MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 9 SOLUTIONS. and g b (z) = eπz/2 1

Transcription

1 MATH 85: COMPLEX ANALYSIS FALL 2009/0 PROBLEM SET 9 SOLUTIONS. Consider the functions defined y g a (z) = eiπz/2 e iπz/2 + Show that g a maps the set to D(0, ) while g maps the set and g (z) = eπz/2 e πz/2 +. Ω a := {z C < Re z < } Ω := {z C < Im z < } to D(0, ). Hence or otherwise, prove the following. Solution. Note that e πz/2 g (z) < e πz/2 + < e πz/2 2 < e πz/2 + 2 [Re(e πz/2 ) ] 2 + [Im(e πz/2 )] 2 < [Re(e πz/2 ) + ] 2 + [Im(e πz/2 )] 2 Re(e πz/2 ) 2 Re(e πz/2 ) + < Re(e πz/2 ) + 2 Re(e πz/2 ) + Re(e πz/2 ) > 0. But Re(e πz/2 ) = Re(exp( π 2 Re z + i π 2 Im z)) = exp( π 2 Re z) cos( π 2 Im z) > 0 iff cos( π 2 Im z) > 0 in particular this holds when < Im z <, ie. when z Ω. To deduce the analogous result for g a and Ω a, just note that g a (z) = g (iz) and Ω a = iω. (a) Let f : D(0, ) C e an analytic function that satisfies f(0) = 0. Suppose Re f(z) < for all z D(0, ). By considering the function g a f or otherwise, prove that f (0) 4 π. Solution. Note that g a maps Ω a onto D(0, ), g a is analytic on Ω a, and g a (0) = 0. The composition F = g a f : D(0, ) C is analytic on D(0, ); F (0) = g a (f(0)) = g a (0) = 0; for any z D(0, ), Re f(z) < and so f(z) Ω a and so g a (f(z)) D(0, ), ie. F (z) = g a (f(z)) < for all z D(0, ). By Schwarz s Lemma, we have F (0). Since and f(0) = 0, we get Date: Novemer 30, 2009 (Version.0). F (z) = g a(f(z))f (z) = πieπif(z)/2 f (z) (e πif(z)/2 + ) 2 f (0) 4 π.

2 () Let S e the set of functions defined y S = {f : Ω C f analytic, f < on Ω, and f(0) = 0}. By considering the function f g or otherwise, Prove that sup f() = eπ/2 f S e π/2 +. Solution. Note that g maps Ω into D(0, ), g is analytic on Ω, and g (0) = 0. Furthermore, note that g is injective on Ω g (z) = g (w) iff (e πz/2 )(e πw/2 + ) = (e πz/2 + )(e πw/2 ) iff e π(z w)/2 = iff Re(z w) = 0 and Im(z w) is an integer multiple of 4; so when z, w Ω, this is only possile if z = w. Hence g is an invertile map and g maps D(0, ) onto Ω, g is analytic on D(0, ), and g (0) = 0. Given any f S, the composition F = f g : D(0, ) C is analytic on D(0, ); F (0) = f(g (0)) = f(0) = 0; for any z D(0, ), g (z) Ω and so f(g (z)) D(0, ), ie. F (z) < for all z D(0, ). By Schwarz s Lemma, we have F (z) z for all z D(0, ). In particular, if we pick z 0 = eπ/2 e π/2 D(0, ), + we get f() = f(g (z 0 )) = F (z 0 ) z 0 = z 0. Since this is true for aritrary f S, we have that But note that equality is attained y for any θ R. So sup f() eπ/2 f S e π/2 +. f(z) = e iθ g (z) sup f() = eπ/2 f S e π/ Let f : D(0, ) C e analytic and f(z) < for all z D(0, ). Recall that ϕ α (z) = (z α)/( αz). (a) By considering the function ϕ f(a) f ϕ a or otherwise, show that f (z) f(z) 2 z 2. Solution. Let a D(0, ). Then f(a) D(0, ). We consider ϕ a and ϕ f(a). Define g = ϕ f(a) f ϕ a : D(0, ) C. Note that g is analytic. If z D(0, ), then ϕ a (z) D(0, ), so f(ϕ a (z)) D(0, ), and so ϕ f(a) (f(ϕ a (z))) D(0, ), ie. g(z) < for all z D(0, ). Also g(0) = ϕ f(a) (f(ϕ a (0))) = ϕ f(a) (f(a)) = 0. Schwartz s Lemma then implies that g (0). But using chain rule and Lemma 4.8 in the lectures, we get g (0) = ϕ f(a) (f(a))f (a)ϕ a(0) = f (a) f(a) 2 ( a 2 ) 2

