and ^i — yi To see this, observe that 2i sin(x + iy) = eixe"y — e"ixey. Hence, if sinCxj + i}^) = sin(x2 + ry2), then g^i^yi _ e~iXieyi = eiXle~yi — eiX2ey2. Thus, eixie~yi — eix2eyi = e~ixievi — ei*2ey2 = e~iXie~iX2eyiey2[eiX2e~y2  eiXle~yi]. lfeiXle~yi  eiX2e~y2 ^ 0, we divide by it and obtain 1 = e~iXle~iX2eyiey2. The absolute value of the left side is 1, and that of the right side is exp^ + y2)\ this implies that yx + y2 = 0. This in turn implies that — 1 = exp[ — ixx — i'x2], which can be true only when xx + x2 = n(2m h 1) for some integer m. However, we have restricted xl9 x2 to lie in [0, {n/2)) and yl9 y2 to lie in [0, oo). The only conclusion, then, is that xx = x2 = 0 and yx = y2 = 0. Otherwise, 1.5 The Exponential, Logarithm, and Trigonometric Functions 51 ei*ieyi = eix*e~y\ so yx = y2 and xx — x2 is an integer multiple of 2n. This again implies that xx = x2. Let us now find the range of f(z) = sin z on the strip 0 ^ x < n/2 and 0 ^ y < oo. First, and Re (sin z) = sin x cosh y ^ 0, Im(sin z) = cos x sinh y ^ 0. Hence, the values of sin z lie in the first quadrant (Fig. 1.25). Next, on the ray 0 ^ y < oo, x = 0, sin(ry) = i sinh y =(ey — e~y). Figure 1.25 The function sinh(y) is zero when y = 0 and is increasing (its derivative is positive), so sin(ry) assumes all the values ij8, 0 ^ ft < oo, as y increases from 0 to oo. Next, sin x increases from 0 to 1 as x increases from 0 to n/2. Finally, sin(! + iy) = cosh y, and this increases from 1 to oo as y increases from 0 to oo. Thus, the boundary of the strip 0 < x ^ n/2, 0 ^ y < oo is carried by sin z onto the boundary of the first quadrant (see Fig. 1.25). We now show that the interior is mapped onto the interior; we already know it is mapped into. Let us write sin z = w = o + ix. The vertical segment z = x0 + iy9 y ^ 0, is mapped onto that portion of the hyperbola1 (sin x0) (cos x0)" = 1, 52 Chapter 1 The Complex Plane which lies in the first quadrant. This shows us how to solve the equation sin z = a + ib9 a, b positive. First choose x such that the point a + ib lies on the hyperbola consisting of those w = a + it with (sin x)2 (cos x)2 then choose y such that a = sin x cosh y and b = cos x sinh y. Consequently, the function f(z) = sin z maps the semiinfinite strip 0 ^ x ^ n/2 and 0 < y < oo in a onetoone fashion onto the whole first quadrant, mapping the boundary of the strip onto the boundary of the first quadrant. This type of information, and this mapping in particular, will be very important in the solution of flow and boundaryvalue problems. Finally, because sin(—z) = — sin z (see Exercise 15), we may conclude that f(z) = sin z maps the semiinfinite strip {— n/2 < x < n/2, y > 0} both onetoone and onto the upper halfplane, {w = s + it: —oo < s < oo, t >0}. Inverse Trigonometric Functions We have just seen that the function w = sin z maps the strip {x + iy: 0 < x < n/2, 0 < y < oo} onto the first quadrant {w = a + it: <t, t > 0}, and distinct zx and z2 in the strip have distinct images vvx and vv2 in the first quadrant. Thus, given a w = a + it with a, t > 0, there is one and only one z in the strip {x + iy: 0 < x < xr/2,0 < y < oo} with w = sin z. Consequently, the function sin z has an inverse function, naturally called arcsin w, defined at least for w in the first quadrant. Let us pursue this a little further. The formula for sin z is w = sin z = — (eiz — e~iz\ so e2iz  2iWz 1=0, which is a quadratic equation in the variable eiz. Solving by the quadratic formula, we obtain eiz = iw + y/\ — w2, so z = —i log(iw + y/\  w2), 1.5 The Exponential, Logarithm, and Trigonometric Functions 53 provided an appropriate branch of the logarithm is chosen. A careful examination of the mapping w = sin z (see the exercises) shows that sin z maps the strip n n x + iy: <x<, co <y < cc' both onetoone and onto the region D obtained from the plane by deleting the two intervals (—00, — l]and[l, oo)(see Fig. 1.26). Thus, on D we can solve uniquely for z in terms of w, and the formula is valid. Interchanging the roles of z and w, write Arcsin z = — i Log(iz + yj\ — z2), z e D. Similarly, and Arccos z = — i Log(z + yjz1 — 1), = 2L°g(^} Arctan z =  Log ( „ ), z # ± U with appropriate interpretations of the resulting logarithms and roots. ^ Figure 1.26 '^>!