^ CHAPTER 1 3 Complex Numbers and Functions Complex numbers and their geometric representation in the complex plane are discussed in Sees. 13.1 and 13.2. Complex analysis is concerned with complex analytic functions as defined in Sec. 13.3. Checking for analyticity is done by the Cauchy-Riemann equations (Sec. 13.4). These equations are of basic importance, also because of their relation to Laplace's equation. The remaining sections of the chapter are devoted to elementary complex functions (exponential, trigonometric, hyperbolic, and logarithmic functions). These generalize the familiar real functions of calculus. Their detailed knowledge is an absolute necessity in practical work, just as that of their real counterparts is in calculus. Prerequisite: Elementary calculus. References and Answers to Problems: App. 1 Part D, App. 2. 13.1 Complex Numbers. Complex Plane Equations without real solutions, such as x2 = — 1 or x2 — lOx + 40 = 0, were observed early in history and led to the introduction of complex numbers.1 By definition, a complex number z is an ordered pair (x, y) of real numbers x and y, written z = (*, y). x is called the real part and y the imaginary part of z, written x = Re z, y = Im z. By definition, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. (0, 1) is called the imaginary unit and is denoted by i, (1) i = (0, 1). 602 1First to use complex numbers for this purpose was the Italian mathematician GIROI.AMO CARDANO (1501-1576), who found the formula for solving cubic equations. The term "complex number" was introduced by CARL H'RIEDRICH GAUSS (see the footnote in Sec. 5.4), who also paved the way for a general use of complex numbers. 603 Addition, Multiplication. Notation z = x + iy Addition of two complex numbers z,\ — (xx, yx) and z2 ~ fej >'2) is defined by (2) Zi + z2 = (xx, yx) + (x2, .y2) = C*i + x2, yx + y2). Multiplication is defined by (3) ZiZ2 = (*i> yi)(x2, y2) = (.v,.y:< - yxy2, .r, v:, + 3C2yi)-In particular, these two definitions imply that (%„ 0) + (X2, 0) = (A'i + x2, 0) and (.rl7 0)fe; 0) = (Xlx2, 0) as for real numbers xx, x2- Hence the complex numbers "extend" the real numbers. We can thus write (x, 0) = x. Similarly, (0, y) = iy because by (1) and the definition of multiplication we have iy = (0, l)y = (0, 1)0-, 0) = (0 • y - 1 • 0, 0 • 0 + 1 • y) = (0, y). Together we have by addition (x, y) = (x, 0) + (0, y) = x + iy: In practice, complex numbers z = (x, y) are written (4) z = x + iy or z = x + yi, e.g., 17 + 4/ (instead of z'4). Electrical engineers often write j instead if i because they need i for the current. If x = 0, then z = iy and is called pure imaginary. Also, (1) and (3) give (5) iz=-\ because by the definition of multiplication, iz = ii = (0, 1)(0, 1) = (—1,0) = —1. For addition the standard notation (4) gives [see (2)J Ui + iyj + (x2 + iyz) = (x1 + x2) + i(yx + y2). For multiplication the standard notation gives the following very simple recipe. Multiply each term by each other term and use iz = — 1 when it occurs [see (3)]: (x1 + iy\)(x2 + iy2) = xxx2 + /.v,y2 + i\\x2 + isyxy2 = (xxx2 - y,y2) + i(xxyz + x2yx). This agrees with (3). And it shows that x + iy is a more practical notation for complex numbers than (x, y). 604 CHAP. 13 Complex Numbers and Functions If you know vectors, you see that (2) is vector addition, whereas the multiplication (3) has no counterpart in the usual vector algebra. EXAMPLE 1 Real Part, Imaginary Part, Sum and Product of Complex Numbers Let z1 = 8 + 3i and z2 = 9 - 2i, Then Re Zi = 8, Im = 3, Re c2 = 9. Im z2 = -2 and ;t + z2 - (8 - 3/) + (9 - 20 = 17 + i. ZiZ2 = (8 - 30(9 - 20 72 + 6 + i'(-16 + 27) = 78 + Hi. ■ Subtraction, Division Subtraction and division arc defined as the inverse operations of addition and multiplication, respectively. Thus the difference .: .-) - z2 is the complex number z for which n = z + "2- Hence by (2), (6) Zi - z2 = (xj - .y2) + /(>>! - y2). The quotient z = zjzz {z2 ^ 0) is the complex number z for which Zi ' zz2. If we equate the real and the imaginary parts on both sides of this equation, setting z = x + iy, we obtain xj = x2x — y2 v, >'i = y2x + x2y. The solution is ~1 , . •*:lJC2 + >'l>'2 -<2.vl — (7*) Z = — = X + >V. X = 22 • >' = 2,2 ■ 42 A2 ^ v2 X2 T )2 The practical rule used to get this is by multiplying numerator and denominator of zi/z2 by x2 — iy2 and simplifiying: ^ = % + I.Vj = C*l + O'l) (*2 ~ 0:2) _ -*l-*2 + VjJa + . X2}'l ~ Xl)'2 x2 + iy2 (x2 + iy2) (x2 - iv2) x2z + y22 x2z + y2 EXAMPLE 2 Difference and Quotient of Complex Numbers For Zl = 8 + 3/ and z2 = 9 - 2; we get :,j - z2 = (8 + 30 - (9 - 2i) = - \ I 5i and £l 8 + 3; _ (8 + 30(9 + 20 _ 66 + 43/ 66 43 z2 ~ 9 - 2i ~ (9 - 20 (9 + 2/') 81+ 4 85 + 85 '' Check the division by multiplication to get 8 + 3/. Complex numbers satisfy the same commutative, associative, and distributive laws as real numbers (see the problem set). Complex Plane This was algebra. Now comes geometry: the geometrical representation of complex numbers as points in the plane. This is of great practical importance. The idea is quite simple and natural. We choose two perpendicular coordinate axes, the horizontal x-axis, called the real axis, and the vertical y-axis, called the imaginary axis. On both axes we choose the same unit of length (Fig. 315). This is called a Cartesian coordinate system. SEC. 13.1 Complex Numbers. Complex Plane 605 We now plot a given complex number z = (x, >') = x + iy as the point P with coordinates x, y. The xy-plane in which the complex numbers are represented in this way is called the complex plane.2 Figure 316 shows an example. Instead of saying "the point represented by z in the complex plane" we say briefly and simply "the point z in the complex plane." This will cause no misunderstandings. Addition and subtraction can now be visualized as illustrated in Figs. 317 and 318. Fig. 317. Addition of complex numbers Fig. 318. Subtraction of complex numbers Complex Conjugate Numbers The complex conjugate z of a complex number z = x + iy is defined by z ~ x — iy. It is obtained geometrically by reflecting the point z in the real axis. Figure 319 shows this for z = 5 + 2i and its conjugate z = 5 — 2i. Fig. 319. Complex conjugate numbers 2Somelimes called the Argand diagram, after the French mathematician JEAN ROBERT ARGAND (1768-1822). born in Geneva and later librarian in Paris. His paper on the complex plane appeared in 1806, nine years after a similar memoir by the Norwegian mathematician CASPAR WESSEL (1745-1818), a surveyor of the Danish Academy of Science. 606 CHAP. 13 Complex Numbers and Functions The complex conjugate is important because it permits us to switch from complex to real. Indeed, by multiplication, zz — xz + y2 (verify!). By addition and subtraction, z + z, = 2x, z — z = 2iy. We thus obtain for the real part x and the imaginary part y (not iy!) of z = x + iy the important formulas (8) Re z = x 1 (Z + z), 1 Tm z = y = — (z 2i z). If z is real, z = x, then z = z by the definition of z, and conversely. Working with conjugates is easy, since we have (zi + Za) -1 + Z2> (9) (Z1Z2) — ZiZ2> (zi - z2) EXAMPLE 3 Illustration of (8) and (9) Let zi = 4 + 3i and z2 = 2 + 5i. Then by (8). Im zi = ~ [(4 + 30 - (4 - 3/)] Also, the multiplication formula in (9) is verified by 3/ + 3/ = 3. (z1z2) = (4 + 30(2 + 50 = (-71 260 = -7 - 26i, zjz2 = (4 - 30(2 - 5/) = -7 - 26/. P ROBLEM S ET 13.1 1, i3 l, I 1, 1. (Powers of i) Show that ;2 = i5 = /,••• and l/i = -i, 1/i2 = -1, L/j3 = /,••■. 2. (Rotation) Multiplication by /' is geometrically a counterclockwise rotation through tt/2 (90=). Verify this by graphing z and iz and the angle of rotation for z. = 2 + 11, z = -1 - 50 z = 4 - 30 3. (Division) Verify the calculation in (7). 4. (Multiplication) If the product of two complex numbers is zero, show that at least one factor must be zero. 5. Show that z = x + iy is pure imaginary if and only if Z = ... 6. (Laws for conjugates) Verify (9) for z, = 24 + 100 z2 = 4 + 60 7-15 COMPLEX ARITHMETIC Let Zl = 2 + 3/ and zz 50 Showing the details of your work, find (in the form x + iy): 7. (5Zl + 3z2)2 8. zlZ2 9. Red/-!2) 10. Re(z22), (Re z2f 11. Vzi 12. zyz2, (Zl/z2) 13. (4Zl - z2)2 15. (Zl + z2)/(z, 14. Zi/zi, Zllzx i) 16-191 Let z = .v + /y. Find: 16. Im z3, (Im z)3 17. Re(l/z) 18. Im [(1 + i)8z2] 19. Re (1 Iz1) 20. (Laws of addition and multiplication) Derive the following laws for complex numbers from the corresponding laws for real numbers. + z. z2Zi (Commutative laws) 1 + (z2 + z3). (Associative laws) (l^Zs = Zi(z2Z3) Zi(z2 + z3) = ZiZ2 + ZiZa (Distributive law) 0 + z = z + 0 = z, + (-z) = (-z) + z = 0, z- 1 = z. 607 13.2 Polar Form of Complex Numbers. Powers and Roots The complex plane becomes even more useful and gives further insight into the arithmetic operations for complex numbers if besides the xv-coordinates we also employ the usual polar coordinates r, 9 defined by (1) x = r cos 0, y= r sin 0. We see that then z = x + iy takes the so-called polar form (2) z = Kcos 6 + i sin 9). r is called the absolute value or modulus of z and is denoted by \z\. Hence (3) \z\ = r = V777 = Vzz . Geometrically, \z\ is the distance of the point z from the origin (Fig. 320). Similarly, ki — Za| is the distance between zi and z2 (Pig- 321). 9 is called the argument of z and is denoted by arg z. Thus (Fig. 320) v (4) 9 = arg z = arctan — (z ¥= 0). x Geometrically, 6 is the directed angle from the positive x-axis to OP in Fig. 320. Here, as in calculus, all angles are measured in radians and positive in the counterclockwise sense. For z = 0 this angle 6 is undefined. (Why?) For a given z # 0 it is determined only up to integer multiples of 27r since cosine and sine are periodic with period 277. But one often wants to specify a unique value of arg z of a given z + 0. For this reason one defines the principal value Arg z (with capital A!) of arg z by the double inequality (5) - 77 < Arg Z=tt. Then we have Arg z = 0 for positive real z = x, which is practical, and Arg z ~ tt (not — tt!) for negative real z, e.g., for z = —4. The principal value (5) will be important in connection with roots, the complex logarithm (Sec. 13.7), and certain integrals. Obviously, for a given z + 0 the other values of arg z are arg z = Arg z ± 2nir(n = ±1, ±2, • • •)• Fig. 320. Complex plane, polar form of a complex number Fig. 321. Distance between two points in the complex plane 608 CHAP. 13 Complex Numbers and Functions EXAMPLE 1 Polar Form of Complex Numbers. Principal Value Arg z ; = 1 — i" (Fig. 322) has the polar form z = V 2 (cos \ir + i sin \tt). Hence we obtain 1 + i |z| = V'2. arg ; = jtt ± 2nir (n = 0. 1, • • •). and Arg z = \tt (the principal value). Similarly. ; = 3 + 3V3i = 6 (cos\tt + i sin j;ir), |;| = 6. and Arg z = \tt. I CAUTION! In using (4), we must pay attention to the quadrant in whieh z lies, since tan 9 has period tt, so that the arguments of z and — z have the same tangent. Example: Fig. 322. Example 1 for öi = arg (i + /} and 02 - arg (_, _ wc nave tan ^ = tan fl2 = 1. Triangle Inequality Inequalities such as x1 < x2 make sense for real numbers, but not in complex because there is no natural way of ordering complex numbers. However, inequalities between absolute values (which are real!), such as \zi\ < |s2| (meaning that Zi is closer to the origin than z2) are of great importance. The daily bread of the complex analyst is the triangle inequality (6) ki + z2| - \zi (Fig. 323) which we shall use quite frequently. This inequality follows by noting that the three points 0, Zi, and Zi + z2 are me vertices of a triangle (Fig. 323) with sides \zx\, |z2|, and \zi + z2\, and one side cannot exceed the sum of the other two sides. A formal proof is left to the reader (Prob. 35). (The triangle degenerates if Zi and z2 lie on the same straight line through the origin.) Fig. 323. Triangle inequality By induction we obtain from (6) the generalized triangle inequality (6*) + that is, the absolute value of a sum cannot exceed the sum of the absolute values of the terms. EXAMPLE 2 Triangle Inequality If zi = 1 + i and z% = ~2 + 3i, then (sketch a figure!) |zi + zal = |-1+ 4i| = Vl7 = 4.123 < V2 + Vl3 = 5.020. ■ Multiplication and Division in Polar Form This will give us a "geometrical" understanding of multiplication and division. Let Zi = ^(cos Ö] + i sin 0X) and r2(cos Ö2 + i sin ö2). SEC. 13.2 Polar Form of Complex Numbers. Powers and Roots 609 Multiplication. By (3) in Sec. 13.1 the product is at first ZiZ2 = r1r2[(cos 6X cos 02 — sin 6l sin 02) + '(sin d1 cos ti2 + cos 6L sin 02)). The addition rules for the sine and cosine [(6) in App. A3.1] now yield (7) Zlz2 = rir2[cos {6% + 02) + i sin (0, + 92)]. Taking absolute values on both sides of (7), we see that the absolute value of a product equals the product of the absolute values of the factors, (8) 1-ZiZal = kiltal- Taking arguments in (7) shows that the argument of a product equals the sum of the arguments of the factors, (9) arg (z.iZ2) = arg zi + arg z2 (up to multiples of 2tt). Division. We have Zl = (zi/z2)z2. Hence = ICZi^fel = ki^H^I ar)d by division by \z2\ (10) Zi fr (z2 * o). K2i Similarly, arg Z\ = arg [(Z}Iz2)z2] = arg (z\lz2) + arg z2 and by subtraction of arg z2 (11) arg — = arg Zi - arg z2 (up to multiples of 2tt). z2 Combining (10) and (11) we also have the analog of (7), (12) — = — [cos (61 - 02) + i sin {0X - 02)1, Zz >2 To comprehend this formula, note that it is the polar form of a complex number of absolute value i\lr2 and argument Qx — 02. But these are the absolute value and argument of Z\lz2, as we can see from (10), (11), and the polar forms of Z\ and z2. EXAMPLE 3 Illustration of Formulas (8)-(ll) Lei zi = -2 + 2i and z2 = 3i. Then ;,;2 = -6 - 6i, z1/z2 = 2/3 + (2/3)/. Hence (make a sketch) Izrtl = 6V2 = 3V8 = \Zl\\z2\, \Zl/z2\ = 2V2/3 = |Zl|/|z2|, and for the arguments we obtain Arg zi = 3-77/4, Arg z2 = it/2, 3lT tt Arg U]c2) = - — = Arg Zl + Arg z2 - 2ir, Arg (Zl/z2) = — = Arg Zl - Arg z2- 610 EXAMPLE 4 Integer Powers of z. De Moivre's Formula From (8) and (9) with z± = Z2 = z we obtain by induction lor n = 0, 1, 2, • • ■ (13) zn = rn (cos n£ + i sin nff). Similarly, (12) withzx = 1 and z2 = --"'gives (13) tor n = - I. -2, • ■ ■ . For |;| = r = 1, formula (13) becomes De Moivre's formula'1 (13*) (cos 9 + i sin f})™ = cos n9 + i sin n8. We can use this to express cos n8 and sin nO in terms of powers of cos $ and sin 0. For instance, for n = 2 we have on the left cos2 0 + 2/ cos ft sin 0 - sin2 ft. Taking the real and imaginary parts on both sides of (13*) with n = 2 gives the familiar formulas cos 20 - cos2 0 - sin2 ft, sin 20 = 2 cos 6 sin ft This shows that complex methods often simplify the derivation of real formulas. Try n = 3. Roots If z = wn (n = 1, 2, ■ ■ ■). then to each value of w there corresponds one value of z. We shall immediately see that, conversely, to a given z # 0 there correspond precisely n distinct values of w. Each of these values is called an nth root of z, and we write (14) w = tyi ■ Hence this symbol is multivalued, namely, n-valued. The n values of V z can be obtained as follows. We write z and w in polar form Z = r(cos 8 + i sin 0) ar|d vv = /?(cos + i sin instead of 6) w'"' = /?"(cos n4> + i sin nT = 1, -£ ± |V3 i, \VT = ±1, ±i, and ^T. If co denotes the value corresponding to k = 1 in (16), then the n values of VI can be written as 1, co, co2, ■ ■ ■ , on-\ More generally, if w1 is any nth root of an arbitrary complex number z 0), then the n values of Vz in (15) are (17) W-yCO2, because multiplying wt by cok corresponds to increasing the argument of wx by Ikiiln. Formula (17) motivates the introduction of roots of unity and shows their usefulness. Fig. 324. V>T y (0 ^\ \ \ \\ TV X \ V / 1 \ ^s / / CO3 CO2 / 1 y CO \\ Vi \ X /1 up / / "(O4 Fig. 325. VI Fig. 326. VI r K v#o L t:M SET I J .2. 1-8 POLAR FORM Do these problems very carefully since polar forms will be needed frequently. Represent in polar form and graph in the complex plane as in Fig. 322 on p. 608. (Show the details of your work.) 1. 3 - 3i 3. -5 1 + i 2. 2i, -2i 4. 1 + l-n-i 3V'2 - -V2 - (2/3)/' 612 CHAP. 13 Complex Numbers and Functions -6 + 5/ 3/ 2 + 3/ 5 + 4/ 9-15 PRINCIPAL ARGUMENT Determine the principal value of the argument. 9. -1 - / 10. -20 + i, -20 - i 11. 4 ± 3/ 12. -tt2 13. 7 ± 7< 14. (1 + i)la 15. (9 + 9if 16-20 CONVERSION TO x + iy Represent in the form x + iy and graph it in the complex plane. 16. cos iir + i sin {±\ir) 17. 3(cos 0.2 + i sin 0.2) 18. 4(cos|i7 ± i sing7r) 19. cos (-1) + / sin(-l) 20. 12(cos|tt + i sin §77) 21-25 ROOTS Find and graph all roots in the complex plane. .8/ 21. V-i 23. 25. TT" 22. VI 24. vTT4/ ■1 26. TEAM PROJECT. Square Root, (a) Show that w = Vz has the values (18) w2 = Vr cos — + / sin — 2 2 cos I — + 77"J + ( sin ^ — - where sign y = 1 if y :' 0, sign y = — 1 if v < 0, and all square roots of positive numbers are taken with positive sign. Hint: Use (10) in App. A3.1 with x = 8/2. (c) Find the square roots of 4/, 16 - 30i, and 9 + 8 Vli by both (18) and (19) and comment on the work involved. (d) Do some further examples of your own and apply a method of checking your results. 27-30 EQUATIONS Solve and graph all solutions, showing the details: 27. z2 ~ (8 - 5i)z + 40 - 20/ = 0 (Use (19).) 28. z4 + (5 - 140z2 - (24 + 10/) = 0 29. 8z2 - (36 - 6i)z + 42 - 11/ = 0 30. z4 + 16 = 0. Then use the solutions to factor z4 + 16 into quadratic factors with real coefficients. 31. CAS PROJECT. Roots of Unity and Their Graphs. Write a program for calculating these roots and for graphing them as points on the unit circle. Apply the program to zn = 1 with n = 2, 3, • ■ •, 10. Then extend the program to one for arbitrary roots, using an idea near the end of the text, and apply the program to examples of your choice. 32-351 INEQUALITIES AND AN EQUATION Verify or prove as indicated. 32. (Re and Im) Prove |Re z| £ |z|, 33. (Parallelogram equality) Prove Im z + z2 2(1 Z!I \z2n (b) Obtain from (18) the often more practical formula (19) Vi = ±[Vi(|z| +x) + (signy)/V±(|z| + x)\ Explain the name. 34. (Triangle inequality) Verify (6) for . z2 = 5 + 2i. 35. (Triangle inequality) Prove (6). 4 + 7/. 13.3 Derivative. Analytic Function Our study of complex functions will involve point sets in the complex plane. Most important will be the following ones. Circles and Disks. Half-Planes The unit circle \z\ = 1 (Fig. 327) has already occurred in Sec. 13.2. Figure 328 shows a general circle of radius p and center a. Its equation is SEC. 13.3 Derivative. Analytic Function 613 because it is the set of all z whose distance \z — a\ from the center a equals p. Accordingly, its interior ("open circular disk") is given by \z - a\ < p, its interior plus the circle itself ("closed circular disk") by \z — a\ = p, and its exterior by \z — a\ > p. As an example, sketch this for a = 1 + i and p = 2, to make sure that you understand these inequalities. An open circular disk \z — a\ < p is also called a neighborhood of a or, more precisely, a p-neighborhood of a. And a has infinitely many of them, one for each value of p (> 0), and a is a point of each of them, by definition! In modern literature any set containing a p-neighborhood of a is also called a neighborhood of a. Figure 329 shows an open annulus (circular ring) p^ < \z — a\ < p2, which we shall need later. This is the set of all z whose distance \z — a\ from a is greater than px but less than p2. Similarly, the closed annulus px ^ \z — a\ '— p2 includes the two circles. Half-Planes. By the (open) upper half-plane we mean the set of all points z = x + iy such that y > 0. Similarly, the condition y < 0 defines the lower half-plane, x > 0 the right half-plane, and x < 0 the left half-plane. For Reference: Concepts on Sets in the Complex Plane To our discussion of special sets let us add some general concepts related to sets that we shall need throughout Chaps. 13-18; keep in mind that you can find them here. By a point set in the complex plane we mean any sort of collection of finitely many or infinitely many points. Examples are the solutions of a quadratic equation, the points of a line, the points in the interior of a circle as well as the sets discussed just before. A set 5 is called open if every point of S has a neighborhood consisting entirely of points that belong to S. For example, the points in the interior of a circle or a square form an open set, and so do the points of the right half-plane Re z = x > 0. A set S is called connected if any two of its points can be joined by a broken line of finitely many straight-line segments all of whose points belong to S. An open and connected set is called a domain. Thus an open disk and an open annulus are domains. An open square with a diagonal removed is not a domain since this set is not connected. (Why'.') The complement of a set S in the complex plane is the set of all points of the complex plane that do not belong to 5. A set S is called closed if its complement is open. For example, the points on and inside the unit circle form a closed set ("closed unit disk") since its complement |z| > 1 is open. A boundary point of a set S is a point every neighborhood of which contains both points that belong to S and points that do not belong to S. For example, the boundary 614 points of an annulus are the points on the two bounding circles. Clearly, if a set S is open, then no boundary point belongs to S; if S is closed, then every boundary point belongs to S. The set of all boundary points of a set S is called the boundary of S. A region is a set consisting of a domain plus, perhaps, some or all of its boundary points. WARNING! "Domain" is the modern term for an open connected set. Nevertheless, some authors still call a domain a "region" and others make no distinction between the two terms. Complex Function Complex analysis is concerned with complex functions that are di ITerentiable in some domain. Hence we should first say what wc mean by a complex function and then define the concepts of limit and derivative in complex. This discussion will be similar to that in calculus. Nevertheless it needs great attention because it will show interesting basic differences between real and complex calculus. Recall from calculus that a real function / defined on a set S of real numbers (usually an interval) is a rule that assigns to every x in S a real number f(x), called the value of / at a-. Now in complex, S is a set of complex numbers. And a function f defined on S is a rule that assigns to every z in S a complex number w, called the value of / at z. We write w = f(z). Here z varies in 5 and is called a complex variable. The set S is called the domain of definition of / or, briefly, the domain of /. (In most cases 5 will be open and connected, thus a domain as defined just before.) Example: w = f(z) — z2 + 3z is a complex function defined for all z; that is, its domain S is the whole complex plane. The set of all values of a function / is called the range off. w is complex, and we write w = u + iv, where u and v are the real and imaginary parts, respectively. Now w depends on z = x + iy. Hence u becomes a real function of x and y, and so does v. We may thus write w = f(z) = u(x, y) + iv(x, y). This shows that a complex function f(z) is equivalent to a pair of real functions u(x, y) and v(x, y), each depending on the two real variables x and y. EXAMPLE 1 Function of a Complex Variable Let w = /(-) = :2 + 3;. Find u and v and calculate the value of / at z — 1 + 3t. Solution, u = Re f(z) = x2 - v2 + 3x and v = Ixy + 3y. Also, f(l + 3<) = (1 + 3/)2 + 3(1 + 3i) = 1- 9 +6i+ 3 + 9/ = -5 + 15/. This shows that h(1, 3) = -5 and o(l, 3) = 15. Check this by using the expressions for u and v. H EXAMPLE 2 Function of a Complex Variable Let w = f(z) = 2iz + 6z. Find u and v and the value of / at z = f + 4i. Solution, fiz) = 2i(x + iy) + 6(jc - iy) gives u(x. y) = 6x - 2y and v(x, y) = 2x - 6v. Also, f{\ + 40 = 2i(& + 4Q + 6(| 40 = i ~ 8 -I 3 - 24i = -5 - 23;'. Check this as in Example 1. SEC. 13.3 Derivative. Analytic Function 615 Remarks on Notation and Terminology 1. Strictly speaking, f{z) denotes the value of / at z, but it is a convenient abuse of language to talk about the function f(z) (instead of the function /), thereby exhibiting the notation for the independent variable. 