7 Taylor and Laurent Series Throughout this text we have compared and contrasted properties of complex func­ tions with functions whose domain and range lie entirely within the reals. There are many similarities, such as the standard differentiation formulas. On the other hand, there are some surprises, and in this chapter we will encounter one of the hallmarks distinguishing complex functions from their real counterparts. It is possible for a function defined on the real numbers to be differentiable everywhere and yet not be expressible as a power series (see Exercise 27 at the end of Section 7.2). In the complex case, however, things are much simpler! It turns out that if a complex function is analytic in the disk Dr(a), its Taylor series about a will converge to the function at every point in this disk. Thus, analytic functions are locally nothing more than glorified polynomials. We shall also see that complex functions are the key to unlocking many of the mysteries encountered when power series are first introduced in a calculus course. We begin by discussing an important property associated with power series— uniform convergence. 7.1 Uniform Convergence Recall that if we have a function/(z) defined on a set T, the sequence of functions {Sn(z}} converges to the function/at the point z = Zo e T provided lim Sn(zo) = n—>« /(zo). Thus, for the particular point zo> this means that for each e > 0, there exists a positive integer A^ro (which depends on both £ and zo) such that (1) if n > N^ then | S„(ZQ) - /fa>) \ < e. c If Sn(z) is the nth partial sum of the series 2^ k(z — ct)\ statement (1) becomes k=0 k (2) if n > N^, then 2 ck(zo - a) - f(z0) < e. It is important to stress that for a given value of £, the integer NEnZQ we need to satisfy statement (1) will often depend on our choice of zo- This is not the case if 208 7.1 Uniform Convergence 209 the sequence {Sn(z)} converges uniformly. For a uniformly convergent sequence, it is possible to find an integer JVe (which depends only on £) that guarantees statement (1) no matter which value for z0 e T we pick. In other words, if n is large enough, the function S„(z) is uniformly close to f(z). Formally, we have the following definition. DEFINITION 7.1 The sequence {Sn(z)} converges uniformly to f(z) on the set T if for every £ > 0, there exists a positive integer N£ (which depends only on £) such that (3) ifn > N€, then \ S„(z) - f(z) | < £ for all z e T. c we sa tnat If in the preceding, Sn(z) is the nth partial sum of a series 2 *(z "" °0*» Y A = 0 the series 2 c*(£ ~~ °0A converges uniformly to/(z) on the set T. £=0 z EXAMPLE 7.1 The sequence {Sn(z)} = \e + -> converges uniformly to f(z) = ez on the entire complex plane because for any £ > 0, statement (3) is satisfied for all z if Nt is any integer greater than 1/e. We leave the details for showing this as an exercise. A good example of a sequence of functions that does not converge uniformly is the sequence of partial sums comprising the geometric series. Recall that the geometric series has S„(z) = 2J zk = — converging to f(z) = for all *=o 1 — z 1 — z z e D\(0). Since the real numbers are a subset of the complex numbers, we can show statement (3) is not satisfied by demonstrating it does not hold when we restrict our attention to the real numbers. In that context, D\(0) becomes the open interval (-1, 1), and the inequality \Sn(z) — f(z)\ < £ becomes \Sn(x) - f(x) | < £, which for real variables is equivalent tof(x) — £ < S„(x) <f(x) + £. If statement (3) were to be satisfied, then given £ > 0, Sn(x) should (for large enough values of n) be within an £-bandwidth of the function f(x) for all x in the interval (— 1, 1). Figure 7.1 illustrates that there is an £ such that no matter how large n is, we can find XQ e (— 1, 1) such that Sn(xo) is outside this bandwidth. In other words, this figure illustrates the negation of statement (3), which in technical terms is (4) there exists £ > 0 such that for all positive integers N, there is some n > N and some zo e T such that | Sn(zo) — f(zo) | ^ £. We leave the verification of statement (4) when applied to the partial sums of a geometric series as an exercise. 