^^:'""**t^^ EXERCISES FOR SECTION 1.5 Find the values(s) of the given expression in Exercises 1 to 14. 1. e""4 4. log(i) 7. e'1"1'3 10. Log(l) 13. log((l  0*) 2. e5nil4 5. (1 + if 8. exp(Log(3 + 2i)) 11. ifi 14 expK^)1 3. log(l + u/3) 6. 21' 9. Log(44i) 12. log^  i) 54 Chapter 1 The Complex Plane 15. Establish the following relations: (a) exp(z) = exp(z) (b) sin(z) = sin(z) (c) cos(z) = cos(z) 16. Establish the formulas cos(x + iy) = cos x cosh y — i sin x sinh y, sin(x + iy) = sin x cosh y + i cos x sinh y, where cosh u = (eM + e~M), w real sinh u = {eu — e u)9 u real. 17. Show that cos z = 0 if and only if z = n/2 + nrc, n = 0, ±1, ± 2,...; show that sin z = 0 if and only if z = nxr, n = 0, ±1, ± 2,.... That is, extending sin z and cos z from the real axis to the whole plane does not introduce any new zeros. 18. Verify that cos(z + w) = cos z cos w — sin z sin w and sin(z + w) = sin z cos w + cos z sin w, for all complex numbers z and w. 19. Show that both cos z and sin z are unbounded if z = iy and y ► oo. Also show that cos(x + iy)\ <,ey if y ^ 0, — cc < x < oc; and sin(x + iy)\ ^ ey if y ^ 0,  oo < x < oo. 20. Prove that cos2 z + sin2 z = 1 for all z. 21. Define cosh z and sinh z by s cosh z = {ez + e~z) sinh z = (ez — e z). 1.5 The Exponential, Logarithm, and Trigonometric Functions 55 Show that the following identities hold: (i) cosh2 (z)  sinh2 (z) = 1 (ii) cosh z = cos(iz) (iii) sinh z = —i sin(iz) (iv) cosh z\2 = sinh2 x + cos2 y (v) sinh z2 = sinh2 x + sin2 y 22. Show that sin( —z) = — sin z and cos z = cos( —z) for all z. Mappings with the Exponential Logarithm, and Trigonometric Functions 23. Show that F(z) = ez maps the strip S = {x + iy: — oo < x < oo, — n/2 ^ y ^ n/2} onto the region Q = {w = s + it: s ^ 0, w # 0} and that F is onetoone on S (see Fig. 1.27). Furthermore, show that F maps the boundary of S onto all the boundary of Q except w = 0. Explain what happens to each of the horizontal lines {Im z = n/2} and {Im z = — n/2}. Figure 1.27 24. Let D be the domain obtained by deleting the ray {x: x ^ 0} from the plane, and let G(z) be a branch of log z on D. Show that G maps D onto a horizontal strip of width of 2n, {x + ry: — oo < x < oo, c0 < y < c0 + 27c}, and that the mapping is onetoone on D. 25. Show that w = sin z maps the strip — n/2 < x < n/2 both onetoone and onto the region obtained by deleting from the plane the two rays ( — oo, — 1] and [1, oo) (see Fig. 1.26). (Hint: Use Exercise 22 and the fact that sin(z) = sin(z).) 26. Show that the function w = cos z maps the strip {0 < x < n} onetoone and onto the region D shown in Figure 1.26. Use this to define the inverse function to cos z. Derive the formula for arccos z given in the text. 27. Let 0 < a < 2. Show that an appropriate choice of log z for f(z) = za = exp[a log z] maps the domain {x + iy: y > 0} both onetoone and onto the domain {w: 0 < arg w < an}. Show that / also carries the boundary to the boundary (see Fig. 1.28). Chapter 1 The Complex Plane Figure 1.28 28. Show that the function w = g(z) = ez2 maps the lines x = y and x = — y onto the circle w = 1. Show further that g maps each of the two pieces of the region {x + iy: x2 > y2} onto the set {w: vv > 1} and each of the two pieces of the region {x + iy: x2 < y2} onto the set {w: w < 1}. Inverse Trigonometric Functions 29. Show directly that if ( is any value of HogCiz + yiz2), then sin £ = z. Likewise, show that if £ is any value of Mir* then tan £ = w. 30. Use the result in Exercise 29 and your knowledge of the branches of the logarithmic function to explain