2. We assume all functions to be single-valued relations, as usual: to each z in S there coircsponds but one value w = f(z) (but, of course, several z may give the same value w = f(z), just as in calculus). Accordingly, we shall not use the term "multivalued function" (used in some books on complex analysis) for a multivalued relation, in which to a z there corresponds more than one w. Limit, Continuity A function f(z) is said to have the limit / as z approaches a point z0. written (1) lim f{z.) = I if / is defined in a neighborhood of z0 (except perhaps at z0 itself) and if the values of / are "close" to / for all z "close" to z0; in precise terms, if for every positive real e we can find a positive real 8 such that for all z + Zn in the disk \z — Zo\ < 8 (Fig. 330) we have (2) \f(z) ~l\z0 Note that by definition of a limit this implies that f(z) is defined in some neighborhood of z0- f(z) is said to be continuous in a domain if it is continuous at each point of this domain. Fig. 330. Limit 616 CHAP. 13 Complex Numbers and Functions Derivative The derivative of a complex function / at a point z0 is written f'(zo) and is defined by (4) f (zo) = i™o-A~z- provided this limit exists. Then / is said to be differentiable at z0. If we write Az = z — za, we have z = z.0 + Az and (4) takes the form ,,. , .. f(z) - f(z0) (4') f (zo) = lim--- . *-*«o Z Zo Now comes an important point. Remember that, by the definition of limit, f(z) is defined in a neighborhood of z0 and z in (4') may approach z0 from any direction in the complex plane. Hence differentiability at z0 means that, along whatever path z approaches z0, the quotient in (4') always approaches a certain value and all these values are equal. This is important and should be kept in mind. EXAMPLE 3 Differentiability. Derivative The function f(z) = z2 is differentiable for all z and has the derivative /'(z) = 2z because (z + Az)2 - z2 z2 + 2zAz + (Az)2 - z2 _ f (z) = lim -:- = lim--:- = lim (2z + Az) = 2z. Az Az^U Az Az »0 The differentiation rules are the same as in real calculus, since their proofs are literally the same. Thus for any analytic functions / and g and constants c we have (cf)' = c/\ (/ + g)' = f + g', (fg)' = f'g + fg', f'g - fg' g2 as well as the chain rule and the power rule (z")' = nzn 1 (n integer). Also, if f(z) is differentiable at z0, it is continuous at z(). (See Team Project 26. EXAMPLE 4 z not Differentiable It may come as a surprise that there are many complex functions that do not have a derivative at any point. For instance, f(z) = z = x - iy is such a function. To see this, we write Az = A.v 4 /Ay and obtain f(z + Az) - f(z) _ (z + Az) - z _ Az _ Ax - iAy (3) Az Az ~ Az ~ Ax + iAy ' If Ay = 0, this is i I. If A.t = 0, this is - 1. Thus (5) approaches +1 along path I in Fig. 331 but 1 along path II. Hence, by definition, the limit of (5) as Az —» 0 does not exist at any z. ■ -Qz + Lz Fig. 331. Paths in (5) SEC. 13.3 Derivative. Analytic Function 617 Surprising as Example 4 may be, it merely illustrates that differentiability of a complex function is a rather severe requirement. The idea of proof (approach of z from different directions) is basic and will be used again as the crucial argument in the next section. Analytic Functions Complex analysis is concerned with the theory and application of "analytic functions," that is, functions that are differentiable in some domain, so that we can do "calculus in complex." The definition is as follows. DEFINITION Analyticity A function f(z) is said to be analytic in a domain D if f(z) is defined and differentiable at all points of D. The function f(z) is said to be analytic at a point Z = Zo in D if f(z) is analytic in a neighborhood of z0. Also, by an analytic function we mean a function that is analytic in some domain. Hence analyticity of f(z) at z0 means that f(z) has a derivative at every point in some neighborhood of z0 (including z0 itself since, by definition, /0 is a point of all its neighborhoods). This concept is motivated by the fact that it is of no practical interest if a function is differentiable merely at a single point zn but not throughout some neighborhood of z0. Team Project 26 gives an example. A more modern term for analytic in D is holomorphic in D. EXAMPLE 5 Polynomials, Rational Functions The nonnegative integer powers 1, z, z , ■ • ■ arc analytic in the entire complex plane, and so are polynomials, that is, functions of the form /(z) = c0 + c, where c0, ■ ■ ■ , cn are complex constants. The quotient of two polynomials g(z) and /?(;), f(z) c2z° + h(z) is called a rational function. This f is analytic except at the points where h{z) = 0; here wc assume that common factors of g and h have been canceled. Many further analytic functions will be considered in the next sections and chapters. The concepts discussed in this section extend familiar concepts of calculus. Most important is the concept of an analytic function, the exclusive concern of complex analysis. Although many simple functions are not analytic, the large variety of remaining functions will yield a most beautiful branch of mathematics that is very useful in engineering and physics. .......mmmmmtmmmmm SE 1-101 CURVES AND REGIONS OF PRACTICAL INTEREST Find and sketch or graph the sets in the complex plane given by 1. |z - 3 - 2/| = | 2. 1 § |z - 1 + 4i| S 5 3. 0 < \z - 1| 5. Im z2 = 2 7. \z + 1| = \z 9. Re z ^ Imz 4. - v < Re z < 6. Re z > - 1 8. |Arg z\ S \ir 10. Re < 1 CHAP. 13 Complex Numbers and Functions 11. WRITING PROJECT. Sets in the Complex Plane. Extend the part of the text on sets in the complex plane by formulating that part in your own words and including examples of your own and comparing with calculus when applicable. COMPLEX FUNCTIONS AND DERIVATIVES 12-151 Function Values. Find Re / and Im f. Also find their values at the given point z. 12. / = 3z2 - 6z + 3/, z = 2 + i 13. / = z/(z + 1), z = 4 - 51 14. / = 1/(1 - z), z = I + \i 15. / - \/z2, z = 1 + i 16-19 Continuity. Find out (and give reason) whether f(z) is continuous at z = 0 if /(()) = 0 and for z + 0 the function / is equal to: 16. [Re (:2)]/|z|2 17. [Im (z2)]/|z| 18. |z|2 Re(l/z) 19. (Tmz)/(1 - |z|) 20-24 Derivative. Differentiate 20. (z2 - 9)/(z2 + 1) 21. (z3 + if 22. (3; + 4i)/(1.5iz - 2) 23. (7(1 - z)2 24. z2Hz + if 25. CAS PROJECT. Graphing Functions. Find and graph Re f, Im /, and |/| as surfaces over the z-plane. Also graph the two families of curves Re f(z) = const and Im f(z) = const in the same figure, and the curves j/(z)| = const in another figure, where (a) f(z) = z2, (b) f(z) = 1/z, (c) f(z) = ;4. 26. TEAM PROJECT. Limit, Continuity, Derivative (a) Limit. Prove that (1) is equivalent to the pair of relations lim Re f(z) = Re /, lim Im /(z) = Im /. (b) Limit. If lint f(z) exists, show that this limit is unique. * (c) Continuity. If z,, z2, • ■ • are complex numbers for which lim zn = a, and if f(z) is continuous at n- >co z = a, show that lim /(?,„.) = f(a). (d) Continuity. If f(z) is differentiable at z0, show that /(z) is continuous at z0. (e) Differentiability. Show that f(z) = Re z = x is not difi'erentiable at any z. Can you find other such functions? (f) Differentiability. Show that f(z) = \z\2 is differentiable only at z = 0; hence it is nowhere analytic. 13.4 Cauchy-Riemann Equations. Laplaces Equation The Cauchy-Riemann equations are the most important equations in this chapter and one of the pillars on which complex analysis rests. They provide a criterion (a test) for the analyticity of a complex function w = f(z) = u(x, y) + iv(x, v). Roughly, f is analytic in a domain D if and only if the first partial derivatives of u and v satisfy the two Cauchy-Riemann equations4 (1) Ux = Vy, Uy = ~VX 4The French mathematician AUGUSTIN-LOUIS CAUCHY (see Sec. 2.5) and the German mathematicians BERNIIARD RIEMANN (1826-1866) and KARL WEIERSTRASS (1815-1897: see also Sec. 15.5) are the founders of complex analysis. Riemann received his Ph.D. (in 1851) under Gauss (Sec. 5.4) at Gottingcn, where he also taught until he died, when he was only 39 years old. He introduced the concept of the integral as it is used in basic calculus courses, and made important contributions to differential equations, number theory, and mathematical physics. He also developed the so-called Riemannian geometry, which is the mathematical foundation of Hinstein's theory of relativity; see Ref. [GR91 in App. 1. SEC. 13.4 Cauchy-Riemann Equations. Laplace's Equation 619 everywhere in D; here ux = du/Bx and uy = du/dy (and similarly for v) are the usual notations for partial derivatives. The precise formulation of this statement is given in Theorems 1 and 2. Example: f(z) z2 = x2 2ixy is analytic for all z (see Example 3 in Sec. 13.3), and u y and v = 2xy satisfy (1), namely, ux More examples will follow. 2x = vy as well as uy -2y THEOREM 1 Cauchy-Riemann Equations Let /(z) = u(x, y) + lv(x, y) be defined and continuous in some neighborhood of a point z — x + iy and differentiable at z. itself. Then at that point, the first-order partial derivatives of it and v exist and satisfy the Cauchy-Riemann equations (1). Hence iff(z) is analytic in a domain D, those partial derivatives exist and satisfy (I) at all points of D. PROOF By assumption, the derivative f'(z) at z exists. It is given by ./<:: + Az) - f(z) (2) f (z) = lim Ä2-^0 A; The idea of the proof is very simple. By the definition of a limit in complex (Sec. 13.3) we can let Az approach zero along any path in a neighborhood of z. Thus we may choose the two paths 1 and II in Fig. 332 and equate the results. By comparing the real parts we shall obtain the first Cauchy-Riemann equation and by comparing the imaginary parts the second. The technical details are as follows. We write Az = Ajc + /Ay. Then z + Az = x + A.y + i(y + Ay), and in terms of u and v the derivative in (2) becomes (3) f (z) = lim [u(x + Ax, y + Ay) + iv(x + Ax, y + Ay)] - [u(x, y) + iv(x, y)] Aj iAy We first choose path I in Fig. 332. Thus we let Ay -> 0 first and then Ax 0. After Ay is zero, Az = Ax Then (3) becomes, if we first write the two w-terms and then the two u-terms, , u(x + Ax, v) — u(x. y) v(x + Ax. v) — v(x, y) f (z) = lim---■--h i lim —---— . ax >o Ax ax^o At - -Q2 + Az Fig. J32. Paths in (2) 620 CHAP. 13 Complex Numbers and Functions Since f'(z) exists, the two real limits on the right exist. By definition, they are the partial derivatives of u and v with respect to x. Hence the derivative f'(z) of /(z) can be written (4) f(t) = ux + ivx. Similarly, if we choose path II in Fig. 332, we let Ax —» 0 first and then Av —» 0. After Ax is zero, Az = iAy, so that from (3) we now obtain , u(x, v + Av) - u(x, v) v(x, y + Av) - v(x, y) f (z) = lim---:- + i lim---. Ay—o /Ay *v-*0 7 A}' Since f'(z) exists, the limits on the right exist and give the partial derivatives of u and v with respect to y; noting that \li = —i, we thus obtain (5) !'{Z] =r -illy + Vi The existence of the derivative f'(z) thus implies the existence of the four partial derivatives in (4) and (5). By equating the real parts ux and vy in (4) and (5) we obtain the first Cauchy-Riemann equation (I). Equating the imaginary parts gives the other. This proves the first statement of the theorem and implies the second because of the definition of analyticity. B Formulas (4) and (5) are also quite practical for calculating derivatives /'(z), as we shall see. EXAMPLE 1 Cauchy-Riemann Equations /(z) = z2 is analytic for all z. It follows that the Cauchy-Riemann equations must be satisfied (as we have verified above). For f(z) — Z = X — iy we have u = x. v = - v and sec that the second Cauchy-Riemann equation is satisfied. uy = —vx = 0, but the first is not: ux = 1 j= vy = -]. We conclude that /(z) = z is not analytic, confirming Example 4 of Sec. 13.3. Note the savings in calculation! The Cauchy-Riemann equations are fundamental because they are not only necessary but also sufficient for a function to be analytic. More precisely, the following theorem holds. THEOREM 2 Cauchy-Riemann Equations If two real-valued continuous functions u(x, y) and v(x, y) of two real variables x and y have continuous first partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the complex function f(z) = u(x, y) + iv(x, y) is anahtic in D. The proof is more involved than that of Theorem 1 and we leave it optional (see App. 4). Theorems 1 and 2 are of great practical importance, since by using the Cauchy-Riemann equations we can now easily find out whether or not a given complex function is analytic. SEC. 13.4 Cauchy-Riemann Equations. Laplace's Equation 621 EXAMPLE 2 Cauchy-Riemann Equations. Exponential Function Is f(z) = u(x, y) + iv(x, y] ex(cosy + i sin v) analytic? Solution. Wc have u = ex cos y, u = ex sin v and by differentiation ux = ex cos y, vy = ex cos y iiy = —ex sin y, ux = ex sin y. Wc sec that the Cauchy-Riemann equations are satisfied and conclude that f(z) is analytic for all z. (flz) will he the complex analog of ex known from calculus.) EXAMPLE 3 An Analytic Function of Constant Absolute Value Is Constant The Cauchy-Riemann equations also help in deriving general properties of analytic functions. For instance, show that if f(z) is analytic in a domain D and \f(z)\ = k = const in D, then /(;) = const in D. (We shall make crucial use of this in Sec. 18.6 in the proof of Theorem 3.) Solution. By assumption, \ff = \u + ivf = u2 + v2 = k2. By differentiation, uux + vvx = 0. Ully + VUy = 0. Now use vx = -Uy in the first equation and vy = ux in the second, to get (a) uur — vu,, = 0, (6) (h) inly + vux = 0. To get rid of uy, multiply (6a) by u and (6b) by v and add. Similarly, to eliminate itx, multiply (6a) by -v and (6b) by u and add. This yields iu2 - u2)ux = 0. in2 + v2)uv 0. If k2 = u2 + v2 = 0, then « v = 0; hence / = 0. If k2 - u2 + v2 i 0, then ux - uy = 0. Hence, by the Cauchy-Riemann equations, also vx = vy 0. Together this implies u = const and u = const; hence / — const. We mention that if we use the polar form z = r(cos 0 + i sin 6) and set f(z) = u(r, 6) + iv{r, ff), then the Cauchy-Riemann equations arc (Prob. 11) Ur = — Vg, (7) (r > 0). 1 Vr =--Llg r Laplace's Equation. Harmonic Functions The great importance of complex analysis in engineering mathematics results mainly from the fact that both the real part and the imaginary part of an analytic function satisfy Laplace's equation, the most important PDE of physics, which occurs in gravitation, electrostatics, fluid flow, heat conduction, and so on (see Chaps. 12 and 18). 622 CHAP. 13 Complex Numbers and Functions THEOREM 3 Laplace's Equation If f(z) — u(x, y) + iv(x, y) is analytic in a domain D, then both u and v satisfy Laplace's equation (8) V2m = uxx + u = 0 (V2 read "nabla squared") and (9) in D and have continuous second partial derivatives in D. O = vxx + vyy = 0, PROOF Differentiating ux = vy with respect to x and uy = —vx with respect to y, we have (10) Uxx — VyX. Uyy — ~ V y-y . Now the derivative of an analytic function is itself analytic, as we shall prove later (in Sec. 14.4). This implies that u and v have continuous partial derivatives of all orders; in particular, the mixed second derivatives are equal: vyx = vxy. By adding (10) we thus obtain (8). Similarly, (9) is obtained by differentiating ux = vy with respect to y and uy — ~ux with respect to x and subtracting, using uxy = uyx. ■ Solutions of Laplace's equation having continuous second-order partial derivatives are called harmonic functions and their theory is called potential theory (see also Sec. 12.10). Hence the real and imaginary parts of an analytic function are harmonic functions. If two harmonic functions u and v satisfy the Cauchy-Riemann equations in a domain D, they are the real and imaginary parts of an analytic function / in D. Then v is said to be a harmonic conjugate function of u in D. (Of course, this has absolutely nothing to do with the use of "conjugate" for z.) EXAMPLE 4 How to Find a Harmonic Conjugate Function by the Cauchy-Riemann Equations Verify thai u = x2 — y2 — y is harmonic in the whole complex plane and find a harmonic conjugate function V of it. Solution. V2u = 0 by direct calculation. Now ux = 2x and uy = -2y — 1. Hence because of the Cauchy-Riemann equations a conjugate v of u must satisfy Vy = ux = 2x, vx = —Uy = 2y + 1. Integrating the first equation with respect to y and differentiating the result with respect to .v, wc obtain dh V = 2xv + h(x). vr = 2v + — . ax A comparison with the second equation shows that dh/dx = I. This gives h(x) = x -r c. Hence i; = 2xy - x + c (c any real constant) is the most general harmonic conjugate of the given it. The corresponding analytic function is f(z) = u + iv = x2 - y2 - y + i(2xy + x + c) = :2 + n + ic. ■ SEC. 13.5 Exponential Function 623 Example 4 illustrates that a conjugate of a given harmonic function is uniquely determined up to an arbitrary real additive constant. The Cauchy-Riemann equations are the most important equations in this chapter. Their relation to Laplace's equation opens wide ranges of engineering and physical applications, as we shall show in Chap. 18. 1-10 CAUCHY-RIEMANN EQUATIONS Are the following functions analytic? [Use (1) or (7).] l. m = z* 3. e2x(cos, y + i sin y) 5. e~x(cos y — i sin y) 7. f(z) = Re z + Im - 9. f(z) = //z8 2. f(z) = lm (z2) 4. f(z) = 1/(1 - Z4) 6. f(z) = Arg ttz 8. f(z) = In |z| + i Arg z 10. f(z) = z2 + 1/z2 11. (Cauchy-Riemann equations in polar form) Derive (7) from (1). 12-211 HARMONIC FUNCTIONS Are the following functions harmonic? If your answer is yes, find a corresponding analytic function f(z) = u{x. y) + iv(x, y). y/(.t2 + y2) 12. u = .vi 14. v = - 16. v = In I -1 18. u = l/(x2 + y2) 20. u = cos x cosh v 13. v = xy 15. u = In I 17. u v3 3.vy: 19. d = (x2 - y2)2 21. u = e~x sin 2y 22-24 Determine «, b. c such lhat the given functions are harmonic and find a harmonic conjugate. 22. u 24. u 23. it = sin x cosh cv ax3 + by3 25. (Harmonic conjugate) Show that if u is harmonic and v is a harmonic conjugate of «, then u is a harmonic conjugate of —v. 26. TEAM PROJECT. Conditions forf(z) = const. Let /(z) be analytic. Prove that each of the following conditions is sufficient for f(z) = const. (a) Re f (z) = const (b) Im f (z) = const (c) f'(z) = 0 (d) |/(z)| = const (see Example 3) 27. (Two further formulas for the derivative). Formulas (4), (5), and (11) (below) are needed from time to time. Derive (11) f'(z) f'(z) Vy + tV, 28. CAS PROJECT. Equipotcntial Lines. Write a program for graphing equipotential lines u = const of a harmonic function u and of its conjugate v on the same axes. Apply the program to (a) u = x v = 2xy. (b) u = x3 - 3.vy2, v = 3x2y (c) ti = ex cos y, v = ex sin v. .2 13.5 Exponential Function In the remaining sections of this chapter we discuss the basic elementary complex functions, the exponential function, trigonometric functions, logarithm, and so on. They will be counterparts to the familiar functions of calculus, to which they reduce when z = x is real. They are indispensable throughout applications, and some of them have interesting properties not shared by their real counterparts. We begin with one of the most important analytic functions, the complex exponential function ez, also written exp z. The definition of ez in terms of the real functions ex, cos y, and sin y is (1) ez = ex(cos y + i sin y). 