Math 22A - Fall 216 - Final Exam Solutions Problem 1. Let f be an entire function and let n 2. Show that there exists an entire function g with g n = f if and only if the orders of all zeroes of f are divisible by n. Answer: If f = g n, then for any zero a of g we have ord(f, a) = ord(g n, a) = n ord(g, a) is divisible by n. Conversely, assume ord (f, a) is divisible by n for any zero a of f. By Weierstraβ, we can construct an entire function h with zeros exactly at the zeros of f and of order 1 n ord(f, a). Clearly f and h n have exactly the same zeros with the same multiplicity, hence H = f/h n is entire and zero free. In particular, by Homework 2, Problem 4 (see solution there), we can write H = G n for some entire function G. Then f = h n H = (hg) n = g n for g = hg.
Problem 2. Calculate the integral x, for n 2. + 1 Answer: We consider γ R the curve consisting of the segment [ R, R] and the half disc S R. We assume R > 1. By the residue theorem, we have z + 1 = 2πi γ R The possible values for the poles are a = ζ k = exp a +1= ( ) 1 Res z + 1, a. ( (2k + 1) πi ), k n 1. The last inequality k n 1 comes from the fact that the poles must be contained in the upper half plane. Using the method of computing residues in class, we have ( ) 1 Res z + 1, ζ 1 k = (z + 1) 1 z=ζ k = ζ 1 k using that ζ k As a result, We show = 1. Thus γ R n 1 z + 1 = 2πi = 2πi exp ( πi = 2πi n R R = ζ k, ( ζ k exp πi ) n 1 ( ) kπi = 2πi exp n k= k= ) 1 exp (πi) 1 exp ( ) = 2πi πi n exp ( ) πi n 1 exp ( ) πi n 1 exp ( πi ) ( exp πi ) = π n 1 ). x + 1 + S R lim R S R z + 1 = π n z + 1 =. 1 ). Indeed, z + 1 R 1, and by the basic estimate z + 1 2πR 1 R 1. In consequence, R lim R R x + 1 = π n 1 S R ) = 2 x + 1 = π n 1 ) = x + 1 = π 1 ).
Problem 3. Consider f(z) = z n + a 1 z n 1 +... + a n. Show that there exists c with c = 1 such that f(c) 1. Answer: By Cauchy s estimates, we have f (n) () n! r n M(r) for M(r) = sup z =r f(z). Making r = 1 and noting f (n) () = n! we obtain Alternate Answer: Let 1 M(1) = sup f(z) = f(c) 1 for some c 1. z =1 g(z) = z n f(1/z) = a n z n +... + a 1 z + 1. Note that g() = 1. By the maximum modulus principle, there exists z = 1, with g(z ) 1. We have Setting c = 1/z, we obtain and c = 1. g(z ) = z n f(1/z ) = f(1/z ) 1. f(c) 1 Alternate Answer: Assume that for all z = 1 we have f(z) < 1. Let G(z) = z n, F (z) = f(z) + g(z). Then for z = 1 we have F (z) G(z) = f(z) < 1 = z n = G(z). By Rouché s theorem, F and G have the same number of zeros inside the unit disc counted with multiplicity. Clearly, G has exactly one zero with multiplicity n. However, F (z) = f(z) + g(z) = a 1 z n 1 +... + a n is either identically equal to or it has at most n 1 zeros. The latter case contradicts Rouché. The former case shows f(z) = g(z) and the assertion of the problem is trivial since f(c) = g(c) = c n = 1 for c = 1.
Problem 4. Suppose f : { < z < 1} C is holomorphic and satisfies Show that f =. note f(z) log z. Answer: We first claim the origin is a removable singularity for f. Indeed, let g(z) = zf(z) and g(z) z log z = lim g(z) = since lim z log z =. z z Thus is a removable singularity for g. Extending g holomorphically across the origin, we have by continuity g() =. Since f(z) = g(z)/z and g() =, it follows that f is holomorphic at. Fix a (, 1) \ {}. Let r be so that a < r < 1. By Cauchy s formula f(a) = 1 f(z) 2π z a 1 log r 2π r a 2πr. z =r Here, we used the basic estimate, the fact that f(z) log z and z a r a. Making r 1 in the above estimate, we conclude conclude f(a) = for all a. Thus f =.