the branches of arcsin z. Line Integrals and Green's Theorem The fundamental theorems of complex variables depend on line integrals, so this section is devoted to that topic and to formulating Green's Theorem, the basic theorem about line integrals. Several consequences of Green's Theorem are also covered. Curves A curve y is a continuous complexvalued function y(t) defined for t in some interval [a, fc] in the real axis. The curve y is simple if y(^) ^ y(t2) whenever a ^tl < t2 <b, and it is closed if y(a) = y(b). The famous Jordan* Curve Theorem asserts that the complement of the range of a curve that is both simple and closed consists of two ♦Camille Jordan, 18381922. 1.6 Line Integrals and Green's Theorem 57 disjoint open connected sets, one bounded and the other unbounded. The bounded piece is the inside of the curve and the unbounded piece the outside. Despite the almost painful obviousness of this statement, the theorem is hard to prove.* We shall accept it as true. Suppose y is a curve; separate the complex number y(t) into its real and imaginary parts and write y(t) = x(t) + iy(t\ a ^t ^b. The functions x(t) and y(t) are realvalued functions of the real variable t, so they may (or may not) be differentiable. If both x(t) and y(t) are differentiable at t0, then we say that y(t) is differentiable at t0 and we set y'(t0) = x'(t0) + iy'(t0). (This is consistent with the usual rules in calculus for differentiating a vectorvalued function of a real variable.) A curve y is smooth if y(t) has the added property that y'(t) not only exists but is also continuous on [a, fc], the derivatives at a and b being taken from the right and left, respectively. A curve is piecewise smooth if it is composed of a finite number of smooth curves, the end of one coinciding with the beginning of the next. That is, the curve y is piecewise smooth if in the interval [a, b] there are points t0,tl9...,tn with a = t0 <tx <" < £„_! < tn = b such that y'(t) is continuous on each closed interval [£k, £k+i]> k = 0,1,..., n — 1. It is not required that y'(t) be continuous on all of fob]. NOTE: It is very common and convenient to refer to the range of y(t) as the curve y and to y(t) itself as the parametrization of the curve. With this use of the word curve, a curve becomes a concrete geometric object such as a circle or a straight line segment and hence is easily visualized. The difficulty with this view is that a particular curve has many different parametrizations. The curves we will use are generally composed of straight line segments and/or arcs of circles. These have standard parametrizations, as illustrated in the examples that follow. Example 1 Fix z0 and zx in the plane and let y(t) = tzx + (1 — t)z0, 0 ^ t ^ 1. This is a smooth simple curve whose range is the straight line segment joining z0 to zx (in that order). □ Example 2 Fix a point p in the plane and a positive number R; the curve y(t) = p + Rel\ 0 ^ t ^ In, is a smooth simple closed curve. The range is precisely the circle of radius R centered at p; the circle is traversed in the counterclockwise direction. □ Example 3 The square with vertices at z0, iz0, — z0, and — iz0 is the range of the piecewise smooth simple closed curve y(t) given by the rule tiz0 + (1  f)z0, 0 ^ t ^ 1 (t l)(z0) + (2t)(iz0), 1 <r<2 (t  2)(iz0) + (3  t)(z0\ 2 ^ t < 3 (t  3)(z0) + (4  t)(iz0\ 3 ^ t^ 4. d 7(0 = i * For a proof, see Newman, M. H. A. Elements of the topology of plane sets of points. Cambridge: Cambridge University Press, 1961. 