624 CHAP. 13 Complex Numbers and Functions This definition is motivated by the fact the ez extends the real exponential function ex of calculus in a natural fashion. Namely: (A) ez = ex for real z = x because cos y = 1 and sin y = 0 when y = 0. (B) e'~ is analytic for all z. (Proved in Example 2 of Sec. 13.4.) (C) The derivative of ez is ez, that is, (2) (ez)' = ez. This follows from (4) in Sec. 13.4, (ez)' = (ex cos y)x + i(ex sin y)x = ex cos y + iex sin y = ez. REMARK. This definition provides for a relatively simple discussion. We could define ez by the familiar series 1 + x + x2/2\ + x3/3\ + • • • with x replaced by z, but we would then have to discuss complex series at this very early stage. (We will show the connection in Sec. 15.4.) Further Properties. A function f(z.) that is analytic for all z is called an entire function. Thus, ez is entire. Just as in calculus the functional relation (3) e = e e holds for any Zl = xx + iy1 and z2 = x2 + iy2- Indeed, by (1), eHeZx = eXl(cosy! + i sin y{) e 2(cos y2 + (' siny2). Since eXleXz = eXl+x,i for these real functions, by an application of the addition formulas for the cosine and sine functions (similar to that in Sec. 13.2) we see that /i/2 = / ":t|eos (_Vl + y2) + i sin (y, + y2)] = /1+"2 as asserted. An interesting special case of (3) is z.i = x, z2 = then (4) ez = exeiv. Furthermore, for z = iy we have from (1) the so-called Euler formula (5) e'v = cos y + i sin y. Hence the polar form of a complex number, z = r(cos t) + i sin 9), may now be written (6) z = re16. From (5) we obtain (7) e2m = 1 as well as the important formulas (verify!) (8) em'2 = i, e^ = -1, e-ml2 = -/, e~m = -1. SEC. 13.5 Exponential Function 625 Another consequence of (5) is (9) \ew\ = |cos y + i sin y\ = Vcos2_y + sin2 cos y + sin y = I. That is, for pure imaginary exponents the exponential function has absolute value 1, a result you should remember. From (9) and (I), (10) \ez\ = ex. Hence arg é = y ± 2mr (n = 0, 1, 2, • • ■), since \er\ = ex shows that (1) is actually ez in polar form. From \ez\ = ex # 0 in (10) we see that (11) é * 0 for all z. So here we have an entire function that never vanishes, in contrast to (nonconstant) polynomials, which are also entire (Example 5 in Sec. 13.3) but always have a zero, as is proved in algebra. Periodicity of ez with period 2m, (12) é12™ = é for all z is a basic property that follows from (1) and the periodicity of cos y and sin y. Hence all the values that w = é can assume are already assumed in the horizontal strip of width 2tt (13) - 77 < y S 77 (Fig. 333). This infinite strip is called a fundamental region of ez. EXAMPLE 1 Function Values. Solution of Equations. Computation of values from (1) provides no problem. For instance, verify that el.4-o.ei = ei-4(coso.6 - j sin 0.6) = 4.055(0.8253 - 0.5646r) = 3.347 - 2.289/ To illustrate (3), take the product of and and verify that it equals eze'i(cosz 1 + sinz \) = e6 = e" Fig. 333. Fundamental region of the exponential function ez in the z-plane 626 CHAP. 13 Complex Numbers and Functions To solve the equation ez = 3 + 4f, note first that \ez\ = ex 5, x = In 5 = 1.609 is the real part of all solutions. Now, since ex = 5, ex cos v = 3, ex sin.y = 4, cosy = 0.6, sin.v = 0.8. y = 0.927. Am. z = 1.609 + 0.927/ ± 2nm in = 0, 1, 2, • • ■). These are infinitely many solutions (due to the periodicity of ez). They lie on the vertical line x = 1.609 at a distance 2tt from their neighbors. To summarize: many properties of ez = exp z parallel those of ex\ an exception is the periodicity of ez with 2m, which suggested the concept of a fundamental region. Keep in mind that ez is an entire function. (Do you still remember what that means?) 1. Using the Cauchy-Riemann equations, show that ez is entire. 2-8 Values of ez. Compute ez in the form u + iv and ez\, where z equals: 2. 3 + 17/ 3. 1 + 2/ 4. V2 - \iri 5. 777/72 6. (1 + i)ir 7. 0.8 - 5; 8. 9it//2 9-12 Real and Imaginary Parts. Find Re and Itn of: 9. e~2z 10. e'3 11. e'Z 12. ellz 13-17 13. Vi Is. V z 17. -9 Polar Form. Write in polar form: 14. 1 + i 16. 3 + 4i 18-21 Equations. Find all solutions and graph some of them in the complex plane. 18. e3z = 4 20. ez = 0 19. ez 21. ez -2 4 - 3/ 22. TEAM PROJECT. Further Properties of the exponential Function, (a) Analyticity. Show that ez is entire. What about e1/z? ez? =° in (7). I EXAMPLE 2 Solutions of Equations. Zeros of cos z and sin z Solve (a) cos z = 5 (which has no real solution!), (b) cos z = 0, (c) sin z = 0. Solution, (a) e2lz - 10eK +1=0 from (1) by multiplication by ezz. This is a quadratic equation in elz, with solutions (rounded off to 3 decimals) eiz = e~y+ix = 5 ± V'25 - 1 = 9.899 and 0.101. Thus e~v = 9.899 or 0.101. e™ = 1, y = =2.292, x = 2nir. Ans. z = ±2nw ± 2.292/ (n -■ 0, l, 2, ■ ■ ■)• C'an you obtain this from (6a)? (b) cos x = 0, sinh y = 0 by (7a), y = 0. Ans. z = ±\(2n + l)ir (it = 0, l, 2, ■ ■ ■)■ (c) sin x = 0, sinh y = 0 by (7b). Ans. z ±nir (n = 0, 1, 2, • • ■)• Hence the only zeros of cos z and sin z are those of the real cosine and sine functions. I General formulas for the real trigonometric functions continue to hold for complex values. This follows immediately from the definitions. We mention in particular the addition rules (9) and the formula cos (z\ ± z2) — cos Z] cos z2 + sin Z\ sin z2 sin (zi ± z2) = sin z\ cos z2 ± sin z2 cos Z\ (10) cos2z + sin2z = 1. Some further useful formulas are included in the problem set. Hyperbolic Functions The complex hyperbolic cosine and sine are defined by the formulas (11) cosh z = \{ez + e~% sinh z = \(ez - e~z). This is suggested by the familiar definitions for a real variable [see (8)]. These functions are entire, with derivatives (12) (coshz)' = sinh z, (sinhz)' = coshz, as in calculus. The other hyperbolic functions arc defined by SEC. 13.6 Trigonometric and Hyperbolic Functions 629 sinh z cosh z tanh z = -:— , coth z cosh z sinh (13) 1 1 sech z. = -;— , csch cosh z ' sinh z Complex Trigonometric and Hyperbolic Functions Are Related. It in (11), we replace z. by iz and then use (1), we obtain (14) cosh iz — cos z, sinh iz = i sin z. Similarly, if in (1) we replace z by iz and then use (11), we obtain conversely (15) cos iz = cosh z, sin iz = i sinh z. Here we have another case of unrelated real functions that have related complex analogs, pointing again to the advantage of working in complex in order to get both a more unified formalism and a deeper understanding of special functions. This is one of the main reasons for the importance of complex analysis to the engineer and physicist. P R O B L E A4 S E T 13.6 1. Prove that cos z, sin z, cosh z, sinh z are entire functions. 2. Verify by differentiation that Re cos z and Im sin z are harmonic. [3^6] FORMULAS FOR HYPERBOLIC FUNCTIONS Show that 3. cosh z = cosh x cos y + i sinh x sin y sinh z = sinhx cosy + i coshx siny. 4. cosh (zx + z2) = cosh Z\ cosh z2 + sinh z\ sinh z2 sinh (zj + z2) = sinhzi cosh z2 + cosh z\ sinh z2. 5. cosrnz - sinh2 I 6. cosh z + sinh z = cosh 2z 17-151 Function Values. Compute (in the form u + iv) 7. cos (1+0 8. sin (1 + i) 9. sin 5i, cos 5/ 10. cos 3tt/ 11. cosh (-2 + 30, cos (-3 - 20 12. — i sinh ( — 77 + 20, sin (2 + 77o 13. cosh (In + 1)77/, n = 1, 2, • • • 14. sinh (4 - 30 15. cosh (4 - 677/) 16. (Real and imaginary parts) Show that sin x cos x Re tan z cos x + sinh y sinh y cosh y 17-21 Im tan z = cos x + sinh v Equations. Find all solutions of the following equations. 17. cosh z = 0 19. cos z = 2/ 21. sinhz = 0 18. sinz = 100 20. cosh z = -1 22. Find all z for which (a) cos z, (b) sin z has real values. 23-25 Equations and Inequalities. Using the definitions, prove: 23. cos z is even, cos (—z) = cos z, and sinz is odd. sin (—z) = —sin z. 24. jsinhyl ' |cos z| — cosh y, |sinh v| ' |sinz| ' coshy. Conclude that the complex cosine and sine are not bounded in the whole complex plane. 25. sin zi cos z2 = g[ sin (Zl + z2) + sin (z, - z2)] 630 CHAP. 13 Complex Numbers and Functions 13.7 Logarithm. General Power We finally introduce the complex logarithm, which is more complicated than the real logarithm (which it includes as a special case) and historically puzzled mathematicians for some time (so if you first get puzzled—which need not happen!—be patient and work through this section with extra care). The natural logarithm of z = x + iy is denoted by In z (sometimes also by log z) and is defined as the inverse of the exponential function; that is, w = In z is defined for z # 0 by the relation (Note that z - 0 is impossible, since ew # 0 for all w; see Sec. 13.5.) If we set w = u + iv and z = re'8, this becomes ew = eu 1= reie. Now from Sec. 13.5 we know that ew+iv has the absolute value e" and the argument v. These must be equal to the absolute value and argument on the right: eu = r, v — 0. eu — r gives u — In r, where In r is the familiar real natural logarithm of the positive number r = \z\- Hence w = u + iv = In z is given by (1) In z = In r + id (r = \z\ > 0, 0= arg z). Now comes an important point (without analog in real calculus). Since the argument of z is determined only up to integer multiples of 2tt, the complex natural logarithm In z (z ^ 0) is infinitely many-valued. The value of In z, corresponding to the principal value Arg z (see Sec. 13.2) is denoted by Ln z (Ln with capital L) and is called the principal value of In z. Thus (2) Ln z = In \z\ + i Arg z (z # 0). The uniqueness of Arg z for given z {¥= 0) implies that Lnz is single-valued, that is, a function in the usual sense. Since the other values of arg z differ by integer multiples of 2tt, the other values of ln z are given by (3) ln z = Ln z ± 2«7ri (n = 1, 2, • • •)• They all have the same real part, and their imaginary parts differ by integer multiples of 277. If z is positive real, then Arg z = 0, and Ln z becomes identical with the real natural logarithm known from calculus. If z is negative real (so that the natural logarithm of calculus is not defined!), then Arg z = tr and Ln z = ln z + iri (z negative real). SEC. 13.7 Logarithm. General Power From (1) and eln r — r for positive real r we obtain (4a) eln z = z as expected, but since arg (ez) = y ± 2mr is multivalued, so is (4b) In (e2) = z ± 2nm, n = 0, 1, EXAMPLE 1 Natural Logarithm. Principal Value In 1 = 0, ±2m, ±4m, ■ ■ ■ In 4 = 1.386 294 ± Irnri ln(-l) = ±1fi, ±3 77"/, ±577"/, ■ ■ ■ In (-4) = 1.386 294 ± (In + l)m In i = 7T//2, -37T/2, 5m/2, ■ ■ ■ In 4/ = 1.386 294 + iri/2 ± 2nvi In (-40 = 1.386 294 - m/2 ± 2«rr/ In (3 - 40 = In 5 — i arg (3 — 4i) = 1.609 438 - 0.927 295/ ± Ln 1 = 0 l.n 4 = 1.386 294 Ln(-I) = 777 Ln(-4) = 1.386 294 I m Ln i = Trill Ln4j = 1.386 294 + jri/2 Ln(-4/) = 1.386 294 - mil Ln (3 - 40 = 1.609 438 - 0.927 295/ (Fig. 334) -0.9 + 67t -0.9 + 4/r -0.9 + 2k 0 0.9 I 1 \ 2 " I 1 -0.9-2tt - ■ Fig. 334. Some values of In (3 — 4i) in Example 1 The familiar relations for the natural logarithm continue to hold for complex values, that is, (5) (a) ln (z±z2) = In zx + In z2. (b) ln (Zi/za) = ln Z] - ln z2 but these relations are to be understood in the sense that each value of one side is also contained among the values of the other side; see the next example. EXAMPLE 2 Illustration of the Functional Relation (5) in Complex 1 ,et If we take the principal values -1. Ln zi = Ln ?2 = tri, then (5a) holds provided we write ln (ziZz) = In 1 = 2m; however, it is not true for the principal value, Ln(z,z2) = Ln 1 = 0. ■ 632 CHAP. 13 Complex Numbers and Functions THEOREM 1 Analyticity of the Logarithm For every n = 0, ±1, ±2, • • • formula (3) defines a function, which is analytic, except at 0 and on the negative real axis, and has the derivative (6) On*)' = 1 (z not 0 or negative real). PROOF We show that the Cauchy-Riemann equations are satisfied. From (1 )-(3) wc have In z = In r + i{6 + c) = ^\ In (x2 + y2) + i I arctan ^ + c where the constant c is a multiple of 2tt. By differentiation. x x2 + V2 1 1 (>'/x)2 ' X x2 + v2 = — v.. 1 + (y/x)2 Hence the Cauchy-Riemann equations hold. [Confirm this by using these equations in polar form, which we did not use since we proved them only in the problems (to Sec. 13.4).] Formula (4) in Sec. 13.4 now gives (6), (In z) = ux + ivx = 1 x2 + y2 1 + (y/x)2 x - ty x2 + v2 1 Each of the infinitely many functions in (3) is called a branch of the logarithm. The negative real axis is known as a branch cut and is usually graphed as shown in Fig. 335. The branch for n = 0 is called the principal branch of In z. Fig. 335. Branch cut for In z General Powers General powers of a complex number z = x + iy are defined by the formula (7) zc = ec In 2 (c complex, z + 0). Since In z is infinitely many-valued, z'' will, in general, be multivalued. The particular value zc = ec L„ , is called the principal value of zc. SEC. 13.7 Logarithm. General Power 633 If c = n = 1, 2, • • ■, then zn is single-valued and identical with the usual nth power of z. If c = — 1, —2, • • ■ , the situation is similar. If c = 1/n, where n — 2, 3, • • • , then zc = Vz = éVrú lnz {z* 0), the exponent is determined up to multiples of 277//« and we obtain the n distinct values of the nth root, in agreement with the result in Sec. 13.2. If c = plq, the quotient of two positive integers, the situation is similar, and zc has only finitely many distinct values. However, if c is real irrational or genuinely complex, then zc is infinitely many-valued. EXAMPLE 3 General Power e1'ln ' = cxp (i In /) = exp All these values arc real, and the principal value (n = 0) is e "l2. Similarly, by direct calculation and multiplying out in the exponent, (1 + i')2~* = exp [(2 - /) ln (1 + = cxp [(2 - i) (ln V2 + \iri ± 2«m'j] = 2g""/4±2n,r[sin (| ln 2) + i cos (| In 2)]. ■ It is a convention that for real positive z — x the expression zc means ec ln x where ln x is the elementary real natural logarithm (that is, the principal value Ln z (z = x > 0) in the sense of our definition). Also, if z = e, the base of the natural logarithm, z° = e° is conventionally regarded as the unique value obtained from (1) in Sec. 13.5. From (7) we see that for any complex number a, (8) az = ez ln a. We have now introduced the complex functions needed in practical work, some of them (ez, cos z. sin z, cosh z, sinh z) entire (Sec. 13.5), some of them (tan z, cotz, tanh z, coth z) analytic except at certain points, and one of them (ln z) splitting up into infinitely many functions, each analytic except at 0 and on the negative real axis. For the inverse trigonometric and hyperbolic functions see the problem set. 1-9 Principal Value Lnz. Find Ln z when z equals: 1.-10 2. 2 + 2/ 3. 2 - 2/ 4. -5 ± 0.1; 5. -3 - 4/ 6. -100 7. 0.6 + 0.8;' 8. -ei 9. 1 - / 10-16 All Values of lnz. Find all values and graph some of them in the complex plane. 10. In 1 11. ln (-1) 12. lne 14. In (4 + 3/) 16. ln(e3i) 13. In (-6) 15. ln(-e"!) 17. Show that the set of values of ln (/2) differs from the set of values of 2 ln /. 18-21 Equations. Solve for z: 18. ln z = (2 - \i)Tt 19. ln z = 0.3 + 0.7/ 20. ln z = e - ni 21. ln z = 2 + \iri 634 22-281 General Powers. Showing the details of your work, find the principal value of: 22. iz\ (2iY 23. 4:i+,: 24. (1 - + 26. (-l)1_2i 28. (3 - 4/)1 25. (1 + 27. i112 iY 29. How can you find the answer to Prob. 24 from the answer to Prob. 25? 30. TEAM PROJECT. Inverse Trigonometric and Hyperbolic Functions. By definition, the inverse sine w = arcsin z is the relation such that sin w = z. The inverse cosine w = arccos z is the relation such that cos w = -. The inverse tangent, inverse cotangent, inverse hyperbolic sine, etc., are defined and denoted in a similar fashion. (Note that all these relations are multivalued.) Using sin w = (e1 e~lw)/(2i) and (a) arccos z - — i In (z + Vz2 (b) arcsin z = —i In (iz + V 1 (c) arccosh z = In (z + Vz2 - (d) arcsinh z = In (z + V;2 + 1) I) 1) similar representations of cos w, etc., show that 1 l + z (e) arctanz = — In- 2 i — z 1 1 + z (f) arctanh z = — In-- 2 1 — z (g) Show that w ~ arcsin z is infinitely many-valued, and if w1 is one of these values, the others are of the form W\ ± 2mr and tt — n\ ± 2«7r, n = 0, 1, • ■ • . (The principal value of w — u + iv = arcsin z is defined to be the value for which — 7i/2 = /./ s= n/2 if v S 0 and - tt/2 < u < n!2 if v < 0.) CHAPTER 13 REVIEW QUESTIONS AND PROBLEMS 1. Add. subtract, multiply, and divide 26 - 7< and 3 + 4i as well as their complex conjugates. 2. Write the two given numbers in Prob. 1 in polar form. Find the principal value of their arguments. 3. What is the triangle inequality? Its geometric meaning? Its significance? 4. If you know the values of 'v' 1, how do you get from them the values of Vz for any z? 5. State the definition of the derivative from memory. It looks similar to that in calculus. But what is the big difference? 6. What is an analytic function? How would you test for analylicity? 7. Can a function be different]able at a point without being analytic there? If yes, give an example. 8. Are |z|, z, Re z, Im z analytic? Give reason. 9. State the definitions of 0 f(z + Az) - f(z) Az everywhere in D. Also, f(z) is analytic at a point z = Zn if it has a derivative in a neighborhood of z0 (not merely at z0 itself). If f(z) is analytic in D, then u(x, y) and u(x, y) satisfy the (very important!) Cauchy-Riemann equations (Sec. 13.4) (3) (in dx dv du dv dv dx everywhere in D. Then u and v also satisfy Laplace's equation (4) It,,,. + Uyy = 0, ^ xx ^ ^yy ^ everywhere in D. If u(x, y) and v(x, y) are continuous and have continuous partial derivatives in D that satisfy (3) in D, then f(z) — u(x, y) + iv(x, y) is analytic in D. See Sec. 13.4. (More on Laplace's equation and complex analysis follows in Chap. 18.) The complex exponential function (Sec. 13.5) (5) exp z = ex (cos y + i sin y) reduces to ex if z = x (y = 0). It is periodic with 2tt/ and has the derivative ez The trigonometric functions are (Sec. 13.6) (6) cos sin z 1 i — (e + e ) = cos x cosh y — / sin x sinh y 2i (e%z — e = sin x cosh y + i cos x sinh y and, furthermore, tan z = (sin z)/cos . cot. 1/tanz, etc. CHAP. 13 Complex Numbers and Functions The hyperbolic functions are (Sec. 13.6) (7) cosh z = \{ez + e~z) — cos iz, sinh z = \{ez — e~z) = —i sin iz etc. The functions (5)-(7) are entire, that is, analytic everywhere in the complex plane. The natural logarithm is (Sec. 13.7) (8) In z = In \z\ + i arg z = In |z| + <' Arg z ± 2nm where z ^ 0 and n = 0, 1, ■ ■ ■ . Argz is the principal value of argz, that is, — 77 < Arg z = 77. We see that In z is infinitely many-valued. Taking n = 0 gives the principal value Ln z of In z; thus Ln z = In |z| + / Arg z. General powers are defined by (Sec. 13.7) (9) zc = ec 1 (c complex, z ^ 0). chapter! 4 Complex Integration Two main reasons account for the importance of integration in the complex plane. The practical reason is that complex integration can evaluate certain real integrals appearing in applications that are not accessible by real integral calculus. The theoretical reason is that some basic properties of analytic functions are difficult to prove by other methods. A striking property of this type is the existence of higher derivatives of an analytic function. Complex integration also plays a role in connection with special functions, such as the gamma function (see [GR1], p. 255), the error function, various polynomials (see [GR10]) and others, and the application of these functions in physics. In this chapter we define and explain complex integrals. The most important result in the chapter is Cauchy's integral theorem or the Cauchy-Goursat theorem, as it is also called (Sec. 14.2). It implies Cauchy's integral formula (Sec. 14.3), which in turn implies the existence of all higher derivatives of an analytic function. Hence in this respect, complex analytic functions behave much more simply than real-valued functions of real variables, which may have derivatives only up to a certain order. A further method of complex integration, known as integration by residues, and its application to real integrals will need complex series and follows in Chap. 16. Prerequisite: Chap. 13 References and Answers to Problems: App. 1 Part D, App. 2. 14.1 Line Integral in the Complex Plane As in calculus we distinguish between definite integrals and indefinite integrals or antiderivatives. An indefinite integral is a function whose derivative equals a given analytic function in a region. By inverting known differentiation formulas we may find many types of indefinite integrals. Complex definite integrals are called (complex) line integrals. They are written f f(z) dz. c Here the integrand f(z) is integrated over a given curve C or a portion of it (an arc, but we shall say "curve" in either case, for simplicity). This curve C in the complex plane is called the path of integration. We may represent C by a parametric representation (1) z(t) = x(t) + iy(t) (a S 1 g b). 637 638 CHAP. 14 Complex Integration The sense of increasing I is called the positive sense on C, and we say that C is oriented by(l). For instance, z{t) = t + 3it (0 § ( g 2) gives a portion (a segment) of the line y = 3x. The function z(t) = 4 cos / + Ai sin t (—tt ^ t S tt) represents the circle \z\ = 4, and so on. More examples follow below. We assume C to be a smooth curve, that is, C has a continuous and nonzero derivative dz Z(t) = — = x(t) + iy(t) dt at each point. Geometrically this means that C has everywhere a continuously turning tangent, as follows directly from the definition z(t + At) - z(t) j(0 = lim--z1-— (Fig. 336). Here we use a dot since a prime ' denotes the derivative with respect to z. Definition of the Complex Line Integral This is similar to the method in calculus. Let C be a smooth curve in the complex plane given by (1), and let /(z) be a continuous function given (at least) at each point of C. We now subdivide (we "partition") the interval a = t S b in (1) by points t0(=a), ?!, fn_x, tn(=b) where t0 < f, <■■• oo. This implies that the greatest SEC. 14.1 Line Integral in the Complex Plane \Azm\ also approaches zero. Indeed, it cannot exceed the length of the arc of C from zm-i to zm and the latter goes to zero since the arc length of the smooth curve C is a continuous function of t. The limit of the sequence of complex numbers S2, S3, ■ ■ ■ thus obtained is called the line integral (or simply the integral) of f{z) over the path of integration C with the orientation given by (1). This line integral is denoted by (3) \f(.z)dz, or by f f(z) dz Jc Jc if C is a closed path (one whose terminal point Z coincides with its initial point z0, as for a circle or for a curve shaped like an 8). General Assumption. All paths of integration for complex line integrals are assumed, to be piecewise smooth, that is, they consist of finitely many smooth curves joined end to end. Basic Properties Directly Implied by the Definition 1. Linearity. Integration is a linear operation, that is, we can integrate sums term by term and can take out constant factors from under the integral sign. This means that if the integrals of fi and f2 over a path C exist, so does the integral of ktfx + k2.f2 over the same path and (4) f \kj,(z) + k2f2(z)\ dz = kx\ f\(z) dz + k2 \ f2(z.) dz, 2. Sense reversal in integrating over the same path, from z0 to Z (left) and from Z to z0 (right), introduces a minus sign as shown, 3. Partitioning of path (see Fig. 338) Fig. 3J8. Partitioning of path [formula (6)] Existence of the Complex Line Integral Our assumptions that f(z) is continuous and C is piecewise smooth imply the existence of the line integral (3). This can be seen as follows. As in the preceding chapter let us write f(z) — u(x, y) + iv(x, y). We also set Cm = Li + 'Vm and AZm = Axm + iAym. 640 Then (2) may be written (7) where u = u(£m, inm), v — v(£m, r\m) and we sum over m from 1 to n. Performing the multiplication, we may now split up Sn into four sums: sn = 2 « A.vm ~ 2 V Aym + i [2 « Av,„ + 2 u Axm] . These sums are real. Since f is continuous, u and V are continuous. Hence, if we let n approach infinity in the aforementioned way, then the greatest Axm and Ay.