MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 7 SOLUTIONS 1. Let β 2 C. (a) Show that for all n = 0; 1; 2;:::; βn 2 1 Z βneβz = n+1 dz: n! 2πi @D(0;1) n!z Solution. Applying generalized Cauchy's integral formula, we get Z n βz n " Z βz # n n n 2 1 β e β 1 e β 1 d βz β n+1 dz = n+1 dz = × n e = : 2πi @D(0;1) n!z n! 2πi @D(0;1) z n! n! dz z=0 n! (b) Show that 1 2 X βn 1 Z 2π = e2β cos θ dθ: n! 2π n=0 0 [Hint: Consider power series expansion of eβ=z and apply (a) on z−1eβ(z+1=z).] Solution. Note that 1 X βn eβ=z = : n!zn n=0 Multiplying by eβz and dividing by z, we get 1 1 X βneβz eβ(z+1=z) = : z n!zn+1 n=0 Integrating about @D(0; 1) and using (a), we get 1 " # 1 2 1 Z 1 X βn 1 Z eβz X βn eβ(z+1=z) dz = dz = 2πi z n! 2πi zn+1 n! @D(0;1) n=0 @D(0;1) n=0 iθ Evaluating the line integral about the path z : [0; 2π] ! C, z(θ) = e and noting that eiθ + e−iθ = 2 cos θ we get 1 Z 1 1 Z 2π eβ(z+1=z) dz = e2β cos θ dθ: 2πi @D(0;1) z 2πi 0 2. Let f : C ! C be an entire function. Let a 2 R be an arbitrary constant. (a) Show that if Re f(z) ≤ a for all z 2 C, then f is constant. (b) Show that if Re f(z) ≥ a for all z 2 C, then f is constant. 2 2 (c) Show that if [Re f(z)] ≤ [Im f(z)] for all z 2 C, then f is constant. 2 2 (d) Show that if [Re f(z)] ≥ [Im f(z)] for all z 2 C, then f is constant. (e) Suppose h is another entire functions and suppose there exists an a 2 R, a > 0, such that Re f(z) ≤ a Re h(z) for all z 2 C. Show that there exist α; β 2 C such that f(z) = αh(z) + β for all z 2 C. Date: November 14, 2009 (Version 1.0). 1 [Hint: if f and g are both entire, then so are f ◦ g and g ◦ f; find an appropriate g so that you may apply Liouville's theorem.] x Solution. Note that e is a monotone increasing function on R. • For (a), we choose g(z) = ez and note that jef(z)j = jeRe f(z)ei Im f(z)j = eRe f(z) ≤ ea. • For (c), we choose g(z) = ez2 and note that jef(z)2 j = je[Re f(z)]2−[Im f(z)]2 e2i Re f(z) Im f(z)j = e[Re f(z)]2−[Im f(z)]2 ≤ e0 = 1. Applying Liouville's theorem then implies that ef(z); ef(z)2 are constant functions. To show that f must also be a constant function, we differentiate ef(z) to get 0 = (ef(z))0 = f 0(z)ef(z): Since ef(z) is never zero, we get f 0(z) = 0 and so f must be a constant function in (a). The same argument shows for (c) that f 2 must be a constant function and therefore f must be a constant function (since it is continuous). (b) could be deduced from (a) and (d) could be deduced from from (c) by applying (a) and (c) to −f. For (e), we just apply (a) to the entire function f − ah, which by assumption satisfies Re f(z) − a Re h(z) ≤ 0 for all z 2 C. 3. Let f : C ! C be an entire function. × (a) Suppose there exists α; β 2 C such that α/β2 = R. Show that if f satisfies the following conditions f(z + α) = f(z); f(z + β) = f(z) for all z 2 C, then f is constant. Solution. Given any real number x 2 R, we will write [x] for the integral part of x and hxi for the fractional part of x. For example [−5:12] = −5 and h−5:12i = 0:12. Note that [x] 2 Z, hxi 2 [0; 1), and x = [x] + hxi for all x 2 R. The condition α/β2 = R implies that α; β span C as a real vector space of dimension 2. In other words, any z 2 C may be written as z = xα + yβ where x; y 2 R. Observe that the two conditions given may be applied recursively to obtain f(z) = f(xα + yβ) = f(αhxi + βhyi + α[x] + β[y]) = f(αhxi + βhyi): for any z = xα+yβ 2 C. Note that for any z = xα+yβ 2 C, αhxi+βhyi 2 [0; α)×[0; β) ⊆ [0; α] × [0; β], ie. the closed parallelogram bounded by the line segments from 0 to α and 0 to β and this is compact, and so supjf(z)j = sup jf(z)j ≤ max jf(z)j: z2C z2[0,α)×[0,β) z2[0,α]×[0,β] The last term is finite by the Extreme Value Theorem in Math 104 (since [0; α] × [0; β] is compact and f is analytic, therefore continuous) and so f is bounded. Liouville's theorem then implies that f is constant. (b) Suppose f(z) lim = 0: jzj!1 z Show that f is a constant function. Solution. Define g : C ! C by 8f(z) − f(0) < z 6= 0; g(z) := z − 0 :f 0(0) z = 0: 2 By Corollary 4.4, g is an entire function too. Now, f(z) lim jg(z)j = lim = 0: jzj!1 jzj!1 z Let " > 0. Then there exists R > 0 such that jg(z)j < " for all jzj ≥ R. In particular, jg(z)j < " for all z 2 @D(0;R). Applying maximum modulus theorem (or Corollary 4.15), we get max jg(z)j = max jg(z)j ≤ ": z2D(0;R) z2@D(0;R) Therefore jg(z)j ≤ " for all z 2 C. Since " is arbitrary, we conclude that g(z) = 0 for all z 2 C. Hence f(z) = f(0) for all z 2 C and so f is a constant function. (a) Find all entire functions f that satisfy 1 1 f 00 + f = 0 n n for all n 2 N. Solution. Note that if f is an entire function, then so is f 00. In particular, f 00 + f is continuous and so 1 1 f 00(0) + f(0) = lim f 00 + f = 0: n!1 n n 00 −1 Hence f + f is zero on a subset of C with limit points, namely, fn j n 2 Ng [ f0g and 00 thus by the uniqueness theorem, f + f ≡ 0 on the whole of C. Now since f is entire, its Taylor series expansion about 0 that converges everywhere in C, and is given by 1 X f (n)(0) f(z) = zn: n! n=0 Likewise for f 00, we have 1 X f (n+2)(0) f 00(z) = zn: n! n=0 Hence, for all z 2 C, 1 " # X f (n+2)(0) f (n)(0) f 00(z) + f(z) = + zn: n! n! n=0 00 (n+2) (n) Now since f + f ≡ 0, we must have f (0) = −f (0) for all n 2 N [ f0g, ie. f(0) = −f 00(0) = ··· = (−1)nf (2n)(0) = ··· f 0(0) = −f 000(0) = ··· = (−1)nf (2n+1)(0) = ··· : Hence 1 X f (n)(0) f(z) = zn n! n=0 1 1 X (−1)n X (−1)n = f(0) z2n + f 0(0) z2n+1 (2n)! (2n + 1)! n=0 n=0 = f(0) cos z + f 0(0) sin z: 3 Note that the `splitting' of the first power series into a sum of two power series is permissible because all three series have infinite radius of convergence. Hence an entire function that satisfies the given condition must be of the form f(z) = f(0) cos z + f 0(0) sin z: (b) Let n 2 N and n ≥ 2. Find all entire functions f that satisfy f(zn) = [f(z)]n for all z 2 C. Solution. By Theorem 4.3, f has a power series expansion 1 X m f(z) = amz m=0 with infinite radius of convergence. By the given condition n n a0 = f(0) = f (0) = a0 2pπi and so either a0 = 0 or a0 = e n−1 for some p 2 f1; : : : ; n − 1g. 2pπi Case I. Suppose a0 = e n−1 for some p 2 f1; : : : ; n − 1g and f is non-constant. Let k 2 N be the smallest positive number such ak 6= 0. Hence n kn f(z ) = 1 + akz + higher order terms and n k [f(z)] = 1 + nakz + higher order terms. n n Since f(z ) = [f(z)] , comparing coefficients tells us that nak = 0 and so ak = 0 | a 2pπi contradiction. In other words, if a0 = e n−1 for some p 2 f1; : : : ; n − 1g, then f must be a 2pπi constant function. Hence f(z) = a0 = e n−1 for all z 2 C. Case II. Suppose a0 = 0 and f is non-constant.