3 and so f (a) f(a) 2 a 2. Now just oserve that this works for aritrary a D(0, ). () Suppose there exist two distinct points a, D(0, ) such that f(a) = a and f() =. By considering the function ϕ a f ϕ a or otherwise, show that f(z) = z for all z D(0, ). Solution. Since a, at least one of them must e non-zero, wlog, let a 0. Define g = ϕ a f ϕ a : D(0, ) C. Note that g is analytic. If z D(0, ), then ϕ a (z) D(0, ), so f(ϕ a (z)) D(0, ), and so ϕ a (f(ϕ a (z))) D(0, ), ie. g(z) < for all z D(0, ). Also g(0) = ϕ a (f(ϕ a (0))) = ϕ a (f(a)) = ϕ a (a) = 0. Schwartz s Lemma then implies that g(z) z for all z D(0, ). Before we proceed further, we oserve that for α D(0, ), ϕ α (ϕ α (z)) = z = ϕ α (ϕ α (z)) for all z D(0, ), ie. ϕ α is an invertile map and ϕ α = ϕ α. Let c = ϕ a () D(0, ). So = ϕ a (c). Now note that g(c) = ϕ a (f(ϕ a (c))) = ϕ a (f()) = ϕ a () = c and so equality is attained in g(z) z y z = c D(0, ). By Schwartz s Lemma again, we get g(z) = e iθ z for all z D(0, ). But e iθ c = g(c) = c and so e iθ =. Hence and so ϕ a f ϕ a = id f = ϕ a id ϕ a = id. (c) Suppose there exist a D(0, ), a 0, such that f(a) = 0 = f( a). By considering the function f(z) g(z) = ϕ a (z)ϕ a (z) or otherwise, show that f(0) a 2. What can you conclude if f(0) = a 2? Solution. Note that g is analytic on D(0, ) except possily at z = a and z = a where the denominator vanishes. However since lim(z a)g(z) = lim z a z a lim (z + a)g(z) = lim z a z a ( az)( + az) (z + a) ( az)( + az) (z a) f(z) = a 4 f(a) = 0, 2a f(z) = a 4 f(a) = 0, 2a g has removale singularity y Theorem 5.2. We thus may assume that g is analytic on D(0, ). If z =, ie. z = e iθ for some θ, then ϕ a (z)ϕ a (z) = ( az)( + az) (z a)(z + a) = a 2 e 2iθ e 2iθ a 2 = e 2iθ (e 2iθ a 2 ) e 2iθ a 2 = e 2iθ a 2 e 2iθ a 2 =. Since f(z) < for all z D(0, ) and f is anlytic, we must have f(z) for z =. Hence F (z) 3

4 for z = and thus for z y Maximum Modulus Theorem. If we take z = 0, F (0) gives f(0) ϕ a (0)ϕ a (0) = a 2. If equality is attained, then F (0) = means that F attains its maximum on D(0, ) at the interior point z = 0. Maximum Modulus Theorem then implies that F is constant throughout D(0, ) and so F e iθ for some θ R. Hence f(z) = e iθ z a z + a az + az for all z D(0, ). 3. Recall the L-shaped path that we used in lectures to estalish the existence of antiderivative for an analytic function, i.e. L(z, z 2 ) is a path that goes from z to z 2 comprising two line segments one parallel to the imaginary axis and the other to the real axis. (a) Let f : Ω C e a continuous function on an open set Ω. Suppose f(z) dz = 0 R along the oundary of any closed rectangle R Ω. Let D(z 0, r) Ω. Show that the function F : D(z 0, r) C defined y F (z) = f(ζ) dζ L(z 0,z) is an antiderivative of f. Hence deduce that f is analytic in Ω. Solution. See proof of Morrera s Theorem on pp in the textook. () Let f : D(0, ) C e analytic and f(z) 0 for all z D(0, ). By considering the function h : D(0, ) C defined y f (ζ) h(z) = f(ζ) dζ, L(0,z) show that there exists an analytic function g : D(0, ) C such that f(z) = e g(z) for all z D(0, ). Solution. Since f(z) 0 for all z D(0, ), f (z)/f(z) is analytic on D(0, ). Let L(0, z) is an L-shaped curve from 0 to z. Recall from the proof of Theorem 3.3 that the function h : D(0, ) C defined y h(z) = L(0,z) f (ζ) f(ζ) dζ is analytic and is an antiderivative of the integrand, i.e. h (z) = f (z) f(z) for all z D(0, ). Now apply chain rule to the analytic function f(z)e h(z) to get d dz f(z)e h(z) = e h(z) [f (z) h (z)f(z)] = e h(z) [f (z) f (z)] = 0. In other words f(z)e h(z) is constant and so f(z)e h(z) = f(0)e h(0) = f(0) (3.) 4