„, will approach zero and each sum on the right becomes a real line integral: (8) lim n^co Sre = I /00 ^< \ udx — \ v dy + i \ udy + \ v dx This shows that under our assumptions on f and C the line integral (3) exists and its value is independent of the choice of subdivisions and intermediate points £„,. First Evaluation Method: Indefinite Integration and Substitution of Limits This method is the analog of the evaluation of definite integrals in calculus by the well-known formula J f(x) dx = Fib) - Fid) [F'ix) = fix)]. a It is simpler than the next method, but it is suitable for analytic functions only. To formulate it, we need the following concept of general interest. A domain D is called simply connected if every simple closed curve (closed curve without self-intersections) encloses only points of D. For instance, a circular disk is simply connected, whereas an annuius (Sec. 13.3) is not simply connected. (Explain!) THEOREM 1 Indefinite Integration of Analytic Functions Lei fiz) be analytic in a simply connected domain D. Then there exists an indefinite integral of fiz) in the domain D, that is, an analytic function F(z.) such that F' (z) = fiz) in D, and for all paths in D joining two points Zo and z\ in D we have (9) / 'fiz) dz = F(Zj) - F{zo) [F'iz) = /<*)]. (Note that we can write z0 and z% instead of C, since we get the same value for all those C from z0 to zj.) SEC. 14.1 Line Integral in the Complex Plane EXAMPLE 1 EXAMPLE 2 EXAMPLE 3 EXAMPLE 4 This theorem will be proved in the next section. Simple connectedness is quite essential in Theorem 1, as we shall see in Example 5. Since analytic functions are our main concern, and since differentiation formulas will often help in finding F(z) for a given f(z) = Fr(z), the present method is of great practical interest. If f(z) is entire (Sec. 13.5), wc can take for D the complex plane (which is certainly simply connected). ■'o f j: dz - (1 - if 2 2 --1--i 3 3 cos z dz = sin ; — 77?. K-3tj-i ^dz = 2S12 = 2 sin m = 2i sinh it = 23.097/ = 2(e4-3«/2 _ ei+^n} = 0 since £'z is periodic with period 2m. f dz ITT I ITT \ J — = Ln < Ln(—i) = -— — ^——I = (17. Here D is D is the complex plane without 0 and the negative real axis (where Ln z is not analytic). Obviously, D is a simply connected domain. Second Evaluation Method: Use of a Representation of a Path This method is not restricted to analytic functions but applies to any continuous complex function. THEOREM 2 Integration by the Use of the Path Let C be a piecewise smooth path, represented by f(z) be a continuous function on C. Then z.(t), where a^ltSb. Let (10) \ f{z)dz - ff[z(t)]z(t)dt dz dt PROOF The left side of (10) is given by (8) in terms of real line integrals, and we show that the right side of (10) also equals (8). We have z = x + iy, hence z = x + iy. We simply write u for u[x(t), y(t)] and v for v[x(i), y(t)]. We also have dx - x dt and dy = y dt. Consequently, in (10) / f[z(t)]z(t)dt = [(« + iv)(x + iy)dt a a = I [udx — V dy + i(u dy + v dx)] c — j (u dx — v dy) + ij (u dy + v dx). 642 CHAP. 14 Complex Integration COMMENT. In (7) and (8) of the existence proof of the complex line integral we referred to real line integrals. If one wants to avoid this, one can take (10) as a definition of the complex line integral. Steps in Applying Theorem 2 (A) Represent the path C in the form z{t) (a ^ t ^ b). (B) Calculate the derivative 'z(t) = dz/dt. (C) Substitute z(t) for every z in f(z) (hence x(t) for x and y(r) for y). (D) Integrate f[z(t)]z{t) over t from a to /?. EXAMPLE 5 A Basic Result: Integral of 1/z Around the Unit Circle We show that by integrating 1/; counterclockwise around the unit circle (the circle of radius 1 and center 0; see Sec. 13.3) we obtain Jc Z counterclockwise). 77hs js a very important result that we shall need quite often. Solution. (A) We may represent the unit circle C in Fig. 327 of Sec. 13.3 by z(t) = cos t + i sin f = elt (O^iS 2ct). so that counterclockwise integration corresponds to an increase of ( from 0 to 2ir. (B) Differentiation gives 'z(t) = ielt (chain rule!). (C) By substitution, f(z(t)) = l/-(r) = e~u. (D) From (10) we thus obtain the result „2

f(z) dz = 0. Sec Fig. 344. Jc D Fig. 344. Cauchy's integral theorem Before we prove the theorem, let us consider some examples in order to really understand what is going on. A simple closed path is sometimes called a contour and an integral over such a path a contour integral. Thus, (1) and our examples involve contour integrals. EXAMPLE 1 No Singularities (Entire Functions) ez dz = 0, cos z dz, = 0, zndz = 0 (n = 0, !,-■■) 'c c Jc for any closed palh, since these functions are entire (analytic for all z). EXAMPLE 2 Singularities Outside the Contour ' sec z dz = 0. dz where C is the unit circle, sec z, = 1/cos z is not analytic at z = ±ir/2, +377/2, • • • , but all these points lie outside C; none lies on C or inside C. Similarly for the second integral, whose integrand is not analytic at z = ±2i outside C. EXAMPLE 3 Nonanalytic Function <|>zrfz = f e~uieudt = 2m where C: z(t) = e1' is the unit circle. This does not contradict Cauchy's theorem because /(z) = z is not analytic. EXAMPLE 4 Analyticity Sufficient, Not Necessary dz where C is the unit circle. This result does not follow from Cauchy's theorem, because /(z) = 1/z2 is not analytical z = 0. Hence the condition that f be analytic in D is sufficient rather than necessary for (I) to be true. I 648 EXAMPLE 5 Simple Connectedness Essential for counterclockwise integration around the unit circle (see Sec. 14.1). C lies in the annulus § < \z\ < § where l/z is analytic, but this domain is not simply connected, so that Cauchy's theorem cannot be applied. Hence the condition that the domain D be simply connected is essential. In other words, by Cauchy's theorem, if f(z) is analytic on a simple closed path C and everywhere inside C, with no exception, not even a single point, then (1) holds. The point that causes trouble here is z = 0 where l/z is not analytic. PROOF Cauchy proved his integral theorem under the additional assumption that the derivative / (z) is continuous (which is true, but would need an extra proof). His proof proceeds as follows. From (8) in Sec. 14.1 we have f(z) dz = f (u dx - v dy) + I f (u dy + v dx). c Jc -c Since f(z) is analytic in D, its derivative f'(z) exists in D. Since f'(z) is assumed to be continuous, (4) and (5) in Sec. 13.4 imply that u and v have continuous partial derivatives in D. Hence Green's theorem (Sec. 10.4) (with u and — v instead of and F2) is applicable and gives r r r / dv du \ 9 (u dx - v dy) = ----dx dy •V ■ V \ dx dy) where R is the region bounded by C. The second Cauchy-Riemann equation (Sec. 13.4) shows that the integrand on the right is identically zero. Hence the integral on the left is zero. In the same fashion it follows by the use of the first Cauchy-Riemann equation that the last integral in the above formula is zero. This completes Cauchy's proof. Goursat's proof without the condition that f'(z) is continuous^ is much more complicated. We leave it optional and include it in App. 4. Independence of Path We know from the preceding section that the value of a line integral of a given function f(z.) from a point Z\ Lo a point z2 will in general depend on the path C over which we integrate, not merely on Z\ and z2. It is important to characterize situations in which this difficulty of path dependence does not occur. This task suggests the following concept. We call an integral of f(z) independent of path in a domain D if for every z%, z2 in D its value depends (besides on f(z), of course) only on the initial point z% and the terminal point ;;•>, but not on the choice of the path C in D [so that every path in D from z.\ to z,2 gives the same value of the integral of /(z)]. 1EDOUARD GOURSAT (1858-1936). French mathematician. Cauchy published the theorem in 1825. The removal of that condition by Goursat (see Transactions Amer. Math. Soc, vol. 1. 1900) is quite important, for instance, in connection with the fact that derivatives of analytic functions are also analytic, as we shall prove soon. Goursat also made important contributions to PDEs. SEC. 14.2 Cauchy's Integral Theorem THEOREM 2 Independence of Path If f(z) is analytic in a simply connected domain D, then the integral of f(z) is independent of path in D. PROOF Let Z\ and z2 be any points in D. Consider two paths Cl and C2 in D from Zi to z,2 without further common points, as in Fig. 345. Denote by C2 the path C2 with the orientation reserved (Fig. 346). Integrate from Z\ over Cx to z2 and over C| back to z\, This is a simple closed path, and Cauchy's theorem applies under our assumptions of the present theorem and gives zero: (2') f fdz + f fdz = 0, thus J fdz = J fdz. But the minus sign on the right disappears if we integrate in the reverse direction, from Z\ to z2, which shows that the integrals of f(z) over C1 and C2 are equal, (2) f /(::) 0. The details are as follows. SEC. 14.2 Cauchy's Integral Theorem 651 We keep z fixed. Then we choose z + Az in D so that the whole segment with endpoints z and z + Az is in D (Fig. 349). This can be done because D is a domain, hence it contains a neighborhood of z- We use this segment as the path of integration in (5). Now we subtract f(z). This is a constant because z is kept fixed. Hence we can write z^Az zlAz J f(z)dz* /<::) J dz* = f(z)Az. Thus 1 ,- ^ m = — J /(z By this trick and from (5) we get a single integral: F(z + Az) - F(z) Az 1 rz+Az f(z) = j- J !./(--) - f(z)] dz*. Since f(z) is analytic, it is continuous. An e > 0 being given, we can thus find a 8 > 0 such that |/(z*) — /(z)| < e when \z* — z.\ < 8. Hence, letting |Az| < 8, we see that the ML-inequality (Sec. 14.1) yields F(z + Az) - F(z) Az - f(z) 1 J U(z*) - ,f(z)] dz* 1 lAzI e|Az| By the definition of limit and derivative, this proves that , F(z + Az) - F(z) Since z is any point in D, this implies that F{z) is analytic in D and is an indefinite integral or antiderivative of f(z) in D, written F(z) = jf(z) Also, if G'(z) = f(z), then F'(z) - G'(z) = 0 in D; hence F(z) - G(z) is constant in D (see Team Project 26 in Problem Set 13.4). That is, two indefinite integrals of f(z) can differ only by a constant. The latter drops out in (9) of Sec. 14.1. so that we can use any indefinite integral of f(z). This proves Theorem 3. ■ D z + Az I \ X „___ Fig. 349. Path of integration 652 CHAP. 14 Complex Integration Cauchy's Integral Theorem for Multiply Connected Domains Cauchy's theorem applies to multiply connected domains. We first explain this for a doubly connected domain D with outer boundary curve C± and inner C2 (Fig- 350). If a function f(z) is analytic in any domain D* that contains D and its boundary curves, we claim that (6) j> f(z) dz = j>f(z) dz (Fig. 350) both integrals being taken counterclockwise (or both clockwise, and regardless of whether or not the full interior of C2 belongs to D*). Fig. 350. Paths in (5) PROOF By two cuts C\ and C2 (Fig. 351) we cut D into two simply connected domains DL and D2 in which and on whose boundaries f(z) is analytic. By Cauchy's integral theorem the integral over the entire boundary of Dx (taken in the sense of the arrows in Fig. 351) is zero, and so is the integral over the boundary of D2, and thus their sum. In this sum the integrals over the cuts and C2 cancel because we integrate over them in both directions—this is the key—and wc are left with the integrals over C1 (counterclockwise) and C2 (clockwise; see Fig. 351); hence by reversing the integration over C2 (to counterclockwise) we have j> fdz - j> fdz = 0 and (6) follows. ■ For domains of higher connectivity the idea remains the same. Thus, for a triply connected domain we use three cuts Cx. C2, C3 (Fig. 352). Adding integrals as before, the integrals over the cuts cancel and the sum of the integrals over Cy (counterclockwise) and C2, C3 (clockwise) is zero. Hence the integral over Cx equals the sum of the integrals over C2 and C3, all three now taken counterclockwise. Similarly for quadruply connected domains, and so on. Fig. 351. Doubly connected domain Fig. 352. Triply connected domain SEC. 14.2 Cauchy's Integral Theorem 653 1-111 CAUCHY'S INTEGRAL THEOREM APPLICABLE? Integrate /(z) counterclockwise around the unit circle, indicating whether Cauchy's integral theorem applies. (Show the details of your work.) 1. /(-) = Re z 3. f(z) = r2'2 5. f(z) = tan z2 7. f(z) = l/(z8 -1.2) 9. f(z) = \/(2\z\s) 11. f(z) = z2 cot- 2. f(z) = I/(3z - iri) 4. f(z) = 1/z 6. f(z) = sec (z/2) 8. f(z.) = l/(4z - 3) 10. f(z) = z 12-17 COMMENTS ON TEXT AND EXAMPLES 12. (Singularities) Can we conclude in Example 2 that the integral of l/(z2 + 4) taken over (a) |z — 2| = 2, (b) \z — 2| = 3 is zero? Give reasons. 13. (Cauchy's integral theorem) Verify Theorem 1 for the integral of z2 over the boundary of the square with vertices 1 + i, — 1 +;',—] — i, and 1 — i (counterclockwise). 14. (Cauchy's integral theorem) For what contours C will it follow from Theorem 1 (hat (a) dz c z 0, (b) (c) + 9 cos z dz = 01 dz = 0, 15. (Deformation principle) Can we conclude from Example 4 that the integral is also zero over the contour in Problem 13? 16. (Deformation principle) If the integral of a function ,f(z) over the unit circle equals 3 and over the circle |z| = 2 equals 5, can we conclude that f(z) is analytic everywhere in the annulus 1 < \z.\ < 2? 17. (Path independence) Verify Theorem 2 for the integral of cos z from 0 to (1 + i)ir (a) over the shortest path, (b) over the x-axis to tt and then straight up to (1 + l)tt. 18. TEAM PROJECT. Cauchy's Integral Theorem. (a) Main Aspects. Each of the problems in Fxamples 1-5 explains a basic fact in connection with Cauchy's theorem. Find five examples of your own, more complicated ones if possible, each illustrating one of those facts. (b) Partial fractions. Write f(z) in terms of partial fractions and integrate it counterclockwise over the unit circle, where (i) f(z) 2z + 3i (ii) f(z) + 2z (c) Deformation of path. Review (c) and (d) of Team Project 34. Sec. 14.1, in the light of the principle of deformation of path. Then consider another family of paths with common endpoints, say, z(t) = t + ia(t — t2), 0§(S 1, and experi ment wi th Ihe integration of analytic and nonanalytic functions of your choice over these paths (e.g., z, Im z, z2. Re z2, Im z2, etc). 19-30 FURTHER CONTOUR INTEGRALS Evaluate (showing the details and using partial fractions if necessary) 19. 20. 21. 23. 24. dz , C the circle \z\ — 3 (counterclockwise) 2z - /' tanh z dz, C the circle |z — \tti\ = | (clockwise) Re 2z dz, C as shown y r ^— -i 1 * Iz - 6 22. 9 —^-— dz, C as shown dz. ,2 , C as shown —o -1 i i X dz, C consists of |z| = 2 (clockwise) and |z| c 4z (counterclockwise) 654 CHAP. 14 Complex Integration 25. cos z dz, C consists of Izl = 1 (counterclockwise) 28. and \z\ = 3 (clockwise) 26. $ Ln (2 + z) dz, C the boundary of the square with Jc vertices ±1, ±i 27. dz C z (counterclockwise) , C: (a) |z| = i (b) \z - i\ dz (counterclockwise) C: (a) \z + i\ = 1, (b) / = 1 29. 30. sin; c Z + 2i tan fe/2) : dz. C: |z — 4 — 2i| = 5.5 (clockwise) fe, C the boundary of the square with c z 16 vertices ±1, ±i (clockwise) 14.3 Cauchy's Integral Formula The most important consequence of Cauchy's integral theorem is Cauchy's integral formula. This formula is useful for evaluating integrals, as we show below. Even more important is its key role in proving the surprising fact that analytic functions have derivatives of all orders (Sec. 14.4), in establishing Taylor series representations (Sec. 15.4), and so on. Cauchy's integral formula and its conditions of validity may be stated as follows. THEOREM 1 Cauchy's Integral Formula Let f(z) be analytic in a simply connected domain D. Then for any point z0 in D and any simple closed path C in D that encloses z0 (Fig. 353), (1) dz = Irrifizo) (Cauchy's integral formula) c z zo the integration being taken counterclockwise. Alternatively (for representing ,f(z0) by a contour integral, divide (1) by 2m), (1*) f(Zo) 1 2tt( Kz) (Cauchy's integral formula). C Z io PROOF By addition and subtraction, f(z) = f(z0) + \f(z) ~ .f(Zo)]- Inserting this into (1) on the left and taking the constant factor /(z0) out from under the integral sign, we have (2) - dz = /(z0) o c dz f(z) - f(z0) The first term on the right equals f(z0)" 2m (see Example 6 in Sec. 14.2 with m = — 1). This proves the theorem, provided the second integral on the right is zero. This is what we are now going to show. Its integrand is analytic, except at z0- Hence by (6) in Sec. 14.2 we can replace Cby a small circle K of radius p and center z0 (Fig. 354), without SEC. 14.3 Cauchy's Integral Formula D Fig. 353. Cauchy's integral formula Fig. 3S4. Proof of Cauchy's integral formula altering the value of the integral. Since f(z) is analytic, it is continuous (Team Project 26, Sec. 13.3). Hence an e > 0 being given, wc can find a 5 > 0 such that \f(z) — /(zo)| < e for all z in the disk \z — z0| < S. Choosing the radius p of K smaller than 5, wc thus have the inequality f(z) ~ f(zq) z - z0 < at each point of K. The length of K is 27rp. Hence, by the /V/L-inequality in Sec. 14.1, m - kzo) dz to < — 2irp = lire. P Since e (> 0) can be chosen arbitrarily small, it follows that the last integral in (2) must have the value zero, and the theorem is proved. ■ EXAMPLE 1 Cauchy's Integral Formula dz = 2-17(6' 46.4268/ for any contour enclosing z0 = 2 (since e~ is entire), and zero for any contour for which ;0 = 2 lies outside (by Cauchy's integral theorem). EXAMPLE 2 Cauchy's Integral Formula dz 2z dz 27ri[^ - 3] z=i/2 6 iri (;0 = \i inside C). EXAMPLE 3 Integration Around Different Contours Integrate + 1 + 1 - 1 (z+ l)(z - 1) counterclockwise around each of the four circles in Fig. 355. 656 CHAP. 14 Complex Integration Solution. g(z) is not analytic at — 1 and 1. These are the points wc have to watch for. We consider each circle separately. (a) The circle \z — l| = 1 encloses the point z0 = 1 where g(z) is not analytic. Hence in (T) wc have to z + \ z + 1 1 (z) = z + 1 thus and (1) gives r + l z2+ 1 z - 1 dz = 2mf(l) = 2m z +- 1 (b) gives the same as (a) by the principle of deformation of path. (c) The function °(z) is as before, but f(z) changes because we must take zo= ~ 1 (instead of 1). This gives a factor z — zq = z + 1 in (1). Hence we must write g(z) + 1 1 z - 1 z + 1 thus f(z) Compare this for a minute with the previous expression and then go on: + 1 dz = 2m/(- 1) = 2m + 1 -2m. (d) gives 0. Why? Fig. 355. Example 3 Multiply connected domains may be handled as in Sec. 14.2. For instance, if f(z) is analytic on C1 and C2 and in the ring-shaped domain bounded by Cx and C2 (Fig. 356) and Zo is any point in that domain, then (3) f(z0) 1 2"777 f(Z) A 4- 1 -dz + — I - Zn 2-rr; f(.z) dz, where the outer integral (over C\) is taken counterclockwise and the inner clockwise, as indicated in Fig. 356. Our discussion in this section has illustrated the use of Cauchy's integral formula in integration. In the next section we show that the formula plays the key role in proving the surprising fact that an analytic function has derivatives of all orders, which are thus analytic functions themselves. 1-4 CONTOUR INTEGRATION Integrate (z2 - 4)/(z2 + 4) counterclockwise around the circle: 1. \z -i\ = 2 3. \z + 3/1 = 2 2. \z - 1| = 2 4. Id = tt/2 5-17 CONTOUR INTEGRATION Using Cauchy's integral formula (and showing the details) integrate counterclockwise (or as indicated) + 2 dz, C: \z - 1| = 2 dz, C: \z\ = 1 t sinh ttz 7.