58 Chapter 1 The Complex Plane In the process of computing certain definite integrals by means of the Residue Theorem (Section 6 of Chapter 2), we shall make use of several contours. As a way of gaining experience in the art of parametrization, we parametrize three of these contours in Examples 4 and 5. Example 4 Parametrize each of the curves in Figure 1.29. Solution The curve in (a) is the semicircle of radius R followed by the real segment from — R to R. A parametrization is 7' z = Re", z = x, R^x^R. The curve in (b) is a variation on the curve of (a), with the segment from — e to e replaced by the indicated semicircle. This curve is parametrized by r> z = Re", z = X, z = eeie9 z = X, O^O^n R^x^ n^9^0 e ^ x ^ R. — e Note that the third portion of the curve is parametrized by requiring that 9 decrease from n to 0; this is because we travel on that portion of y in the clockwise direction. □ (a) <b) Figure 1.29 Example 5 Find a parametrization of the "keyhole" contour shown in Figure 1.30. Solution The outer portion is on the circle \z\ = R and so is parametrized by z = Re", where 9 increases from 8 to In — <5. This takes us from A to B. We go from B to C on the ray with argument In — 5, so that segment is parametrized by z = tei{2n~8\ where t decreases from R to e. The inner portion of the keyhole contour is on the circle \z\ = e and so is parametrized by z = eeie9 where 9 decreases from 2n — 5 to 8. This takes us from C to D. Finally, to complete the contour, we go from D to A along the ray with argument 5. This segment is parametrized by z = teid, 1.6 Line Integrals and Green's Theorem 59 Figure 1.30 where t increases from e to R. In summary, then, r z = Re*, z = te*2lt8\ z = sew, z = te»9 5^9^2n5 R^t^e 2n8^0^S e ^ t < R. Each curve y is oriented by increasing t. The curve y begins at y(a), is traversed as t increases from a to b, and y ends at y(b). The reverse orientation is given to y by beginning at y(b) and ending at y(a); this curve is denoted by — y and is given by — y(t) = y(a + b — t\ a ^ £ ^ fc. A simple closed curve y is positively oriented if, for each point p on the inside of y, the argument of y(t) — p increases by 2n as t increases from a to b. Equivalently, y is positively oriented if, as you walk along y in the direction of the orientation of y, the inside of y is on your left. For example, a circle is positively oriented when it is traversed counterclockwise; the same is true for a triangle or a rectangle. Suppose g(t) = <r(t) + ix(t) is a continuous complexvalued function on the interval [a, b]. We define the integral of g over [a, b] by g(t) dt = a{t) dt + i\ x(t) dt. This definition, like the definition of the derivative of a complexvalued function of the real variable f, is consistent with the definition of the integral of a vectorvalued function of the real variable t studied in calculus. Note, for instance, that Re IMf {Re g(t)} dt, since both these expressions equal j* o{t) dt. 60 Chapter 1 The Complex Plane Suppose now that 7 is a smooth curve and u is a continuous function on the range of y. We define the line integral of u along y by u(z) dz = u(y(0)y'(0 dt9 where the right side is the integral of the complexvalued function w(y(0)y'(0 from a to b and is computed as discussed above. For a piecewise smooth curve y9 we define the line integral of u along y by f v;1 CtJ+l u(z)dz = £ u(y(s))y'(s)ds. The points t09tl9...9tH come from the definition of "piecewise smooth"; by assumption y'(s) is continuous on each segment [tj9 tj+1]9 j — 0, 1,..., n — 1. Line integrals have the familiar properties of definite integrals studied in calculus. For example, {Au(z) + Bv(z)} dz = A u(z) dz + B \ v(z) dz if A and B are complex numbers and w, v are continuous functions on the range of y. For the curve —y9 u(z) dz = u(y(a + b  t)){y(a + b  t)}' dt 1 = — u(y(s))y'(s) ds — — \ u(z) dz. =  I u{y{a + b  ij)y'{a + bt)dt; put s = a + b  t Suppose yx and y2 are two curves with parameter intervals \_al9 b{] and \al9 fc2], respectively. If y^b^ = y2(a2\ tnen the sum of yx and y2 is the curve Further, a simple computation shows that u(z) dz = u(z) dz + u(z) dz9 Jvi+y2 Jvi Jv2 provided, of course, that u is continuous on the range of both yx and y2. Let # be a complexvalued continuous function on [a9 b]; we shall show the 1.6 Line Integral