210 Chapter 7 Taylor and Laurent Series y = Six) 0.5 x l y=f(x)-E FIGURE 7.1 The geometric series does not converge uniformly on (-1, 1). There is a useful procedure known as the Weierstrass M-test, which can help determine whether an infinite series is uniformly convergent. Theorem 7.1 (Weierstrass Af-Xest): Suppose the infinite series 2 uk(z) k = 0 has the property that for each k, | uk(z) | ^ Mkfor all zeT. If2jk = 0 Mk converges, u then /:=2 0 k(z) converges uniformly on T. Proof Let S„(z) = 2 uk(z) be the nth partial sum of the series. If n > m, k = 0 n-\ M \S„(z) - Sm(z)\ = \um(z) + w,„, \(z) + ••• + MW_I(Z)| < 2 k- Since the series 2 Mk converges, the last expression can be made as small as we k = 0 wish by choosing m large enough. Thus, given £ > 0, there is a positive integer N£ such that if n, m > N£i then | Sn(z) — Sm(z) \ < £. But this means that for all z e T, {Sn(z)} is a Cauchy sequence. According to Theorem 4.2, this sequence must con­ verge to a number which we might as well designate by f(z). That is, f(z) = lim S„(z) = 2 uk(z). 7.1 Uniform Convergence 211 This gives us a function to which the series 2 uk(z) converges; it remains to be shown that the convergence is uniform. Let e > 0 be given. Again, since 2 Mk converges, there exists NE such that if n > NZ9 then ^ Mk< z. Thus, if n > 7Ve, we have for all z e T that M \f(z)-Sn(z)\ 2 «A(Z) - 2 *(Z) *=0 * = 0 2 uk(z) k=n < e, which completes the argument. As an application of the Weierstrass M-test, we have the following. Theorem 7.2 Suppose the power series 2J Ck(z — a)* has radius of con- vergence p > 0. Then for each r, 0 < r < p, the series converges uniformly on the closed disk Dr(a) = {z: | z — a | < r}. Proof Given r with 0 < r < p, choose z0 c £>P(o0 such that j zo — oc | = r. c Since ^ *(z " oO* converges absolutely for z e Dp(a) (Theorems 4.9 and 4.11, k r part ii), we know that ^] I ck(zo - a) \ = 2 1^1 * converges. For all z e Dr(a) it is clear that k k \ck(z- a) \ = \ck\ \z- a|*< \ck\r . k The conclusion now follows from the Weierstrass M-test with Mk = \ck\ r . Corollary 7.1 For each r, 0 < r < 1, the geometric series converges uni­ formly on the closed disk Dr(0). The following theorem gives important properties of uniformly convergent sequences. Theorem 7.3 Suppose {Sk} is a sequence of continuous functions defined on a set T containing the contour C. If {Sk} converges uniformly to f on the set T, then (i) f is continuous on T, and 212 Chapter 7 Taylor and Laurent Series (ii) lim Sk(z) dz = lim Sk(z) dz = f(z) dz. £_»«, JC JC £_>«, JC Proof (i) Given zo e T, we must prove lim/(z) = f(zo). Let e > 0 be given. Z-*ZQ Since Sk converges uniformly on T, there exists a positive integer N£ such that i i £ for all z e T, \f{z) — Sk(z) \ < - whenever k > N£. Since SNg is continuous at zo, £ there exists 8 > 0 such that if | z — Zo \ < 8, then | SN((z) - SM£(ZQ) \ < ~z • Hence, if | z — Zo I < 8, we have | f(z) - /(zo) | = | f(z) - SNt(z) + SNt(z) - SNg(zo) + SNe(zo) - /(¾) | * \f(z) ~ SN((z)\ + \SNe(z) - SNe(z0)\ + \SN£(z0) -/(¾) | e e e <- + - + - = £. 3 3 3 (ii) Let e > 0 be given, and let L be the length of the contour C. Since {Sk} converges uniformly to/on T, there exists a positive integer Ne such that if k > JVe, then | S*(z) - /(z) I < - for all z e T. Since C C T, max I Sk(z) - f(z)\ < 7 , so if N£, jc Sk(z) dz - \c f(z) dz\ = \\c [Sk(z) ~ f(z)] dz < max I Sk(z) ~ f(z) I L (by Lemma 6.2) ZeC Corollary 7.2 7/" the series 2 c„(z ~~ oc)w converges uniformly to f(z) on n = 0 the set T, and C is a contour contained in T, then n = 0 JC JC „ = Q JC ^ 1 EXAMPLE 7.2 Show that -Log(l - z) = X " z" for a11 * G D' (°)- Solution Given z0 e Di(0), choose r such that 0 < | zo | < r < 1, thus rt ensuring that zo e £>r(0). By Corollary 7.1, the geometric series 2 z converges n=0 7.1 Uniform Convergence 213 uniformly to on Dr(0). If C is any contour contained in Dr(0), Corollary 7.2 gives dz = 2 \ zn dz.