—2 - j dz, C consists of |z - 2/| = 2 (counterclockwise) and \z — 2/| = | (clockwise) 15. Ln (z dz, C: I sin z T ~i-zzr dz, Cconsistsof |z| = 3 (counterclockwise) c2 ~ 2,z and Id 17. = 1 (clockwise) cosh2 z dz, C as in Prob. 16 >c(z- 1 -i)z< 18. Show that J (z - Zt)~\z ~ Z2)_1 0. Being analytic, the function f(z) is continuous on C, hence bounded in absolute value, say, |/(z)| = K. Let d be the smallest distance from z0 to the points of C (see Fig. 357). Then for all z on C, <-ol hence 1 Furthermore, by the triangle inequality for all z on C we then also have d Š ]z - z0| = \z - Zo - Az + Az| Š \z - zo - Az| + |Az|, We now subtract |Az| on both sides and let |Az| S d/2, so that -|Az| ' -d/2. Then \d^l d - |Az| ^ jz - z0 - Az|. Hence 1 \z - z0 - Aj Let L be the length of C. If |Az| ž d/2, then by the ML-inequality 2 /fa)Az (z - zo - Az)(z - z0)2 ^ ^L Az 2 1 660 CHAP. 14 Complex Integration This approaches zero as Az —> 0. Formula (1') is proved. Note that wc used Cauchy's integral formula (1*), Sec. 14.3, but if all we had known about /(zq) is the fact that it can be represented by (1*), Sec. 14.3, our argument would have established the existence of the derivative /'(z0) of /(z). This is essential to the continuation and completion of this proof, because it implies that (1 ) can be proved by a similar argument, with f replaced by and that the general formula (1) follows by induction. EXAMPLE 1 Evaluation of Line Integrals From (l'), for any contour enclosing the point jri (counterclockwise) (z - 777) dz = 2m'(cos z) = —2m sin m = 2-7T sinh tt. EXAMPLE 2 From (1"), for any contour enclosing the point i we obtain by counterclockwise integration (z + 0 dz = ttHz* - 3z2 + 6)' iri\l2zf - 61. ■18m. EXAMPLE 3 By (1'), for any contour for which 1 lies inside and ±2i lie outside (counterclockwise). (z - l)2(z2 + 4) dz = 2m ez(z2 + 4) - ez2z (z2 + 4)2 6t'7f ^25 i « 2.0501. Cauchy's Inequality. Liouville's and Morera's Theorems As a new aspect, let us now show that Cauchy's integral theorem is also fundamental in deriving general results on analytic functions. Cauchy's Inequality. Theorem I yields a basic inequality that has many applications. To get it, all we have to do is to choose for c in (1) a circle of radius r and center Zq and apply the ML-inequality (Sec. 14.1); with |/(z)| = M on c we obtain from (1) \f(n\zo)\ = 277 /(z) c (z - z0) n+T * 2tt M 1 2irr. This gives Cauchy's inequality (2) l/ R and \m\ > m. (h) Growth of polynomials. If f(z) is a polynomial of degree n > 0 and M is an arbitrary positive real number (no matter how large), show that there exists a positive real number R such that \f(z)\ > M for all \z\>R- /■ (c) Exponential function. Show that f(z) = ez has the property characterized in (a) but docs not have that characterized in (b). (d) Fundamental theorem of algebra. If f(z) is a polynomial in z, not a constant, then f(z) = 0 for at least one value of z. Prove this, using (a). 15. (Proof of Theorem 1) Complete the proof of Theorem 1 by performing the induction mentioned at the end. CHAPTER 14 REVIEW QUESTIONS AND PROBLEMS 1. What is a path of integration? What did we assume about paths? 2. State the definition of a complex line integral from memory. 3. What do we mean by saying that complex integration is a linear operation? 4. Make a list of integration methods discussed. Illustrate each with a simple example. 5. Which integration methods apply to analytic functions only? 6. What value do you gel if you integrate 1/z counterclockwise around the unit circle? (You should memorize this basic result.) If you integrate 1/z2, 1/z3, • • • ? 7. Which theorem in this chapter do you regard as most important? Stale it from memory. 8. What is independence of path? What is the principle of deformation of path? Why is this important? 9. Do not confuse Cauchy's integral theorem and Cauchy's integral formula. State both. How are they related? 10. How can you extend Cauchy's integral theorem to doubly and triply connected domains? 11. If integrating f(z) over the boundary circles of an annulus D gives different values, can /(z) be analytic in D? (Give reason.) 12. jf(z)dz = \\f(z)\dz ? How would you find a 13. Is Re I /(z) dz = Re f(z) dzP. Give examples. Jc c 14. How did we use integral formulas for derivatives in integration? 15. What is Liouville's theorem? Give examples. State consequences. 16-301 INTEGRATION Integrate by a suitable method: 16. 4z3 + 2z from — i to 2 + / along any path 17. 5z — 3/z counterclockwise around the unit circle 18. z + 1/z counterclockwise around \z + 3i| = 2 19. eZz from -2 + 3t7í along the straight segment to -2 + 5-iri 20. ez2/(z — I)2 counterclockwise around |z| = 2 21. z/(z2 + 1) clockwise around |z + i\ = 1 22. Re z from 0 to 4 and then vertically up to 4 + 3i 23. cosh 4z from 0 to 2/ along the imaginary axis 24. ezlz over C consisting of |z| = 1 (counterclockwise) and \z\ = \ (clockwise) 25. (sin z)/z clockwise around a circle containing z = 0 in its interior 26. Imz counterclockwise around |z] = r 27. (Ln z)/(z — 2/)2 counterclockwise around |z — 2i| = 1 28. (tan irz)t(z — l)2 counterclockwise around |z - 11 = 0.2 29. |z| + z clockwise around the unit circle 30. (z /) 3(z3 + sin z) counterclockwise around any- bound for the integral on the left? circle with center /' Summary of Chapter 663 SUMMARY OF CHAPTER 14 Complex Integration The complex line integral of a function f(z) taken over a palh C is denoted by (1) \ f(t) dz or, if C is closed, also by j> f(z) (Sec. 14.1). c c If fix) is analytic in a simply connected domain D, then we can evaluate (1) as in calculus by indefinite integration and substitution of limits, that is, (2) / fiz) dz = F(Zl) - F(z0) [F'iz) = fiz)] c for every path C in d from a point z\q to a point Zi (see Sec. 14.1). These assumptions imply independence of path, that is, (2) depends only on zo and zx (and on fiz), of course) but not on the choice of C (Sec. 14.2). The existence of an Fiz) such that F'iz) = fiz) is proved in Sec. 14.2 by Cauchy's integral theorem (see below). A general method of integration, not restricted to analytic functions, uses the equation z = z(t) of C, where a S t S b, (3) J fiz)dz = f f(zit)Yzit) dt [z = . Cauchy's integral theorem is the most important theorem in this chapter. It states that if fiz) is analytic in a simply connected domain D, then for every closed path C in D (Sec. 14.2), (4) ffiz)dz = 0. c Under the same assumptions and for any z0 in D and closed palh C in D containing z0 in its interior we also have Cauchy's integral formula (5) fiZo) = — f *. 2m Jc z - z0 Furthermore, under these assumptions fiz) has derivatives of all orders in D that are themselves analytic functions in D and (Sec. 14.4) 2m Jc (z - ZoT+1 (6) f'Xzo) = — f :w+1 dz in = 1, 2, • ■ •) This implies Morera's theorem (the converse of Cauchy's integral theorem) and Cauchy's inequality (Sec. 14.4), which in turn implies Liouville's theorem that an entire function that is bounded in the whole complex plane must be constant. chapter! 5 Power Series, Taylor Series Complex power series, in particular, Taylor series, are analogs of real power and Taylor series in calculus. However, they are much more fundamental in complex analysis than their real counterparts in calculus. The reason is that power series represent analytic functions (Sec. 15.3) and, conversely, every analytic function can be represented by power scries, called Taylor series (Sec. 15.4). Use Sec. 15.1 for reference if you are familiar with convergence tests for real series— in complex this is quite similar. The last section (15.5) on uniform convergence is optional. Prerequisite: Chaps. 13, 14. Sections that may be omitted in a shorter course: 14.1, 14.5. References and Answers to Problems: App. 1 Part D, App. 2. 15.1 Sequences, Series, Convergence Tests In this section we define the basic concepts for complex sequences and scries and discuss tests for convergence and divergence. This is very similar to real sequences and series in calculus. If you feel at home with the latter and want to take for granted that the ratio test also holds in complex, skip this section and go to Sec. 15.2. Sequences The basic definitions are as in calculus. An infinite sequence or, briefly, a sequence, is obtained by assigning to each positive integer n a number zn, called a term of the sequence, and is written Zi, z2, ■ • ■ or {zi, z2, ■ ■ •} or briefly {zn}. We may also write z0, zlt • • ■ or z2, Z%, ■ • • or start with some other integer if convenient. A real sequence is one whose terms are real. Convergence. A convergent sequence z1( z2, ■ ■ ■ is one that has a limit c, written lim zn = C or simply zn —» c. n—*cc By definition of limit this means that for every e > 0 we can find an N such that (1) |zB - c\< e for all n > N; 664 SEC. 15.1 Sequences, Series, Convergence Tests 665 geometrically, all terms zn with n > N lie in the open disk of radius e and center c (Fig. 358) and only finitely many terms do not lie in that disk. [For a real sequence, (1) gives an open interval of length 2e and real midpoint c on the real line; see Fig. 359. | A divergent sequence is one that does not converge. Fig. 358. Convergent complex sequence Fig. 359. Convergent real sequence EXAMPLE 1 Convergent and Divergent Sequences The sequence {inln} = [i, The sequence ji") = [i. -1/2, —i/3, 1/4, • • •} is convergent with limit 0. -1, —i, 1, ■ ■ ■) is divergent, and so is {zn} with (l + 0" EXAMPLE 2 Sequences of the Real and the Imaginary Parts THEOREM 1 1 - 1/h2 + i(2 + Alii) is 6i, 3/4 + 4/. 8/9 + 10/73, 15/16 + 3i, ■ • •. . = 1 + 2i. Observe that {xn} has the limit 1 = Rec and {yn\ has the limit 2 = Imc. This is typical. It illustrates the following theorem by which the convergence of a complex sequence can be referred back to that of the two real sequences of the real parts and the imaginary parts. I The sequence {zn} with zn = xn + iyn (Sketch it.) It converges with the limit c Sequences of the Real and the Imaginary Parts A sequence z%, z2, ■ ' ' , Zn, ' ' ■ of complex numbers zn = xn + iyn (where n = 1, 2, • • •) converges to c = a + ib if and only if the sequence of the real parts xlt x2, • • • converges to a and the sequence of the imaginary parts y\, y2, ■ ■ ■ converges to b. PROOF Convergence zn —* c = a + ib implies convergence xn —» a and yn b because if \zn — c| < e, then z„n lies within the circle of radius e about c = a + ib, so that (Fig. 360a) \xn - a\ < e, \yn - b\ < e. a-iE a a +e x (a) / a X Ibl Fig. 360. Proof of Theorem 1 CHAP. 15 Power Series, Taylor Series Conversely, if xn —> a and yn —> b as n —> then for a given e > 0 we can choose so large that, for every n > N, e e K - a\< — , |y„ - fc| < — . These two inequalities imply that zn = x.„ + iyn lies in a square with center c and side e. Hence, zn must lie within a circle of radius e with center c (Fig. 360b). ■ Series Given a sequence Zi, z2, • • • , ■ ■ ■ , we may form the sequence of the sums *1 — Zli s2 = + Z2, S3 = Z\ + Z2 + ^3' and in general (2) sn = Zl + z2 + • ■ • + zn («=1,2, ••■)• s„ is called the nth partial sum of the infinite series or series (3) 2 Zm = Zi + z2 + • • • • The zi, Za» " ' " are called the terms of the series. (Our usual summation letter is «, unless we need « for another purpose, as here, and we then use m as the summation letter.) A convergent series is one whose sequence of partial sums converges, say, 00 lim sn = s. Then we write s = 2 Zm = Zi + z2 + ' 4 ' m=l and call s the sum or value of the.series. A series that is not convergent is called a divergent series. If we omit the terms of sn from (3), there remains (4) Rn — Zn+1 + Zn+2 + Zn i 3 + This is called the remainder of the series (3) after the term z,r Clearly, if (3) converges and has the sum s, then s = sn + Rn, thus Rn = s - s,n. Now sn —» s by the definition of convergence; hence /(n —* 0. In applications, when s is unknown and we compute an approximation sn of s, then \Rn\ is the error, and Rn 0 means that we can make as small as we please, by choosing n large enough. An application of Theorem 1 to the partial sums immediately relates the convergence of a complex series to that of the two series of its real parts and of its imaginary parts: SEC. 15.1 Sequences, Series, Convergence Tests 667 THEOREM 2 Real and Imaginary Parts A series (3) with z.m = xm + iym converges and has the sum s — u + iv if and only if Xj + x2 + ' ' ' converges and has the sum u and >'i + y2 + • ■ ■ converges and has the sum v. Tests for Convergence and Divergence of Series Convergence tests in complex are practically the same as in calculus. We apply them before we use a series, to make sure that the series converges. Divergence can often be shown very simply as follows. THEOREM 3 Divergence If a series Zi + z2 + ' ' ' converges, then lim z„ the series diverges. 0. Hence if this does not hold. PROOF If Zi + Zz + " ' ' converges, with the sum s, then, since zm = sm — 5t„_1, lim Zm = lim (s. .]_) = lim sm - lim sm_, 0. CAUTION! 0 is necessary for convergence but not sufficient, as we see from the harmonic series 1+2 + 3 + 4-1"'"'' which satisfies this condition but diverges, as is shown in calculus (see, for example, Ref. [GR11] in App. 1). The practical difficulty in proving convergence is that in most cases the sum of a scries is unknown. Cauchy overcame this by showing that a series converges if and only if its partial sums eventually get close to each other: THEOREM 4 Cauchy's Convergence Principle for Series A series Z\ + Z2 + ' ' ' w convergent if and only if for every given e > 0 (no matter how small) we can find an N (which depends on e, in general) such that (5) Zn+p < e for every n > N and p = 1,2, The somewhat involved proof is left optional (see App. 4). Absolute Convergence. A series z\ + z2 + ' " ' is called absolutely convergent if the series of the absolute values of the terms 2 \zm\ ~ \zi\ + l^zl + • • • m= 1 is convergent. If Z] + Zi + • • • converges but |zj] + |z2| + • • • diverges, then the series z\ + z2 + is called, more precisely, conditionally convergent. 668 CHAP. 15 Power Series, Taylor Series EXAMPLE 3 A Conditionally Convergent Series The series 1 — 2 + s — i1^ — ■- - converges, but only conditionally since the harmonic series diverges, as mentioned above (after Theorem 3). If a series is absolutely convergent, it is convergent. This follows readily from Cauchy's principle (sec Team Project 30). This principle also yields the following general convergence test. THEOREM 5 Comparison Test If a series Zi + Z2 + ' * ' is given and we can find a convergent series bx + b2 + • ■ ■ with nonnegalive real terms such that |zi| = bx, |z2| = b2, ' ' • , then the given series converges, even absolutely. PROOF By Cauchy's principle, since b± + b2 + • • • converges, for any given e > 0 we can find an N such that bn+i + ■ ■ ■ + bn_, p < e for every n > N and p = 1, 2, ■ • • . From this and |zi| = bx, \z2\ = b2, ■ ■ ■ we conclude that for those n and p, kn + il + ■ ■ ' + |ztl+p| = bn+-[ + • ■ ■ + bn+p < e. Hence, again by Cauchy's principle, |zx| + \z2\ + ■ • • converges, so that zx + z2 + ■ ■ • is absolutely convergent. ■ A good comparison series is the geometric series, which behaves as follows. THEOREM 6 Geometric Series The geometric series (6*) CO 2 qm = 1 + q + q2 + ' ' • m-0 converges with the sum 1/(1 — q) if\q\ < 1 and diverges if\q\ WWW 1. PROOF If \q\ S 1, then \qm'\ s 1 and Theorem 3 implies divergence. Now let |^| < 1. The «th partial sum is sn = 1 + q + ■ ■ ■ + q" From this, qsr q + ■ • • + qn + qn+1. On subtraction, most terms on the right cancel in pairs, and we are left with $n ~ asn = (1 ~ AO {where q < 1 is fixed), this series converges absolutely. If for every n > N, (8) the series diverges. '-n I 1 (n > AO, PROOF If (8) holds, then \zn+i\ = \zn\ f°r n > N, so that divergence of the series follows from Theorem 3. If (7) holds, then |z„+1| ": \zn\ q for n > N, in particular, I I — l-Zjv-t-al — I -Zat i 2I — kw+ik2' erc-- and in general, |zjv+p| = |zjy 1 ll?13-1- Since q < 1, we obtain from this and Theorem 6 I I I I < I I 2 < 1 1 1 - q Absolute convergence of Zi + Z2 + " ' ' now follows from Theorem 5. ■ CAUTION! The inequality (7) implies \zn+\lzn\ < 1, but this does not imply convergence, as we see from the harmonic series, which satisfies zn+1/zn - n/(n + 1) < 1 for all n but diverges. If the sequence of the ratios in (7) and (8) converges, we get the more convenient CHAP. 15 Power Series, Taylor Series Ratio Test If a series Zi + Z% + • • • with zn # 0 (n = 1, 2, • • •) is such that lim = L, then: (a) If L < 1, the series converges absolutely. (b) If L > 1, the series diverges. (c) If L = 1, f«e series may converge or diverge, so that the lest fails and permits no conclusion. PROOF (a) We write kn = \zn+ilzn\ and let L = I - b < 1. Then by the definition of limit, the kn must eventually get close to 1 - b, say, kn S= q = 1 — \b < 1 for all n greater than some A'. Convergence of Zj + z2 + ' ' ' now follows from Theorem 7. (b) Similarly, for L = 1 + c > 1 we have § 1 + \c > 1 for all n > N* (sufficiently large), which implies divergence of Z\ + z2 + ■ ■ ■ by Theorem 7. (c) The harmonic series 1 + 5 + g + • • ■ has zn, \lzn = nl(n + 1), hence L = 1, and diverges. The series 1 + 1 1 1 "9 + T6 25 I has hence also L = 1, but it converges. Convergence follows from (Fig. 361) 1 + fn dx 1 • • • + o — X so that slt «2, • • • is a bounded sequence and is monotone increasing (since the terms of the series are all positive); both properties together are sufficient for the convergence of the real sequence slt s2, • ■ • . (In calculus this is proved by the so-called integral test, whose idea we have used.) ■ 0 12 3 4 Fig. 361. Convergence of the series 1 + \ + 5 + fj + • • • EXAMPLE 4 Ratio Test Is the following series convergent or divergent? (First guess, then calculate.) » (100 + 75/)" I .. 2 -;- = 1 + (100 + 75/) + — (100 + 75/)2 + „ nl 2! ■ SEC. 15.1 Sequences, Series, Convergence Tests Solution. By Theorem 8. the series is convergent, since 100 + 75i\a+1/(n + 1)! 1100 + 75i| 125 1100 + 75i\n/n\ EXAMPLE 5 Theorem 7 More General than Theorem 8 n + 1 n + 1 1 = 0. Let an = ;723"' and bn = l/23n+i. Is the following series convergent or divergent? 1 i I i 1 an + bD + a, +&-, + ••» = H---1---1--H--+ - f 11 0 1 1 2 8 16 64 128 Solution. The ratios of the absolute values of successive terms arc; . Hence convergence follows from Theorem 7. Since the sequence of these ratios has no limit. Theorem 8 is not applicable. Root Test The ratio test and the root test are the two practically most important tests. The ratio test is usually simpler, but the root test is somewhat more general. THEOREM 9 Root Test If a series Z\ + Zs + " " 1 is such that for every n greater than some N, (9) V|z„J =S q < 1 in > AO (where q < 1 is fixed), this series converges absolutely. If for infinitely many n, (10) Cls 1 the series diverges. PROOF If (9) holds, then \zn\ ^ qn < 1 for all n > N. Hence the series \zi\ + IZ2I -)-*■• converges by comparison with the geometric series, so that the series Zi + Zz + - * ' converges absolutely. If (10) holds, then \zn\ S 1 for infinitely many n. Divergence of zx + z2 + ' ' ' now follows from Theorem 3. ■ CAUTION! Equation (9) implies V |zj < 1, but this does not imply convergence, as we see from the harmonic series, which satisfies Vl/n < 1 (for n > 1) but diverges. If the sequence of the roots in (9) and (10) converges, we more conveniently have THEOREM 10 Root Test If a series Z\ + Z2 + ' ' ' is such that lim v\zn\ = L, then: (a) The series converges absolutely if L < 1. (b) The series diverges if L > 1. (c) If L = 1, the test fails; that is, no conclusion is possible. CHAP. 15 Power Series, Taylor Series PROOF The proof parallels that of Theorem 8. (a) Let L = 1 — a* < 1. Then by the definition of a limit we have V |z„| < q = 1 - 5a* < 1 for all n greater than some (sufficiently large) N*. Hence kJ < N*' Absolute convergence of the series z, + z2 + • " ' now follows by the comparison with the geometric series. (b) If L > 1, then we also have V \zn\ > 1 for all sufficiently large n. Hence |z„| > 1 for those n. Theorem 3 now implies that z.\ + Z2 + ' " " diverges. (c) Both the divergent harmonic series and the convergent series i+5 + f> + f6 + 2^+''" give L - 1. This can be seen from (In n)ln —» 0 and JT 1 1 _L_ n/T = J_ = 1 J_ Vn nVn ealn)\nn ~* g0 ' y ^2 W2M g(2M)ln n ~^ g0- 1-10 SEQUENCES Are the following sequences Zi, z2. • • • , zH, • ■ • bounded? Convergent? Find their limit points. (.Show the details of your work.) 1. z„ = (-1)" + U2n 2- Zn _ —niri/4 3. Z„ = (-l)"/(ii + i) 4. z„ = (1 + ,■)* 5. z„ = Ln ((2 + /)") 6. z„ = (3 + 4i)B/n! 7. Zn = sill(«7T/4) + i" 8. zn = [(1 + 3i)/vTo 9. z„ = (0.9 + 0.1/)2n 10. zn = (5 + 5i)~n 11. Illustrate Theorem 1 by an example of your own. 12. (Uniqueness of limit) Show that if a sequence converges, its limit is unique. 13. (Addition) If zi, z2, ' ' • converges with the limit / and Zi*, Z2*, ' ' " converges with the limit / , show that Zi + Zi*, z2 + z2*, " • ' converges with the limit I + I . 14. (Multiplication) Show that under the assumptions of Prob. 13 the sequence ZiZt*, z2z2*, 1 ' ■ converges with ihe limit 11*. 15. (Boundedness) Show that a complex sequence is bounded if and only if the two corresponding sequences of the real parts and of the imaginary parts are bounded. 16-24 SERIES Are the following scries convergent or divergent? (Give a reason.) 16. 2 n=0 18. 1 n=u cc 20. 2 (10 - 15i)" n2 - 2i 17.2 n=0 oo 19.2 (-1)™(T + 202 (2n + 1)! Vn 1 In n CO -1 21. 2 - 22. 2 (nlf (3»)l (1 + 0" 24. 2 | 25. What is the difference between (7) and just stating IWzJ < 1? 26. Illustrate Theorem 2 by an example of your choice. 27. For what n do we obtain the term of greatest absolute value of the series in Example 4? About how big is it? First guess, then calculate it by the Stirling formula in Sec. 24.4. 28. Give another example showing that Theorem 7 is more general than Theorem 8. 29. CAS PROJECT. Sequences and Series, (a) Write a program for graphing complex sequences. Apply it to sequences of your choice that have interesting "geometrical" properties (e.g., lying on an ellipse, spiraling toward its limit, etc.). (b) Write a program for computing and graphing numeric values of the first n partial sums of a series of complex numbers. Use the program to experiment with the rapidity of convergence of series of your choice. 30. TEAM PROJECT. Series, (a) Absolute convergence. Show that if a series converges absolutely, it is convergent. (b) Write a short report on the basic concepts and properties of series of numbers, explaining in each case whether or not they earn' over from real series (discussed in calculus) to complex series, with reasons given. SEC. 15.2 Power Series (c) Estimate of the remainder. Let \zn h xlz,\ S q < I, so that the series Zi + z2 + •' ■ converges by the ratio test. Show that the remainder Rn = zn+1 + z„+2 + ' ' ' satisfies the inequality \Rn\ = |zn+1|/(l — q). (d) Using (c), find how many terms suffice for computing the sum s of the scries V - + I with an error not exceeding 0.05 and compute s to this accuracy. (e) Find other applications of the estimate in (c). 15.2 Power Series Power series arc the most important series in complex analysis because we shall see that their sums are analytic functions, and every analytic function can be represented by power series (Theorem 5 in Sec. 15.3 and Theorem 1 in Sec. 15.4). A power series in powers of z — z0 is a series of the form cc (1) 2 an(z - z0T = «0 + ai(z ~ z0) + "2(z - %)2 + • " n-0 where z is a complex variable, a0, ax, • • ■ are complex (or real) constants, called the coefficients of the series, and z0 is a complex (or real) constant, called the center of the series. This generalizes real power series of calculus. If z0 = 0, we obtain as a particular case a power series in powers of z: oo (2) 2 an no. EXAMPLE 3 Convergence Only at the Center. (Useless Series) The following power series converges only at z = 0, but diverges for every z ¥■ 0, as we shall show. X "\zn = 1 + z + 2z2 + 6z3 + n=u In fact, from the ratio test we have (n + !)