5 for all z D(0, ). Since f(0) 0, we have f(0) = re iθ for some r > 0 and some θ [0, 2π). Since r > 0, we may write r = e ln r and thus f(0) = e ln r+iθ. Let c = ln r + iθ C. Then (3.) ecomes f(z) = f(0)e h(z) = e c+h(z). Hence g(z) = h(z) + c is an analytic function satisfying f(z) = e g(z) for all z D(0, ). Note that (a) is a converse to Cauchy s theorem while () estalishes the existence of logarithms. 4. Let Ω C e a region and let f : Ω C. Suppose there exists n N such that g(z) = [f(z)] n and h(z) = [f(z)] n+ are oth analytic on Ω. Show that f is analytic on Ω. Solution. Note that g(z) = 0 if and only if h(z) = 0. So the zeros of g and h are in common. Let Z Ω e these common zeros. If Z contains a limit point, then g and h are identically zero y the identity theorem and f 0 is of course analytic. We will assume that all zeros in Z are isolated. Let z 0 Z. So there exist analytic functions g and h and some ε > 0 such that for all z D(z 0, ε), g(z) = (z z 0 ) k g (z), g (z 0 ) 0, h(z) = (z z 0 ) l h (z), h (z 0 ) 0, where k and l are the orders of zero of g and h respectively at z 0. But g n+ = f n(n+) = h n and so So we get (z z 0 ) (n+)k g (z) n+ = (z z 0 ) nl h (z) n. g (z) n+ h (z) n = (z z 0) (n+)k nl. Since the lhs is non-zero for z = z 0, the rhs is also non-zero for z = z 0 and this is only possile if (n + )k nl = 0. In other words, l > k, i.e. the order of zero of h is larger than the order of zero of g. So h/g has a removale singularity at z 0. Since this is true for all z 0 Z, h/g may e extended to an analytic function f on Ω. But since for all z Ω\Z, f(z) = h(z) g(z) = f(z), and Ω\Z clearly has limit points, f f y the identity theorem. 5. Show that each of the following series define a meromorphic function on C and determine the set of poles and their orders. sin(nz) f(z) = n!(z 2 + n 2 ), g(z) = [ (z z n ) 2 ] zn 2 where {z n } C is a sequence of distinct complex numers (i.e. z n z m if n m) such that <. (5.2) z n 3 Solution. We claim that f is meromorphic on C with poles of order at z P f := iz = {..., 3i, 2i, i, i, 2i, 3i,...}, 5

6 and a removale singularity at z = 0. For any fixed R > 0, there exists an integer N such that N > 2R. We will consider f on the disc D(0, R). We may write N sin(nz) f(z) = n!(z 2 + n 2 ) + sin(nz) n!(z 2 + n 2 ). (5.3) Note that the zeroes of sin(nz) are {..., 2π n, π n, 0, π n, 2π } n,... and so the zeroes of sin(nz) and z 2 + n 2 = (z + in)(z in) are disjoint for all n. So the first sum in (5.3) has poles of order for each ±in where 0 < n N. The second sum defines an analytic function for all z D(0, R) since it is uniformly convergent y the Weierstrass M-test: For z < R and n > N, we have sin(nz) e n z y considering the power series expansions of sin and exp; and we also have z 2 + n 2 n 2 z 2 n 2 R 2 = n 2 (R/n) 2 4 3n 2 since n > N > 2R (and so R 2 /n 2 /4 = 3/4). Hence sin(nz) n!(z 2 + n 2 ) 4 e n z 3 n!n 2 =: M n. Since M n = 4 3 e nr n!n 2 is convergent converge y the ratio test, the series of function sin(nz) n!(z 2 + n 2 ) is uniformly convergent. Since this argument is true for aritrary R > 0, f is a meromorphic on C with poles of order at z P f. Let R > 0 e fixed ut aritrary. We will show that g defines a meromorphic function on the D(0, R). By (5.2), we must have lim n z n 3 = 0 and therefore lim n z n =. So since the z n s are all distinct, there can only e finitely many n such that z n D(0, R). Hence there exists N N such that z n > 2R for all n > N. Each z n D(0, R) is a pole of order 2 of g. We write g(z) = N [ (z z n ) 2 z 2 n ] + [ (z z n ) 2 zn 2 The first sum contains all the poles of g in D(0, R). We claim that the second sum is analytic on D(0, R). For n > N, (z z n ) 2 = z 2 + 2zz n zn(z 2 z n ) 2 z 2 n = z 2 /z n + 2z z n 3 z/z n 2 R 2 /2R + 2R z n 3 R/2R = 5R z n 3 =: M n. ]. 6

7 Since M n = 5R z n 3 <. The Weierstrass M-test implies that the second sum converges uniformly and is thus analytic on D(0, R). So g is meromorphic on D(0, R). Since R is aritrary, g is meromorphic on C. 7