Problem 5. Let f(z) = P (z) Q(z) be a rational function with deg P deg Q 2 such that Q has no zeros along the non-negative real axis. Show that f(x) = Res z=a (f(z) log z) a C\R where for z C \ R we set log z = log r + iθ and θ (, 2π). Answer: We consider the keyhole contour consisting of a portion of a circle C r of radius r, a portion of a circle C R of radius R > r, and two line segments L + and L at distance δ and δ away from the positive x-axis, going from [r, R]. We let By the residue theorem γ γ = L + C R ( L ) ( S r ). f(z) log z = 2πi a Res z=a (f(z) log z). This gives R(z) log z R(z) log z + R(z) log z R(z) log z = 2πi C R C r L + L a Res z=a (R(z) log z). We first make δ, and then we make r, R. We claim the limits of the integrals over C R and C r equal. We first consider C R. Indeed, since Hence deg P deg Q 2 = f(z) M z 2, lim z z2 f(z) <. for z sufficiently large. If z = R, then log z = log R+iθ log R +2π. By the basic estimate, we have f(z) log z 2πR M (log R + 2π). C R R2 The estimates for C r are similar, but using that f(z) is bounded near the origin by continuity. Then, f(z) log z 2πr M ( log r + 2π) asr. C r Here, we used r log r which can be seen using the change of variables r = 1/y with y. By continuity arguments (f is continuous, and log is continuous provided we stay in compact regions either above or below the non-negative real axis), we have R lim f(z) log z = f(x) log x δ L + r
R R R lim f(z) log z = f(x)(log x + 2πi) = f(x) log x + 2πi f(x). δ L r r r Substituting in the residue theorem, after taking limits, first for δ, and then r, R, we obtain This completes the proof. 2πi f(x) = 2πi a C\R Res z=a (f(z) log z).
Problem 6. Let f : C C be entire, and assume that (Re f) 216 + (Im f) 216 = 1. Show that f is constant. Answer: We present first a solution that works for functions f : U C and can be generalized to polynomial identities P (Re f, Im f) =, for any polynomial P. If f is not constant, the image of f must be an open set in C so it must contain a disc. By assumption, the image is contained in the set u 216 + v 216 = 1. However, for each fixed value of u, there are at most 216 possible values of v, so (u, v) for finitely many v s. The contradiction comes because if (u, v), then (u, v ) for all v close to v. This shows f must be constant. Alternate Answer: For the problem at hand, a simpler second solution is as follows. We note (Re f) 216 + (Im f) 216 = 1 = Re f 1, Im f 1 = f 2. By Liouville s theorem, f is bounded, hence constant.
Problem 7. Let f : U \ {a} C be a holomorphic function with an isolated singularity at a. Let P be a non-constant polynomial. Show that f has a removable singularity, or a pole, or an essential singularity at a respectively if and only if P f has a removable singularity, or a pole, or an essential singularity at a. Answer: Write P (w) = a w n +... + a n. (i) Assume first that a is a removable singularity for f. Then f is bounded in a neighborhood of a, say for z (a, ɛ) \ {a}. In this case f(z) M P (f(z)) = a f(z) n +... + a n a M n +... + a n is also bounded so the singularity of P f is also removable at a. (ii) Assume a is a pole for f. We show a is a pole for P f by proving lim P (f(z)) =. z a To this end, let R >. We seek to show P (f(z)) > R for all z (a, ɛ) \ {a} for a suitable ɛ. Note that for w real, we have lim a w n a 1 w n 1... a n =, w we can find r > such that if w > r, we have a w n a 1 w n 1... a n > R. Since a is a pole, we have lim z a f(z) =, so we can find ɛ > such that Thus f(z) > r for all z (a, ɛ) \ {a}. P (f(z)) = a f(z) n +... + a n a f(z) n... a n > R where we used the triangle inequality, and the fact that w = f(z) > r. This shows that a is a pole for P f. (iii) Assume a is an essential singularity for f. Fix α and β two complex numbers such that We claim that we can find a sequence P (α) P (β). x n a such that f(x n ) α.
Indeed, by Casoratti-Weierstrass, for n sufficiently large so that ( a, n) 1 U, we have f ( ( a, n) 1 ) \ {a} is dense, hence we can find xn ( a, n) 1 \ {a} such that f(x n ) α < 1 n. In a similar fashion we can find y n a such that f(y n ) β. Thus P (f(x n )) P (α), P (f(y n )) P (β). If a were removable for P f, then P (α) = P (β) by continuity. This is however a contradiction. If a were a pole for P f, then the two limits would have to be infinite. Thus the only option is that a is an essential singularity. The reverse implication follows by putting (i)-(iii) together.