!;" = (n + 1) \z\ as n —» oc (z fixed and # Ü). THEOREM 1 Convergence of a Power Series (a) Every power series (1) converges at the center z0. (b) If (I) converges at a point z = Zi ^ Zo> '< converges absolutely for every z closer to z0 /ftan zx, j/?af is, |z - z0| < |zi ~~ Zol- See Fig. 362. (c) //(l) diverges at a z = Z2, ft diverges for every z farther away from z0 //ictn z2. See Fig. 362. - Divergent / „ V1 * Conv. I o , i ? *2 I / _ Fig. 362. Theroem 1 PROOF (a) For z = z() the series reduces to the single term a0. (b) Convergence at z = z-> gives by Theorem 3 in Sec. 15.1 an(zi _ ZoT —» 0 as n —> This implies boundedncss in absolute value, |fln(Zi - z0)n| < M for every n = 0, 1, • • • . Multiplying and dividing an(z - z0)n by (zt - z0)n we obtain from this 2n(Zl Zo)" I j \ ^1 ?0 / s M Z Zo Zl — Zo SEC. 15.2 Power Series 675 Summation over n gives O) 2 k(z-zo)1=^2 n=l n=l Now our assumption \z - z0\ < [zi _ Zq| implies that |(z - z0)/(zi - z0)| < 1- Hence the series on the right side of (3) is a converging geometric series (see Theorem 6 in Sec. 15.1). Absolute convergence of (1) as stated in (b) now follows by the comparison test in Sec. 15.1. (c) If this were false, we would have convergence at a z3 farther away from z0 than z2. This would imply convergence at z2, by (b), a contradiction to our assumption of divergence at z2. ' Radius of Convergence of a Power Series Convergence for every z (the nicest case, Example 2) or for no z + z0 (the useless case. Example 3) needs no further discussion, and we put these cases aside for a moment. We consider the smallest circle with center z0 that includes all the points at which a given power series (1) converges. Let R denote its radius. The circle I* ~ Zol = R (Fig. 363) is called the circle of convergence and its radius R the radius of convergence of (1). Theorem 1 then implies convergence everywhere within that circle, that is, for all z for which (4) \z ~ z0\ R. No general statements can be made about the convergence of a power series (1) on the circle of convergence itself. The series (1) may converge at some or all or none of these points. Details will not be essential to us. Hence a simple example may just give us the idea. X Fig. 363. Circle of convergence 676 CHAP. 15 Power Series, Taylor Series EXAMPLE 4 Behavior on the Circle of Convergence On the circle of convergence (radius R= 1 in all three series), 2 zn/n2 converges everywhere since 2 1/n2 converges, 2 z"ln converges at 1 (by Leibniz's test) but diverges at I, 2 zn diverges everywhere. Notations R = 00 and R = 0. To incorporate these two excluded cases in the present notation, we write r = oc if the series (1) converges for all z (as in Example 2), r = 0 if (1) converges only at the center z — Zc, (as in Example 3). These are convenient notations, but nothing else. Real Power Series. In this case in which powers, coefficients, and center are real, formula (4) gives the convergence interval \x — x0\ < r of length 2r on the real line. Determination of the Radius of Convergence from the Coefficients. For this important practical task we can use THEOREM 2 Radius of Convergence R Suppose that the sequence |an+1/an|, n — 1, 2, • ■ • , converges with limit L*. If L* = 0, then /? = <»; that is, the power series (1) converges for all z. If L* ¥= 0 (hence L* > 0), then (6) r = 1 lim ■'n + 1 (Cauchy-Hadamard formula1). If \an+1lan\ —> oc, then r = 0 (convergence only at the center z0). PROOF For (1) the ratio of the terms in the ratio test (Sec. 15.1) is an I iU ^o; a„{z - z0Y' The limit is L = L*\z - z0\- Let l* # 0, thus L* > 0. We have convergence if L = L*\z - z0| < 1, thus |z - z0\ < 1/L*, and divergence if \z — Zq\ > 1/L*. By (4) and (5) this shows that ml* is the convergence radius and proves (6). If l* = 0, then l = 0 for every z, which gives convergence for all z by the ratio test. If \an+1/an\ —> °°, then \anl i/an\\z — Zq\ > 1 for any z + Zn and all sufficiently large n. This implies divergence for all z ¥= Zn by the ratio test (Theorem 7, Sec. 15.1). ■ earned after the French mathematicians A. L. CAUCHY (see Sec. 2.5) and JACQUES HADAMARD (1865-1963). Hadamard made basic contributions to the theory of power scries and devoted his lifework to partial differential equations. SEC. 15.2 Power Series Formula (6) will not help if L* does not exist, but extensions of Theorem 2 are still possible, as we discuss in Example 6 below. EXAMPLE 5 Radius of Convergence By (6) the radius of convergence of the power series 2 Qn)\ W.f (z - 30" is R = lim n—*oc {2n)\ I (In + 2)1 (rt!)2 / ((« + l)!)2 lim (2n)! ((n + l)!f (2n + 2)1 (n\f lim {n + If 1 ,oo (2« + 2)(2n + 1) 4 ' The series converges in the open disk \z — 3/| < J of radius j and center 3i. EXAMPLE 6 Extension of Theorem 2 Find the radius of convergence R of the power series l + r-rf I 3 + h)' i 16 Solution. The sequence of the ratios 1/6, 2(2 + \), 1/(8(2 + \)), ■ • ■ does not converge, so that Theorem 2 is of no help. It can be shown that (6*) R = Ml. L = lim V leu, This still does not help here, since {V|a„|) does not converge because vjflj = V 1/2" = 1/2 for odd n, whereas for even n we have VW = V2 + l/2n-» 1 as n —> «, so that V \av\ has the two limit points 1/2 and 1. It can further be shown that 5**) R = III, I the greatest limit point of the sequence Here / = 1. so that R = 1. Answer. The series converges for z < 1. Summary. Power series converge in an open circular disk or some even for every z (or some only at the center, but they are useless); for the radius of convergence, see (6) or Example 6. Except for the useless ones, power series have sums that are analytic functions (as we show in the next section); this accounts for their importance in complex analysis. 1. (Powers missing) Show that if Ea„?" has radius of convergence R (assumed finite), then 2 anz2n has radius of convergence \rR. Give examples. 2. (Convergence behavior) Illustrate the facts shown by Examples 1-3 by further examples of your own. 3-181 RADIUS OF CONVERGENCE Find the center and the radius of convergence of the following power series. (Show the details.) (z + i) . ^ n 3.2 5- 2 ^7 (z + IT cc 7-2 n=0 CO 9. 2 (" " i)nzn cc n.2 00 (-l)n+1 6.2 22>!)2 * 00 CT-\2n io. 2 (2zf (2n)l n-0 4«- 12.2 (1 + if (Z - 5)" 678 such that all three formulas (6), (6*), and (6**) will come up. 20. TEAM PROJECT. Radius of Convergence, (a) Formula (6) for R contains \anlan , ]J, not \an , ylan\. How could you memorize this by using a qualitative argument? (b) Change of coefficients. What happens to R ft) < R < x) if you (i) multiply all an by k # 0, (ii) multiply an by kn i= 0, (iii) replace an by 1 lanl (c) Example 6 extends Theorem 2 to nonconvergent cases of an/an + l. Do you understand the principle of "mixing" by which Example 6 was obtained? Use this principle for making up further examples. (d) Does there exist a power series in powers of z that converges at z = 30 + 10/ and diverges at z = 31 — 6/? (Give reason.) 15.3 Functions Given by Power Series The main goal of this section is to show that power series represent analytic functions (Theorem 5). Along our way we shall see that power series behave nicely under addition, multiplication, differentiation, and integration, which makes these series very useful in complex analysis. To simplify the formulas in this section, we take z0 = 0 and write GO (1) 2 anzn. 71 = 0 This is no restriction because a series in powers of £ — z0 with any zn can always be reduced to the form (1) if we set £ — z0 = z. Terminology and Notation. If any given power series (1) has a nonzero radius of convergence R (thus R > 0), its sum is a function of z, say f(z). Then we write cc (2) fit) = 2 anzn = fl0 + axz + a2z.2 + • • ■ (|z| < R). n-0 We say that f(z) is represented by the power series or that it is developed in the power series. For instance, the geometric series represents the function /(z) = 1/(1 — z) in the interior of the unit circle |z| = 1. (See Theorem 6 in Sec. 15.1.) Uniqueness of a Power Series Representation. This is our next goal. It means that a function f(z) cannot be represented by two different power series with the same center. We claim that if fix) can at all be developed in a power series with center z0, the development is unique. This important fact is frequently used in complex analysis (as well as in calculus). We shall prove it in Theorem 2. The proof will follow from 13. 2 »(« - 1)(z - 3 + 2/)" •re - 2 ~-° /_i \n 14- 2 -7Z—r <2n 15- 2 2"(z - ifn (2«)! T7=0 16.2 17. 2^z2» 18. 2 ^ ft + .if n=0 19. CAS PROJECT. Radius of Convergence. Write a program for computing R from (6), (6*), or (6**), in this order, depending on the existence of the limits needed. Test the program on series of your choice and SEC. 15.3 Functions Given by Power Series 679 THEOREM 1 Continuity of the Sum of a Power Series If a function f(z) can be represented by a power series (2) with radius of convergence R > 0, then f{z) is continuous at z = 0. PROOF From (2) with z = 0 we have /(0) = a0. Hence by the definition of continuity we must show that limz^,0 f(z) = /(0) = a0. That is, wc must show that for a given e > 0 there is a 8 > 0 such that |z| < 8 implies |/(z) — a0\ < e. Now (2) converges absolutely for |z| r with any r such that 0 < r < R, by Theorem 1 in Sec. 15.2. Hence the series DC j CC 2 kkn_1 = - 2 kk" n=l ' n=l converges. Let S =fr 0 be its sum. (5 = 0 is trivial.) Then for 0 < \z\ ' r, \Kz) - Oo\ 2 0 is less than r and less than eAS'. Hence \z\S < 8S < (e/S)S = e. This proves the theorem. ■ From this theorem we can now readily obtain the desired uniqueness theorem (again assuming z0 = 0 without loss of generality): THEOREM 2 Identity Theorem for Power Series. Uniqueness Let the power series a0 + axz + a2Z + • • • and b0 + bxz + b2z2 + • • • both be convergent for \z\ < R, where R is positive, and let them both have the same sum for all these z. Then the series are identical, that is, a0 = b0, ax = bx, a2 = b2, • • • . Hence if a function f(z) can be represented by a power series with any center z0, this representation is unique. PROOF We proceed by induction. By assumption, a0 + axz + a2z2 + ••■ = b0 + blZ + b2z2 + • • • (\z\ < R). The sums of these two power series are continuous at z = 0, by Theorem 1. Hence if we consider \z\ > 0 and let z —» 0 on both sides, we see that a0 = b0: the assertion is true for n = 0. Now assume that an = bn for n = 0, 1, • • • , m. Then on both sides we may omit the terms that are equal and divide the result by zm+1 0); this gives am+i + am+2z + am+3z2 + • • • = hm+1 + bm+2Z + bm+[iz2 + • • • . Similarly as before by letting z —* 0 we conclude from this that am+1 — bm+1. This completes the proof. ■ 680 CHAP. 15 Power Series, Taylor Series Operations on Power Series Interesting in itself, this discussion will serve as a preparation for our main goal, namely, to show that functions represented by power series are analytic. Termwise addition or subtraction of two power series with radii of convergence Rx and R2 yields a power series with radius of convergence at least equal to the smaller of R1 and R2. Proof. Add (or subtract) the partial sums sn and s% term by term and use lim (sn ± s*) = lim sn ± lim s%. Termwise multiplication of two power scries CO f(z) = 2 «feZfc = a0 + axz + • • ■ and Riz) = 2 bmzm = b0 + b^ + m=0 means the multiplication of each term of the first series by each term of the second series and the collection of like powers of z. This gives a power series, which is called the Cauchy product of the two series and is given by a0b0 + (a0b1 + axb0)z + (a0b2 + arbx + a2b0)z2 + • • ■ co = 2 («rA,. + «A,,-i + ' ■ • + anb0)z.n. n=0 We mention without proof that this power series converges absolutely for each z within the circle of convergence of each of the two given series and has the sum s(z) = f(z)g(z). For a proof, see [D5] listed in App. 1. Termwise differentiation and integration of power series is permissible, as we show next. We call derived scries of the power series (1) the power series obtained from (1) by termwise differentiation, that is, (3) ^ nan: n-l _ = a1 + 2a2z. + 3a-jZ + THEOREM 3 Termwise Differentiation of a Power Series The derived series of a power series has the same radius of convergence as the original series. PROOF This follows from (6) in Sec. 15.2 because n\an\ n lim--—|-r = lim - lim ?wcc (n + 1) |an+1| n-^x n + 1 m~»oc On = lim On an+l or, if the limit does not exist, from (6**) in Sec. 15.2 by noting that ^n —> 1 as n^> ». SEC. 15.3 Functions Given by Power Series EXAMPLE 1 Application of Theorem 3 Find the radius of convergence R of the following series by applying Theorem 3. QO 2 Solution. Differentiate the geometric scries twice term by term and multiply the rcsu It by ;2/2. This yields the given series. Hence R = 1 bv Theorem 3. THEOREM 4 Termwise Integration of Power Series The power series n + 1 a0z + a. -2 + ü2 r3 obtained by integrating the series a0 + ajz + a2Z2 + same radius of convergence as the original series. term by term has the The proof is similar to that of Theorem 3. With Theorem 3 as a tool, we are now ready to establish our main result in this section. Power Series Represent Analytic Functions THEOREM 5 Analytic Functions. Their Derivatives A power series with a nonzero radius of convergence R represents an analytic function at every point interior to its circle of convergence. The derivatives of this function are obtained by differentiating the original series term by term. All the series thus obtained have the same radius of convergence as the original series. Hence, by the first statement, each of them represents an analytic function. PROOF (a) We consider any power series (1) with positive radius of convergence R. Let f(z.) be its sum and fL(z) the sum of its derived series; thus (4) Kz) = 2 anzn n,=0 and fi(z) We show that /(z) is analytic and has the derivative /x(z) in the interior of the circle of convergence. We do this by proving that for any fixed z with \z\ < R and Az —» 0 the difference quotient [f(z + Az) — j'(z)]/Az approaches /i(z). By termwise addition we first have from (4) (5) ,f(z + Az) - f(z) Az fiiz) = 2 an (z + AzT A: Note that the summation starts with 2, since the constant term drops out in taking the difference f(z + Az) - f(z), and so does the linear term when we subtract /j(z) from the difference quotient. 682 CHAP. 15 Power Series, Taylor Series (b) We claim that the series in (5) can be written oo (6) 2 anAz[(z + A-)""2 + 2z(z + Az)n~3 + ••■ + («- 2)zn~\z + Az) + (n-])zn-2]. The somewhat technical proof of this is given in App. 4. (c) We consider (6). The brackets contain n — I terms, and the largest coefficient is n - 1. Since (n - l)2 S n(n - 1), we see that for \z\ = R0 and |z + Az| ' tf0, #0 < the absolute value of this series (6) cannot exceed (7) |Az| 2 kl«(« " 1)<~2 This series with an instead of \an\ is the second derived series of (2) at z — Ro and converges absolutely by Theorem 3 of this section and Theorem 1 of Sec. 15.2. Hence our present series (7) converges. Let the sum of (7) (without the factor |Az|) be K(R0). Since (6) is the right side of (5), our present result is ' |Az| K(R0). Letting Az —> 0 and noting that RQ (< R) is arbitrary, we conclude that f(z) is analytic at any point interior to the circle of convergence and its derivative is represented by the derived series. From this the statements about the higher derivatives follow by induction. Summary. The results in this section show that power series are about as nice as we could hope for: we can differentiate and integrate them term by term (Theorems 3 and 4). Theorem 5 accounts for the great importance of power series in complex analysis: the sum of such a scries (with a positive radius of convergence) is an analytic function and has derivatives of all orders, which thus in turn are analytic functions. But this is only part of the story. In the next section we show that, conversely, every given analytic function f(z) can be represented by power series, called Taylor series and being the complex analog of the real Taylor series of calculus. f(z + Az) - ,f(z) Az fviz) PROBLEM SET 15.3 1—101 RADIUS OF CONVERGENCE BY DIFFERENTIATION OR INTEGRATION Find the radius of convergence in two ways: (a) directly by die Cauchy-Hadamard formula in Sec. 15.2, (b) from a series of simpler terms by using Theorem 3 or Theorem 4. ^ n(n — 1) I. 2 - (z - 20" 2-2 4" \ n(n + 1) 3.2 (z + if 4-2 n-0 so in (-1)" In + 1 \ 77 5.2 Vln(n + 1) (z - if \tG)(fF GO 7-2 (-7)" ^2n 8.2 n(n + l)(n + 2) 2n(2n - 1) 9„_ SEC. 15.4 Taylor and Maclaurin Series 9-2 CG / 10.2 n-0 V In + k \ k x In + in 11. (Addition and subtraction) Write out the details of the proof on tcrmwise addition and subtraction of power series. 12. (Cauchy product) Show that (1 - -r2 = 2';,^0 (n + l)zn (a) by using the Cauchy product, (b) by differentiating a suitable series. 13. (Cauchy product) Show that the Cauchy product of 2^=0 znln\ multiplied by itself gives 2^=0 (2z)n/n\. 14. (On Theorem 3) Prove that vti —» 1 as n —> x (as claimed in the proof of Theorem 3). 15. (On Theorems 3 and 4) Find further examples of your own. 16-20 APPLICATIONS OF THE IDENTITY THEOREM State clearly and explicitly where and how you are using Theorem 2. 16. (Bionomial coefficients) Using (1 + z)p(\ + z)q = (1 + zf 9, obtain the basic-relation P + q\ „-o \nl \r - n 17. (Odd function) If/(;) in (I) is odd (i.e.. /( — z) = —f(z)), show that an = 0 for even n. Give examples. 18. (Even functions) If /(-) in (1) is even (i.e., /(—z) = f(z)), show that an = 0 for odd n. Give examples. 19. Find applications of Theorem 2 in differential equations and elsewhere 20. TEAM PROJECT. Fibonacci numbers.2 (a) The Fibonacci numbers are recursively defined by do - a1 = 1, an + 1 = an + an_t if n = 1, 2, • ■ • . Find the limit of the sequence (an_L/an). (b) Fibonacci's rabbit problem. Compute a list of flj, • • ■ , «12. Show that a12 = 233 is the number of pairs of rabbits after 12 months if initially there is 1 pair and each pair generates 1 pair per month, beginning in the second month of existence (no deaths occurring). (c) Generating function. Show that the generating function of the Fibonacci numbers is f(z) = 1/(1 - z - J2); that is, if a power series (1) represents this f(z), its coefficients must be the Fibonacci numbers and conversely. Hint. Start from f(z) (1 — z — z2) = I and use Theorem 2. 15.4 Taylor and Maclaurin Series The Taylor series3 of a function f(z), the complex analog of the real Taylor series is 1 (1) /(z) = 2 Onfe - Zof where an = — fn\z0) n n=l or, by (1), Sec. 14.4, 1 r Hz*) 2m Jc (z{ - Zo) In (2) we integrate counterclockwise around a simple closed path C that contains z0 in its interior and is such that f(z) is analytic in a domain containing C and every point inside C. A Maclaurin series3 is a Taylor series with center z0 = 0. 2LEONARDO OF PISA, called FIBONACCI (= son of Bonaccio), about 1180-1250, Italian mathematician, credited with the first renaissance of mathematics on Christian soil. 3BROOK TAYLOR (1685-1731), English mathematician who introduced real Taylor series. COI.IN MACLAURIN (1698-1746), Scots mathematician, professor at Edinburgh. 684 CHAP. 15 Power Series, Taylor Series The remainder of the Taylor series (1) after the term an(z — z0)n is (Z - Zq)" + 1 f /(z*) 2m Jc (z* - Zo)n+\z* - z) (3) Ruiz) = —^3- ? ~,--3T dz* (proof below). Writing out the corresponding partial sum of (1), we thus have (4) f(z) = /(zo) + ^ f'(zo) + /"to) + i--_ y» This is called Taylor's formula with remainder. We sec that Taylor series are power series. From the last section we know that power series represent analytic functions. And we now show that every analytic function can be represented by power series, namely, by Taylor series (with various centers). This makes Taylor series very important in complex analysis. Indeed, they are more fundamental in complex analysis than their real counterparts are in calculus. THEOREM 1 Taylor's Theorem Let f(z) be analytic in a domain D, and let z = z0 be any point in D. Then there exists precisely one Taylor series (1) with center z0 that represents f(z). This representation is valid in the largest open disk with center za in which f(z) is analytic. The remainders Rn(z) of (I) can be represented in the form (3). The coefficients satisfy the inequality (5) M where M is the maximum of \f(z)\ on a circle \z also in D. r in D whose interior is PROOF The key tool is Cauchy's integral formula in Sec. 14.3; writing z and z* instead of z0 and z (so that z* is the variable of integration), we have 1 f f(z*) (6) /(z) = — f J^-L dz*. 2m Jc z* - z z lies inside C, for which we take a circle of radius r with center z0 and interior in D (Fig. 364). We develop l/(z* — z) in (6) in powers of z — z0. By a standard algebraic manipulation (worth remembering!) we first have 1 1 1 (7) -I- = " 685 For later use we note that since z* is on C while z is inside C, we have (7*) < 1 (Fig. 364). Fig. 364. Cauchy formula (6) To (7) we now apply the sum formula for a finite geometric sum (8*) 1 + q + ■ ■ ■ + qn 1 - qn+1 1 qn+1 (4 * D. 1 — q 1 — q 1 — q which we use in the form (take the last term to the other side and interchange sides) (8) 1 ~n+l 1 + q 1 — q 1 — q Applying this with q = (z - Z0)/(z* ~ z0) to the right side of (7), we get 1 1 7* — •o L 1 + co Z Zo 1 + <-0 n+l We insert this into (6). Powers of z ~ z0 do not depend on the variable of integration z*, so that we may take them out from under the integral sign. This yields 1 I fiz*) f(z) = — f dz* + 2m Jc z* ~ Zo 2 m fiz*) (2* - ZoY dz* + ■ (z - zpT 2-777 f(z*) ,n +1 + Rn(z) (z* - Zo) with Rn(z.) given by (3). The integrals are those in (2) related to the derivatives, so thai we have proved the Taylor formula (4). Since analytic functions have derivatives of all orders, we can take n in (4) as large as we please. If we let n approach infinity, we obtain (1). Clearly, (1) will converge and represent f(z) if and only if (9) lim Rn{z) = 0. 686 CHAP. 15 Power Series, Taylor Series THEOREM 2 We prove (9) as follows. Since z* lies on C, whereas z lies inside C (Fig. 364), we have |z* — zj > 0. Since f(z) is analytic inside and on C, it is bounded, and so is the function /(z*)/(z* - z), say, f(z*) - M for all z* on C. Also, C has the radius r = \z* ML-inequality (Sec. 14.1) we obtain from (3) (10) 277 _ \n I 1 Zol ■ z0| and the length 27rr. Hence by the f(z*) c (z* - zu)"+1(z* dz" M 2m- = /W z i0 Now |z — Zo| < r because z lies inside C. Thus |z — z„|/r < 1, so that the right side approaches 0 as n; —» °°. This proves the convergence of the Taylor series. Uniqueness follows from Theorem 2 in the last section. Finally, (5) follows from (1) and the Cauchy inequality in Sec. 14.4. This proves Taylor's theorem. ■ Accuracy of Approximation. We can achieve any preassinged accuracy in approximating f(z.) by a partial sum of (1) by choosing n large enough. This is the practical aspect of formula (9). Singularity, Radius of Convergence. On the circle of convergence of (1) there is at least one singular point of /(z), that is, a point z = c at which /(z) is not analytic (but such that every disk with center c contains points at which f(z) is analytic). We also say that /(z) is singular at c or has a singularity at c. Hence the radius of convergence R of (I) is usually equal to the distance from zq to the nearest singular point of f(z). (Sometimes jf? can be greater than that distance: Ln z is singular on the negative real axis, whose distance from zn = ~ 1 + is 1, but the Taylor series of Ln z with center Z0 = — 1 + i has radius of convergence V2.) Power Series as Taylor Series Taylor series are power series—of course! Conversely, we have Relation to the Last Section A power series with a nonzero radius of convergence is the Taylor series of its sum. PROOF Given the power series f(z) = fl0 + fliCz - Zo) + «2(~ " z0f + a3(z - zQf + ■■■ . Then /(z0) = aQ. By Theorem 5 in Sec. 15.3 wc obtain f'iz) = «i + 2a2(z - z0) + 3fl3(z - zof + ■ ■ ■ , thus f'(zo) = 0 we can find an A'j(e) such that Kzi) - sn(zi)\ < e for all n > N^e). If we pick a z2 in G, keeping e as before, we can find an N2(e) such that \s(z2) - sn(z,2)\ < e for all n > N2(e), and so on. Hence, given an e > 0, to each z in C there corresponds a number Nz(e). This number tells us how many terms we need (what sn we need) at a z to make \s(z) — sn(z)\ smaller than e. Thus this number Nz{e) measures the speed of convergence. Small Nz(e) means rapid convergence, large Nz(e) means slow convergence at the point z considered. Now, if w'c can find an N(e) larger than all these Nz(e) for all z in G, we say that the convergence of the series (1) in G is uniform. Hence this basic concept is defined as follows. DEFINITION Uniform Convergence A series (1) with sum s(z) is called uniformly convergent in a region G if for every e > 0 we can find an N = Me), not depending on z, such that Hz) - sn(z)\ < for all n > N(e) and all z in G. Uniformity of convergence is thus a property that always refers to an infinite set in the z-plane, that is, a set consisting of infinitely many points. 692 CHAP. 15 Power Series, Taylor Series EXAMPLE 1 Geometric Series Show that the geometric series 1 + z + z2 + ■ ■ - is (a) uniformly convergent in any closed disk \z\ = r < 1, (b) not uniformly convergent in its whole disk of convergence |z| < 1. Solution, (a) For z in that closed disk we have 1 - z| £ 1 - r (sketch it). This implies that 1/|1 - -| Š 1/(1 - r). Hence (remember (8) in Sec. 15.4 with q = z) r«+i ~ I - r ' Since r < 1, we can make the right side as small as we want by choosing n large enough, and since the right side does not depend on z (in the closed disk considered), this means that the convergence is uniform. (b) For given real K (no matter how large) and n we can always find a z in the disk |z| < 1 such that n i 1 1 - Z simply by taking z close enough to 1. Hence no single A'(e) will suffice to make |.s(z) - sn(z)\ smaller than a given € > 0 throughout the whole disk. By definition, this shows that the convergence of the geometric series in |z| < 1 is not uniform. Hz) - sn(z)\ II -z\ This example suggests that for a power series, the uniformity of convergence may at most be disturbed near the circle of convergence. This is true: THEOREM 1 Uniform Convergence of Power Series A power series (2) 2 am(Z ZoT with a nonzero radius of convergence R is uniformly convergent in every circular disk \z — Zq\ = r <>f radius r < R. PROOF For \z — z.0\ = r and any positive integers n and p we have (3) \an+1(z - ZoT+1 + ■■■+ an+p{z - z0)reiP! ^ k+il'-"'+1 + ■ ■ • + \an+P\rn+p. Now (2) converges absolutely if |z — z0| = r < 7? (by Theorem 1 in Sec. 15.2). Hence it follows from the Cauchy convergence principle (Sec. 15.1) that, an e > 0 being given, we can find an N(e) such that \an+l\rn'1 + • • ■ + \an+p\rn+p < e for n > N(e) and p = 1, 2, • • • . From this and (3) we obtain K+i(z - ZoT'1 + ■■■ + an+p(z - z0rlp| < e for all z. in the disk \z — z0\ = r, every n > N(e), and every p = 1, 2, • • • . Since /V(e) is independent of z, this shows uniform convergence, and the theorem is proved. ■ Theorem 1 meets with our immediate need and concern, which is power series. The remainder of this section should provide a deeper understanding of the concept of uniform convergence in connection with arbitrary series of variable terms. SEC. 15.5 Uniform Convergence. Optional 693 Properties of Uniformly Convergent Series Uniform convergence derives its main importance from two facts: 1. If a series of continuous terms is uniformly convergent, its sum is also continuous (Theorem 2, below). 2. Under the same assumptions, termwisc integration is permissible (Theorem 3). This raises two questions: 1. How can a converging series of continuous terms manage to have a discontinuous sum? (Example 2) 2. How can something go wrong in term wise integration? (Example 3) Another natural question is: 3. What is the relation between absolute convergence and uniform convergence? The surprising answer: none. (Example 5) These are the ideas we shall discuss. If we add finitely many continuous functions, we get a continuous function as their sum. Example 2 will show that this is no longer true for an infinite series, even if it converges absolutely. However, if it converges uniformly, this cannot happen, as follows. THEOREM 2 Continuity of the Sum Let the series 2 fm(z) = fo(z) + fiiz) + •■■ ■m-0 be uniformly convergent in a region G. Let F(z) be its sum. Then if each term fm(z) is continuous at a point Z\ in G, the function F(z) is continuous at zv PROOF Let sn(z) be the nth partial sum of the series and Rn(z) the corresponding remainder: sn = /o + /l + " ' ' + fw R-n ~ fn+l + fn + 2 + ' " ' - Since the series converges uniformly, for a given e > 0 we can find an N = N(e) such that \Rn(z)\ < 3 for all z in G. Since sN(z) is a sum of finitely many functions that are continuous at Z\, this sum is continuous at z-\. Therefore, we can find a 5 > 0 such that \sN(z) - sN{z-\)\ < ~ for all z in G for which \z - z\\ < S. Using F = sN + RN and the triangle inequality (Sec. 13.2), for these z we thus obtain \F(z) - F(Zl)\ = \sN(z) + RN(z) - [sN(Zl) + Rn(z!)]\ ^ kvfc) - %(zi)l + \Rn(z)\ + \RM\ N(e) and all x, sav, in the interval OIj:? 1. -1 0 1 * Fig. 365. Partial sums in Example 2 695 Termwise Integration This is our second topic in connection with uniform convergence, and we begin with an example to become aware of the danger of just blindly integrating term-by-term. EXAMPLE 3 Series for which Termwise Integration is Not Permissible Let um(x) = mxe~mx and consider the series oc 2 fmW where fm(x) = um(x) - um_i(x) m-Q in the interval 0 ^ x ^ I. The nth partial sum is Sn = Ui - u0 + M2 -«! + ••■ + un - «„_! - U„ - U0 = Un. Hence the series has the sum F(x) = lim sn(x) = lim u.a(x) = 0 (Og^s l). From this we obtain J Fix) dx = 0. On the other hand, by integrating term by term and using f1 + /2 + • • ■ + fn = sn, we have .1 . „1 I dx. 2 /,„(»") dx = lim 2 /mW ^ = 1™ Jr, n—>äo j.. n—»oc Jn Now sn = u„ and the expression on the right becomes lim I un(x) dx = lim J nxe nx2 dx = lim - (1 - e '"') I 2" ' but not 0. This shows that the series under consideration cannot be integrated term bv term from x = 0 to x = 1. ■ The series in Example 3 is not uniformly convergent in the interval of integration, and we shall now prove that in the case of a uniformly convergent series of continuous functions we may integrate term by term. THEOREM 3 Termwise Integration Let oc F(Z) = 2 fnSX) = U(Z) + A(Z) + " - • be a uniformly convergent series of continuous functions in a region G. Let C be any path in G. Then the series (4) 2 jfmdd dz = / fo(z) dz + j h(z) dz+ ■■■ m=0 c c c is convergent and has the sum j F(z.) dz. 696 CHAP. 15 Power Series, Taylor Series PROOF From Theorem 2 it follows that F(z) is continuous. Let sn(z) be the nth partial sum of the given series and Rn(z) the corresponding remainder. Then F — sn + Rn and by integration, THEOREM 4 j F(z) dz = j sjz) dz + \ Rn(z) dz. Let L be the length of C. Since the given scries converges uniformly, for every given e > 0 we can find a number N such that < e/L for all n > N and all z in G. By applying the ML-inequality (Sec. 14.1) we thus obtain j et ) dz < — L = e L Since Rr s.„, this means that f F(z) dz - f s„(z) dz < e for all n > N. for all n > N. 'c c Hence, the series (4) converges and has the sum indicated in the theorem. Theorems 2 and 3 characterize the two most important properties of uniformly convergent series. Also, since differentiation and integration are inverse processes, Theorem 3 implies Termwise Differentiation Let the series f0(z) + j\(z) + /2(2) + ' • ' be convergent in a region G and let F(z) be its sum. Suppose that the series f'Q(z) + f[(z) + f2(z) + • ■ • converges uniformly in G and its terms are continuous in G. Then F'(z) = fo(z) + f[(z) + f2(z) + for all z in G. Test for Uniform Convergence Uniform convergence is usually proved by the following comparison test. THEOREM 5 Weierstrass5 M-Test for Uniform Convergence Consider a series of the form (1) in a region G of the z-plane. Suppose that one can find a convergent series of constant terms, (5) M0 + M, + M2 such that \fm(z)\ ' Mrn for all z. in G and every in = 0. 1, uniformly convergent in G. Then (1) is 5KARL WEIERSTRASS (1815-1897), great German mathematician, whose life-work was the development of complex analysis based on the concept of power series (see the footnote in Sec. 13.4). He also made basic contributions to the calculus, the calculus of variations, approximation theory, and differential geometry. He obtained the concept of uniform convergence in 1841 (published 1894, sic!); the first publication on the concept was by G. G. STOKES (see Sec 10.9) in 1847. 697 The simple proof is left to the student (Team Project 18). EXAMPLE 4 Weierstrass M-Test Does the following series converge uniformly in the disk \z\ = I ? | zm + 1 m=i '"2 + coshm|z| Solution. Uniform convergence follows by the Weierstrass M-test and the convergence of 21/m2 (see Sec. 15.1, in the proof of Theorem 8) because it! + cosh m\z No Relation Between Absolute and Uniform Convergence We finally show the surprising fact that there are series that converge absolutely but not uniformly, and others that converge uniformly but not absolutely, so that there is no relation between the two concepts. EXAMPLE 5 No Relation Between Absolute and Uniform Convergence The series in Example 2 converges absolutely but not uniformly, as we have shown. On the other hand, the series 2 m=1 (-if1-1 i + 3 (x real) converges uniformly on the whole real line but not absolutely. Proof. By the familiar Leibniz test of calculus (see App. A3.3) the remainder Rn does not exceed its first term in absolute value, since we have a series of alternating terms whose absolute values form a monotone decreasing sequence with limit zero. Hence given e > 0, for all x we have \Rn(x)\ 1 1 + n + 1 if n > N(e) g - . e This proves uniform convergence, since N(e) does not depend on x. The convergence is not absolute because for any fixed x we have (-If1" where k is a suitable constant, and kZl/m diverges. 1-8 UNIFORM CONVERGENCE Prove that the given series converges uniformly in the indicated region. i. 2 (z n-0 1ifr\ 2i\ s 0.999 2-2 3.2 (In + 1)! g 101 S 0.56 698 CHAP. 15 Power Series, Taylor Series 4-2 ~ "{n + 1) Z S 105 5.2 "i n-i DO 6.X 7-2 8.2 n cosh n\: tanh" z = 1 9-16 POWER SERIES Find the region of uniform convergence. (Give reason.) 9-2 (Z + 1 - 2i) 10. 2 n-0 (z - 0 (2n)! li. 2 2nn 13. 2 is. 2 -= 12. 2 (2z - 0" 14. 2 (3"tanhn).-2" 5/i 16. 2 n=0 (~l)TCZ2n (2n)! 17. CAS PROJECT. Graphs of Partial Sums, (a) Figure 365. Produce this exciting figure using your software and adding further curves, say. those of s2se, $1024* etc. (b) Power series. Study the nonuniformity of convergence experimentally by plotting partial sums near the endpoints of the convergence interval for real z = x. 18. TEAM PROJECT. Uniform Convergence. (a) Weierstrass /W-test. Give a proof. (b) Term wise differentiation. Derive Theorem 4 from Theorem 3. (c) Subregions. Prove that uniform convergence of a series in a region G implies uniform convergence in any portion of G. Is the converse true? (d) Example 2. Find the precise region of convergence of the series in Example 2 with x replaced by a complex variable z. (e) Figure 366. Show that x2 (1 + x2)~m = 1 if x + 0 and 0 if x = 0. Verify by computation that the partial sums st, s2, s3, look as shown in Fig. 366. 19-20 -1 0 Fig. 366. Sum s and partial sums in Team Project 18(e) HEAT EQUATION Show that (9) in Sec. 12.5 with coefficients (10) is a solution of the heat equation for t > 0, assuming that f(x) is continuous on the interval 0£i§L and has one-sided derivatives at all interior points of that interval. Proceed as follows. 19. Show that \Bn\ is bounded, say \Bn\ < K for all n. Conclude that kJ < Ke if *n > 0 and, by the Weierstrass test, the series (9) converges uniformly with respect to x and t for t 5 f0, 0 S x == L. Using Theorem 2, show that u(x, f) is continuous for t = t0 and thus satisfies the boundary conditions (2) for t S t0. 20. Show that \duJBt\ < An2 Ke^A"H" if t s tu and the series of the expressions on the right converges, by the ratio test. Conclude from this, the Weierstrass test, and Theorem 4 that the series (9) can be differentiated term by term with respect to t and the resulting series has the sum du/dt. Show thai (9) can be differentiated twice with respect to x and the resulting series has the sum d2u/dx2. Conclude from this and the result to Prob. 19 that (9) is a solution of the heat equation for all t =2 10. (The proof that (9) satisfies the given initial condition can be found in Rcf. [C10] listed in App. 1.) £H3a^HKET3EER£EVIEW QUESTIONS AND PROBLEMS 1. What are power series? Why are these series very important in complex analysis? 2. State from memory the ratio test, the root test, and (he Cauchy-Hadamard formula for the radius of convergence. 3. What is absolute convergence? Conditional convergence? Uniform convergence? What do you know about the convergence of power series? What is a Taylor series? What was the idea of obtaining it from Cauchy's integral formula? Give examples of practical methods for obtaining Taylor series. What have power series to do with analytic functions? Summary of Chapter 15 8. Can properties of functions be discovered from their Maclaurin series? If so, give examples. 9. Make a list of Maclaurin series of ez, cos z, sin z, cosh z, sinh z, Ln (1 — z) from memory. 10. What do you know about adding and multiplying power scries? 11-201 RADIUS OF CONVERGENCE Find the radius of convergence. Can you identify the sum as a familiar function in some of the problems? (Show the details of your work.) 11.2 (3z)" i3-2-1 In 12. 2 n=l oo 14.2 (-2)" 15. 2 (-1)" . (2/1)! Z (-If 16- 2 ,,, —, (z ~ 2)2" + 1 (2n + 1)! 17. 2 ^n(z 2ifn i9. 2 e 71= 1 18. 2 20. 2 (2z> -\2n (2n)! (z - 0" (3 + 4/)TC 21-301 TAYLOR AND MACLAURIN SERIES Find the Taylor or Maclaurin series with the given point as center and determine the radius of convergence. (Show details.) 21. e\ 77/ 22. Lnz, 2 23. 1/(1 - z), -1 24. 1/(4 - 3z), 1 + i 25. 1/(1 - z)3. 0 26. 1/z2, i 27. 1/z, 29. cos z, 577 28. 1 30. sin2 z, 0 31. Does every function f(z) have a Taylor series? 32. Docs there exist a Taylor series in powers of z — 1 — / that diverges at 5 + 5i but converges at 4 + 6/? 33. Do we obtain an analytic function if we replace x by z in the Maclaurin series of a real function f(x)l 34. Using Maclaurin series, show that if f(z) is even, its integral (with a suitable constant of integration) is odd. 35. Obtain the first few terms of the Maclaurin series of tan z by using the Cauchy product and sin z = cos z tan z. SUMMARY OF CHAPTER 1 5 Power Series, Taylor Series Sequences, series, and convergence tests are discussed in Sec. 15.1. A power series is of the form (Sec. 15.2) 3D (1) 2 an(z - z0)n = a0 + «,(■ - z0) + a2(z - z0f + ■ •■ ; z0 is its center. The series (1) converges for \z — z0| < R and diverges for \z — Zo| > R, where R is the radius of convergence. Some power series converge for all z (then we write R = °°). In exceptional cases a power series may converge only at the center; such a series is practically useless. Also, R = lim \an/an+1\ if this limit exists. The series (1) converges absolutely (Sec. 15.2) and uniformly (Sec. 15.5) in every closed disk \z — z0| = r < R (R > 0). It represents an analytic function f(z) for |z — z0| < R. The derivatives f'(z). f"(z), ■ • • are obtained by termwise differentiation of (1), and these series have the same radius of convergence R as (1). See Sec. 15.3. 700 CHAP. 15 Power Series, Taylor Series Conversely, every analytic function /(z) can be represented by power series. These Taylor series of f(z) are of the form (Sec. 15.4) (2) f(z) = 2 -T fn\z0)(z - z0)n (\z ~ Zo| < R), as in calculus. They converge for all z in the open disk with center z0 and radius generally equal to the distance from z0 to the nearest singularity of /(z) (point at which f(z) ceases to be analytic as defined in Sec. 15.4). If/(z) is entire (analytic for all z; see Sec. 13.5), then (2) converges for all z. The functions ez, cos z, sin z, etc. have Maclaurin series, that is, Taylor series with center 0, similar to those in calculus (Sec. 15.4). CHAPTER 1 6 Laurent Series. Residue Integration Laurent series generalize Taylor series. Indeed, whereas a Taylor series has positive integer powers (and a constant term) and converges in a disk, a Laurent series (Sec. 16.1) is a series of positive and negative integer powers of z — Zo and converges in an annulus (a circular ring) with center z0. Hence by a Laurent series we can represent a given function f(z) that is analytic in an annulus and may have singularities outside the ring as well as in the "hole" of the annulus. We know that for a given function the Taylor series with a given center 70 is unique. We shall see that, in contrast, a function f(z) can have several Laurent series with the same center z0 and valid in several concentric annuli. The most important of these series is that which converges for 0 < \z — z0\ < R, that is, everywhere near the center z0 except at Zo itself, where Zo is a singular point of f(z). The series (or finite sum) of the negative powers of this Laurent series is called the principal part of the singularity of f(z) at z0, and is used to classify this singularity (Sec. 16.2). The coefficient of the power l/(z — z0) of this series is called the residue of f(z) at z0. Residues are used in an elegant and powerful integration method, called residue integration, for complex contour integrals (Sec. 16.3) as well as for certain complicated real integrals (Sec. 16.4). Prerequisite: Chaps. 13, 14, Sec. 15.2. Sections that may be omitted in a shorter course: 16.2, 16.4. References and Answers to Problems: App. 1. Part D, App. 2. Laurent series generalize Taylor series. If in an application we want to develop a function f(z) in powers of z — Zo when f(z) is singular at Zo (as defined in Sec. 15.4), we cannot use a Taylor series. Instead we may use a new kind of series, called Laurent series,1 consisting of positive integer powers of z — Zo (and a constant) as well as negative integer powers of z — Zo', this is the new feature. Laurent series are also used for classifying singularities (Sec. 16.2) and in a powerful integration method ("residue integration", Sec. 16.3). A Laurent series of f(z) converges in an annulus (in the "hole" of which f(z) may have singularities), as follows. PIERRE ALPHONSE LAURENT (1813-1854), French military engineer and mathematician, published the theorem in 1843. 16.1 Laurent Series 701 702 CHAP. 16 Laurent Series. Residue Integration THEOREM 1 Laurent's Theorem Let f(z) be analytic in a domain containing two concentric circles Cx and C2 with center z0 and the annulus between them (blue in Fig. 367). Then f(z) can be represented by the Laurent series KZ) = E an(z - Zof + E (1) (Z - Zo)™ n=0 n=l v^ = a0 + ctiiz - z0) + a2(z - z0)2 + • + (z - z0) ,^2 consisting ofnonnegative and negative powers. The coefficients of this Laurent series are. given by the integrals (2) r 9 —-^rr dz:% bn = — 2tTl ■lc (Z* - Zn) 27TÍ (z* - zof-1 f(z*) dŕ taken counterclockwise around any simple closed path C that lies in the annulus and encircles the inner circle, as in Fig. 367. [The variable of integration is denoted by z* since z is used in (l).l This series converges and represents f(z) in the enlarged open annulus obtained from the given annulus by continuously increasing the outer circle Ca and decreasing C2 until each of the two circles reaches a point where f(z) is singular. In the important special case that z0 is the only singular point of /(z) inside C2. this circle can be shrunk to the point Zn, giving convergence in a disk except at the center. In this case the series (or finite sum) of the negative powers of (I) is called the principal part of the singularity of f(z) at z0- Fig. 367. Laurent's theorem COMMENT. Obviously, instead of (1), (2) we may write (denoting bn by a_n) ď) fiz) = 2 ajz-zof1 SEC. 16.1 Laurent Series where all the coefficients are now given by a single integral formula, namely. (2') an =^-.j> /(^n+1 dz* (n = 0, ±1, ±2, ■ ■ ■)• PROOF We prove Laurent's theorem, (a) The nonnegative powers are those of a Taylor scries. To see this, wc use Cauchy's integral formula (3) in Sec. 14.3 with z* (instead of z) as the variable of integration and z instead of z0. Let g(z) and h(z) denote the functions represented by the two terms in (3), Sec. 14.3. Then (3) /(z) = *(z) + *(z) = -^- (z* - z0)n-lf(z*) dz* ' ,M, i <£ (z* - Zo)n/(z*) [ + ~ Zo) Jc, J (z - Zo) -c + with the last term on the right given by I (7) K*(z) 2m(z - z0) n + l 7 Y< As before, we can integrate over C instead of C2 in the integrals on the right. We see that on the right, the power l/(z - ZoT is multiplied by b.n as given in (2). This establishes Laurent's theorem, provided (8) lim R't(z) = 0. (c) Convergence proof of (8). Very often (l) will have only finitely many negative powers. Then there is nothing to be proved. Otherwise, we begin by noting that f(z*)/(z — z*) in (7) is bounded in absolute value, say. f(z*) < M for all z* on C2 because /(z*) is analytic in the annulus and on C2, and z* lies on C2 and z outside, so that z - z* + 0. From this and the ML-inequality (Sec. 14.1) applied to (7) we get the inequality (L = length of C2, |z* — Zo\ ~ radius of C2 = const) SEC. 16.1 Laurent Series 705 From (6b) we see that the expression on the right approaches zero as n approaches infinity. This proves (8). The representation (1) with coefficients (2) is now established in the given annulus. (d) Convergence of (1) in the enlarged annulus. The first series in (1) is a Taylor series [representing g(z)]', hence it converges in the disk D with center z0 whose radius equals the distance of the singularity (or singularities) closest to z0- Also, g(z) must be singular at all points outside C% where f(z) is singular. The second series in (1), representing h(z), is a power series in Z = l/(z — z0). Let the given annulus be r2 < \z — Zq\ < rl5 where rx and r2 are the radii of Cx and C2, respectively (Fig. 367). This corresponds to l/r2 > |Z| > Urx. Hence this power series in Z must converge at least in the disk |Z| < l/r2. This corresponds to the exterior \z. — z0\ > r2 of C2, so that h(z) is analytic for all z outside C2. Also, h(z) must be singular inside C2 where f(z) is singular, and the series of the negative powers of (1) converges for all z in the exterior E of the circle with center z0 and radius equal to the maximum distance from z0 to the singularities of f(z) inside C2. The domain common to D and E is the enlarged open annulus characterized near the end of Laurent's theorem, whose proof is now complete. ■ Uniqueness. The Laurent series of a given analytic function f(z.) in its annulus of convergence is unique (see Team Project 24). However, f(z) may have different Laurent series in two annuli with the same center; see the examples below. The uniqueness is essential. As for a Taylor series, to obtain the coefficients of Laurent series, we do not generally use the integral formulas (2); instead, we use various other methods, some of which we shall illustrate in our examples. If a Laurent series has been found by any such process, the uniqueness guarantees that it must be the Laurent series of the given function in the given annulus. EXAMPLE 1 Use of Maclaurin Series Solution. By (14), Sec. 15.4, we obtain - =2 72^^=7- 6? + -lio - ^2 + ---' 'I- n=0 Here the "annulus" of convergence is the whole complex plane without the origin and the principal part of the scries at 0 is z 4 - -* 2 EXAMPLE 2 Substitution Find the Laurent series of z2e with center 0. Solution. From (12) in Sec. 15.4 with z replaced by 1/z we obtain a Laurent series whose principal part is an infinite series, zVfe = z* (] + Th + ~h + ''') = z"+ z + I + i + i + '' ■ (|z| > ox " EXAMPLE 3 Development of 1/(1 - z) Develop 1/(1 — z) (a) in nonnegative powers of z, (b) in negative powers of z. Solution. 1 x (a) —7 = 2 z" (valid if |z| < 1). 1 _1 ^,111 (b)--- = —-— = -2 -jZZT = - 7 - S---- (valid if |zj > 1). ■ 1 - z z(l - z ) n=0 z z z CHAP. 16 Laurent Series. Residue Integration EXAMPLE 4 Laurent Expansions in Different Concentric Annuli Find all Laurent series of l/(z3 - z4) with center 0. Solution. Multiplying by l/z3, we get from Example 3 1 (1) (II) °° 111 2 = "3 + "2 + 7 ^ + i + . V 1 _ 1 1 n+4 4 _5 re=0 4 EXAMPLE 5 Use of Partial Fractions Find all Taylor and Laurent series of f(z) Solution. In terms of partial fractions, - 3; + 2 with center 0. (0 < |z| < 1). (W>d. (a) and (b) in Example 3 take care of the first fraction. For the second fraction, (c) (d) 1 1 l - 2 do " 2 m+l (I) From (a) and (c), valid for |z| < 1 (see Fig. 368), m = 2 (i + (II) From (c) and (b), valid for 1 < |z| < 2. 2 + 4 2 4 DD 1 DC [ I, f(z) = 2 ?n+l J ~ 2 _n+1 "7 + n=0 n-0 z (III) From (d) and (b), valid for \z\ > 2, 1 2 1 — z + ■ ■ ■-- /w = -2 (2" + i) — n=0 z 2 3 JL 7 ~ 72 ~~ ~Jl (|z| < 2), (W > 2>- 3' III n _/ Fig. 368. Regions of convergence in Example 5 If f(z) in Laurent's theorem is analytic inside C2, the coefficients bn in (2) are zero by Cauchy's integral theorem, so that the Laurent series reduces to a Taylor series. Examples 3(a) and 5(1) illustrate this. 707 1-61 LAURENT SERIES NEAR A SINGULARITY ATO Expand the given function in a Laurent series that converges for 0 < \z\ < R and determine the precise region of convergence. (Show the details of your work.) 1. 3. I 5. z~öe 1 2. z cos — cosh 2z ez 6. —5-0 7-14 LAURENT SERIES NEAR A SINGULARITY AT z0 Expand the given function in a Laurent series that converges for 0 < \z — z0l " and determine the precise region of convergence. (Show details.) 9. z - 1 I 1 sin z /_ 1_\3 > «0 4' 10. Z0 = 7T 11. 12. (z + iY - (z + i) „3 (z + 0" " 14. z2 sinh - , z0 = 0 13. - 4 15-23 TAYLOR AND LAURENT SERIES Find all Taylor and Laurent series with center z = z0 and determine the precise regions of convergence. 15. , z0 = 0 16. y^-j • ;<> = 1 17. 19. 21. 23. z2 1 --« , z0 = 0 18. - , z0 1 — z z 2iz' (z - if 4z - 1 , z0 = i 20. z0 = 0 22. sinh: (z - \T .«0=1 ,2 • ^0 sin z Z + ä7T 24. TEAM PROJECT. Laurent Series, (a) Uniqueness. Prove that the Laurent expansion of a given analytic function in a given annulus is unique. (b) Accumulation of singularities. Does tan (1/z) have a Laurent series that converges in a region 0 < |z| < R? (Give a reason.) (c) Integrals. Expand the following functions in a Laurent scries that converges for |z| > 0: fz el - 1 1 fz sin i -dt, — - ■>n I z Jn r dt. 25. CAS PROJECT. Partial Fractions. Write a program for obtaining Laurent series by the use of partial fractions. Using the program, verify the calculations in Example 5 of the text. Apply the program to two other functions of your choice. 16.2 Singularities and Zeros. Infinity Roughly, a singular point of an analytic function f(z) is a z0 at which f(z) ceases to be analytic, and a zero is a z at which f(z) = 0. Precise definitions follow below. In this section we show that Laurent series can be used for classifying singularities and Taylor series for discussing zeros. Singularities were defined in Sec. 15.4, as we shall now recall and extend. We also remember that, by definition, a function is a single-valued relation, as was emphasized in Sec. 13.3. We say that a function f(z) is singular or has a singularity at a point z = z0 if /(z) is not analytic (perhaps not even defined) at z — Zo> but every neighborhood of z, = z0 contains points at which f(z.) is analytic. We also say that z — Zo is a singular point of /(z). We call z = z0 an isolated singularity of f(z.) if z = z0 has a neighborhood without further singularities of /(z). Example: tan z has isolated singularities at ±tt/2, ±3tt/2, etc.; tan (1/z) has a nonisolated singularity at 0. (Explain!) 708 CHAP. 16 Laurent Series. Residue Integration Isolated singularities of f(z) at z = Zo can be classified by the Laurent series (1) f(z) = 2 ««(* - z0r + 2 —(Sec- 16-D n=0 n = l ^ Z°' valid in the immediate neighborhood of the singular point z = Zo, except at z0 itself, that is, in a region of the form 0 < \z - z0\ < R- The sum of the first scries is analytic at z = "0> as we know from the last section. The second series, containing the negative powers, is called the principal part of (1), as we remember from the last section. If it has only finitely many terms, it is of the form b-i bm (2) +■■■+ - _m (bm * 0). Z <,o (Z Zo) Then the singularity of f(z) at z — z0 is called a pole, and in is called its order. Poles of the first order are also known as simple poles. If the principal part of (1) has infinitely many terms, we say that f(z) has at z = z0 an isolated essential singularity. Wc leave aside nonisolated singularities. EXAMPLE 1 Poles. Essential Singularities The function has a simple pole at z = 0 and a pole of fifth order at z = 2. Examples of functions having an isolated essential singularity at z = 0 are ^ = f-L = 1 + I + -L and 1 ^, (-1)" 1 J_ 1 sin Section 16.1 provides further examples. For instance, Example 1 shows that z~° sin z has a fourth-order pole at 0. Example 4 shows that l/(z3 - z4) has a third-order pole at 0 and a Laurent series with infinitely many negative powers. This is no contradiction, since this series is valid for [z > 1; it merely tells us that in classifying singularities it is quite important to consider the Laurent series valid in the immediate neighborhood of a singular point. In Example 4 this is the series (I), which has three negative powers. The classification of singularities into poles and essential singularities is not merely a formal matter, because the behavior of an analytic function in a neighborhood of an essential singularity is entirely different from that in the neighborhood of a pole. EXAMPLE 2 Behavior Near a Pole ,f(z) = l/z2 has a pole at z = 0, and |/(z) —» x as z —> 0 in any manner. This illustrates the following theorem. SEC. 16.2 Singularities and Zeros. Infinity THEOREM 1 Poles fff(z) is analytic and has a pole at. Zo, then \f(z)\ —> °° as z Zn in any manner. The proof is left to the student (see Prob. 12). EXAMPLE 3 Behavior Near an Essential Singularity The function /(z) = ellz has an essential singularity at z = 0. It has no limit for approach along the imaginary axis; it becomes infinite if <: —* 0 through positive real values, but it approaches zero if z —» 0 through negative real values. It takes on any given value c = c0e"" 0 in an arbitrarily small e-neighborhood of z = 0. To see the letter, we set z = re1", and then obtain the following complex equation for and 0, which we must solve: l/z _ (cos 0 — z sin ti)lr ia e — e. — c^e . Equating the absolute values and the arguments, we have t>(cos = c0, that is cos 0 = r In c0, and -sin 0 = ctr respectively. From these two equations and cos2 0 - sin2 6 = r2(ln c0)2 + a2r2 = 1 we obtain the formulas 2 1 a r = -%-7z and tan 0 =--. (In c0) - a In c0 Hence r can be made arbitrarily small by adding multiples of 2tt to a. leaving c unaltered. This illustrates the very famous Pkard's theorem (with z = 0 as the exceptional value). For the rather complicated proof, sec Ref. [D4]. vol. 2, p. 258. For Picard, see Sec. 1.7. ■ "HEOREM 2 Picard's Theorem If f(z) is analytic and has an isolated essential singularity at a point Zq, it takes on every value, with at most one exceptional value, in an arbitrarily small e-neighborhood ofz0. Removable Singularities. We say that a function /(z) has a removable singularits? at z = zo if /(z) is not analytic at z = Zo- but can be made analytic there by assigning a suitable value f(z0). Such singularities are of no interest since they can be removed as just indicated. Example: /(z) = (sin z)/z becomes analytic at z = 0 if we define ,f(0) = I. Zeros of Analytic Functions A zero of an analytic function f(z) in a domain D is a z = zn in D such that f(z0) = 0. A zero has order n if not only / but also the derivatives /", • • • , f(n~1) are all 0 at z = z0 but fn)(zo) # 0. A first-order zero is also called a simple zero. For a second-order zero, /(z0) = /'(zo) = 0 but /"(z0) # 0. And so on. E X A M P L E 4 Zeros The function 1 + z2 has simple zeros at ±i. The function (1 - ;4)2 has second-order zeros at ±1 and ±i. The function (z - a)3 has a third-order zero at z = a. The function ez has no zeros (see Sec. 13.5). The function sin z has simple zeros at 0, ± tt, ±2tt, ■ ■ ■ , and sin2 z has second-order zeros at these points. The function 1 - cos z has second-order zeros at 0, ±2-n\ ±4tt, • ■ ■ . and the function (1 - cos z)2 has fourth-order zeros at these points. 710 CHAP. 16 Laurent Series. Residue Integration Taylor Series at a Zero. At an nth-order zero z = zn of /(z), the derivatives /'(z0), • • •, .f("_1)(zo) are zero, by definition. Hence the first few coefficients a0, ■ ■ • , an_x of the Taylor series (1), Sec. 15.4, are zero, too, whereas an + 0, so that this series takes the form f(z) = an(z - ZoT + an+1(z - z0T+1 + • ■ ■ (3) = (z - Zo)n [On + an+t(.z - z0) + an+2(z - z0) + •« •] (a„ # 0). This is characteristic of such a zero, because if f(z) has such a Taylor series, it has an nth-order zero at z = z0, as follows by differentiation. Whereas nonisolated singularities may occur, for zeros we have THEOREM 3 Zeros The zeros of an analytic function f(z) 0) are isolated; that is, each of them has a neighborhood that contains no further zeros of /(z). PROOF The factor (z - z0)n in (3) is zero only at z = Zo- The power series in the brackets [• • •] represents an analytic function (by Theorem 5 in Sec. 15.3), call it g(z). Now g(Zrj) = an ^ 0, since an analytic function is continuous, and because of this continuity, also g(z) + 0 in some neighborhood of z = z0- Hence the same holds of /(z). This theorem is illustrated by the functions in Example 4. Poles are often caused by zeros in the denominator. (Example: tan z has poles where cos z is zero.) This is a major reason for the importance of zeros. The key to the connection is the following theorem, whose proof follows from (3) (see Team Project 24). THEOREM 4 Poles and Zeros Let f(z) he analytic at z — Zo and have a zero of nth order at z = Zo- Then \lf(z) has a pole of nth order at z = Zo, and so does h(z)lf(z), provided h(z) is analytic at z — Zo and h(zo) £ 0. Riemann Sphere. Point at Infinity Wrhen we want to study complex functions for large |z|, the complex plane will generally become rather inconvenient. Then it may be better to use a representation of complex numbers on the so-called Riemann sphere. This is a sphere S of diameter 1 touching the complex z-plane at z = 0 (Fig. 369), and we let the image of a point P (a number z in the plane) be the intersection P* of the segment PN with S, where TV is the "North Pole" diametrically opposite to the origin in the plane. Then to each z there corresponds a point on S. Conversely, each point on ,S' represents a complex number z, except for N, which does not correspond to any point in the complex plane. This suggests that we introduce an additional point, called the point at infinity and denoted °° ("infinity") and let its image be N. The complex plane together with ^ is called the extended complex plane. The complex plane is often called the finite complex plane, for distinction, or simply the 711 JV Fig. 369. Riemann sph complex plane as before. The sphere S is called the Riemann sphere. The mapping of the extended complex plane onto the sphere is known as a stereographic projection. (What is the image of the Northern Hemisphere? Of the Western Hemisphere? Of a straight line through the origin?) Analytic or Singular at Infinity If we want to investigate a function f(z) for large |z|, we may now set z = 1/w and investigate f(z) - i'(l/w) = g(w) in a neighborhood of to = 0. We define f(z) to be analytic or singular at infinity if g(w) is analytic or singular, respectively, at w = 0. We also define (4) g(0) = lim g(w) if this limit exists. Furthermore, we say that f(z) has an nth-order zero at infinity if /(1/vv) has such a zero at w = 0. Similarly for poles and essential singularities. EXAMPLE 5 Functions Analytic or Singular at Infinity. Entire and Meromorphic Functions The function f(z) = I/z2 is analytic at oo since g(w) = /(1/vv) = w2 is analytic at w = 0, and /(z) has a second-order zero at oo. The function f(z) = z3 is singular at oc and has a third-order pole there since the function g(w) = f(llw) = 1/vv3 has such a pole at w = 0. The function ez has an essential singularity at ao since eUm has such a singularity at w = 0. Similarly, cos z and sin z have an essential singularity at oc. Recall that an entire function is one that is analytic everywhere in the (finite) complex plane. Liouville's theorem (Sec. 14.4) tells us that the only bounded entire functions are the constants, hence any nonconstant entire function must be unbounded. Hence it has a singularity at oc, a pole if it is a polynomial or an essential singularity if it is not. The functions just considered arc typical in this respect. An analytic function whose only singularities in the finite plane arc poles is called a meromorphic function. Examples arc rational functions with nonconstant denominator, tan cot z, sec z, and esc z. In this section we used Laurent series for investigating singularities. In the next section we shall use these series for an elegant integration method. 1-10 SINGULARITIES Determine the location and kind of the singularities of the following functions in the finite plane and at infinity. In the case of poles also state the order. 1. tan2 77z 3. cotz2 5. cos z — sin : 2-z + 1, ~ 7 4. 73e1/ 1 at z = z0, then / (z) has a zero of order n - 1 at z0. (b) Poles and zeros. Prove Theorem 4. (c) Isolated fc-points. Show that the points at which a nonconslant analytic function f(z) has a given value k are isolated. (d) Identical functions. If fL(z) are analytic in a domain D and equal at a sequence of points zn in D that converges in D, show that /x(z) = f2(z) in D. 25. (Ricmann sphere) Assuming that we let the image of the x-axis be meridians 0° and 180°, describe and sketch (or graph) the images of the following regions on the Riemann sphere: (a) |z| > 100, (b) the lower half-plane, (c) § S |z| S 2. 16.3 Residue Integration Method The purpose of Cauchy's residue integration method is the evaluation of integrals $ m dz Jc taken around a simple close path C. The idea is as follows. If f(z) is analytic everywhere on C and inside C, such an integral is zero by Cauchy's integral theorem (Sec. 14.2), and we are done. If f(z) has a singularity at a point z = z0 inside C, but is otherwise analytic on C and inside C, then f(z) has a Laurent series m = 2 aJz - zof + + : - -2 + • ' • n=0 Z Zo (z z0) that converges for all points near z = z0 (except at z = Zo itself), in some domain of the form 0 < \z — Zol < R (sometimes called a deleted neighborhood, an old-fashioned term that we shall not use). Now comes the key idea. The coefficient bL of the first negative power l/(z — Zo) of this Laurent series is given by the integral formula (2) in Sec. 16.1 with n = 1, namely, Now, since we can obtain Laurent series by various methods, without using the integral formulas for the coefficients (see the examples in Sec. 16.1), we can find /?1 by one of those methods and then use the formula for bx for evaluating the integral, that is, (1) !> f(z) dz = 2m&!. c SEC. 16.3 Residue Integration Method 713 Here we integrate conuntcrclockwise around a simple closed path c that contains z = z0 in its interior (but no other singular points of f(z) on or inside ci). The coefficient bx is called the residue of f(z) at z = z0 and we denote it by (2) bx = Res f(z). EXAMPLE 1 Evaluation of an Integral by Means of a Residue Integrate the function /(z) = z-4 sin z counterclockwise around the unit circle C. Solution. From (14) in Sec. 15.4 we obtain the Laurent series ,3 sin z 1 z 3!z + si ~~ 7! m- 4 - 3 which converges for |z| > 0 (that is, for all z + 0). This series shows that /(;) has a pole of third order at z — 0 and the residue bx = —1/3!. From (1) we thus obtain the answer sin z 3 4 dz = 2irib-[ = — EXAMPLE 2 CAUTION! Use the Right Laurent Series! Integrate /(z) = l/(z3 - z4) clockwise around the circle C: \z\ = 1/2. Solution, z3 - z4 = z3(l - z) shows that J(z) is singular at z = 0 and z = 1. Now z = 1 lies outside C. Hence it is of no interest here. So we need the residue of f(z) at 0. We find it from the Laurent series that converges for 0 < \z\ < 1. This is series (I) in Example 4. Sec. 16.1, 1 111 _ ^ = -j + + ~ + 1 + z + ' • ■ (0 < |z| < 1). We see from it that this residue is 1. Clockwise integration thus yields dz ■g-j 2tt( Res/(z) = —2m. C Z - Z z=0 CAUTION.' Had we used the wrong series (II) in Example 4, Sec. 16. 1 111 .4 -b _6 :kl > i). we would have obtained the wrong answer. 0, because this series has no power 1/z. Formulas for Residues To calculate a residue at a pole, we need not produce a whole Laurent scries, but, more economically, we can derive formulas for residues once and for all. Simple Poles. Two formulas for the residue of f(z) at a simple pole at z0 are (3) Res f(z) = b, = lim (z - z0)/(z) and, assuming that f(z) = p(z)/q(z), p(Zo) # 0, and q(z) has a simple zero at Zo (so that f(z) has at z0 a simple pole, by Theorem 4 in Sec. 16.2), p(z) P(zo) (4) Resf(z) = Res^ = -7— . z-z„ z=z0 q{z) q (Zo) 714 CHAP. 16 Laurent Series. Residue Integration PROOF For a simple pole at z = z0 the Laurent series (1), Sec. 16.1, is f(z) a0 + ciiiz - Z0) + a2(z - Zo) (0 < \z - zol < r). Here ^ 0. (Why?) Multiplying both sides by z - Z0 ar>d then letting z —» z0, we obtain the formula (3): lim (z - z0)/(z) = fti + lim (z - z„)[ao + - Zd) + • ■ ■] = fei where the last equality follows from continuity (Theorem 1, Sec. 15.3). We prove (4). The Taylor series of q(z) at a simple zero z0 is (z - z f q(Z) = (Z - Zo)^'(Zo) + q"(Zo) + ■■■■ Substituting this into / = p/q and then f into (3) gives Res/(z) = hm (z - z0) — = lim -—---: • (/') = 9< + / and 5 (z) = 3z + 1 we confirm the result, Res 9z + i 9z + 1 10/ -5i. *=i z(zz I 1) Poles of Any Order. The residue of f(z) at an mth-order pole at z0 is (5) 1 I d"1'1 Res /(-) =--— lim (m — I)! «-»zu [ d; (z - z0)"7(z) In particular, for a second-order pole (m = 2), (5*) Res/(z) = lim {[(z - z0)2/(z)]'] Z = Z0 Z->2„ PROOF The Laurent series of ,f(z) converging near z0 (except at z0 itself) is (Sec. 16.2) ,f(z) + «o + öite - zo) + • • (z Zo) (z Zo) Z Zo where &m + 0. The residue wanted is bv Multiplying both sides by (z — z0)m gives (Z - Z0)m/(Z) = &m + &m_,(z " Z0) + • ' • + fei(z " Zo)"'"1 +