ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

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$Math 33B/1 S00 K. Solutions to Midterm Exam G1, Version B If you ...$

$Math 3C Homework 3 Solutions$

48 <strong>ANALYSIS</strong> QUALS Let f = f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1. Then f is a conformal map taking D to a strip, which takes J to the line Im(z) = π/2, and takes R to the line Im(z) = −π/2. Let u(z) = 1 1 2 + π Im(f(z)). Then since the imaginary part of a holomorphic function is harmonic, we have that u is harmonic on D, and u(x + iy) → 0 as y ↓ 0 and u(z) → 1 as z → J. Problem 11: Prove that a meromorphic function f(z) in the extended complex plane C ∗ = C ∪ {∞} is the sum of the principal parts at its poles. Solution. In a neighborhood of each pole a of f, we have f(z) = c−m (z − a) c−1 + · · · + (z − a) m−1 z − a + c0 + c1(z − a) + · · · , m + c−m+1 c−m where the principal part is (z−a) m + · · · + c−1 z−a . Either f vanishes at infinity or not. That is, either f(1/z) has a pole at 0 or a removable singularity at 0. If f(1/z) has a pole at 0, then f has a pole at ∞. Including possibly ∞, let z1, . . . , zm be the poles of f. Near each non-infinite zk, as above, we may write f(z) = fk(z) + gk(z), where fk is the principal part and gk is holomorphic If zk = ∞, we can write f(1/z) = ˜ f∞(z)+g∞(z), where ˜ f∞(z) is the principal part at 0 of f(1/z), and g∞ is holomorphic in a neighborhood of 0. Define fk(z) = ˜ f∞(1/z). Let H = f − m fk. Near each pole, we have subtracted the principal k=1 part of f, so we see that H is entire (the singularities at the points zk are all removable). But since we also subtracted ˜ f∞(1/z), we have that H is bounded. Hence by Liousville, H is constant. But by the definition of H, we must have that H = 0 everywhere. Thus, f is the sum of the principal parts of its poles. Problem 12: Let D be a domain (connected open set) in C and let (un) be a sequence of harmonic functions un : D → (0, ∞). Show that if un(z0) → 0 for some z0 ∈ D, then un → 0 uniformly on compact subsets of D. Solution. As un(z0) → 0, we have that {un(z0)} is Cauchy. Thus, for every ε > 0 there exists N such that 0 ≤ un(z0) − um(z0) < ε for all N ≤ n ≤ m. Let K ⊂ D be compact. Harnack’s inequality states that there exists a constant CK > 0 so that maxx∈K u(x) ≤ CK minx∈K u(x) for u : D → (0, ∞) harmonic. Applying this to un − um, we have that max x∈K (un(x) − um(x)) ≤ CK min x∈K (un(x) − um(x)). If z0 ∈ K, then minx∈K(un(x) − um(x)) < ε, and thus un → 0 uniformly on K.
<strong>ANALYSIS</strong> QUALS 49 If z0 ∈ K, choose K ′ such that K ′ ∩ K = ∅ and z0 ∈ K ′ , and choose z1 ∈ K ′ ∩K. Since un(z1) → 0, the argument can be repeated to show that un → 0 uniformly on K. Thus, un → 0 uniformly on compact subsets of D.
Page 1 and 2: ANALYSIS QUALIFYING EXAM PROBLEMS B
Page 3 and 4: ANALYSIS QUALS 3 Proof. (a) First,
Page 5 and 6: ∂∆n ANALYSIS QUALS 5 0, 1, 2, .
Page 7 and 8: ANALYSIS QUALS 7 Thus, as |h ′ (0
Page 9 and 10: ANALYSIS QUALS 9 But by the geometr
Page 11 and 12: Then Γ (j) k ANALYSIS QUALS 11 f(
Page 13 and 14: ANALYSIS QUALS 13 = 1 √ |f(w)|
Page 15 and 16: ANALYSIS QUALS 15 Then B has the s
Page 17 and 18: ∂Q ANALYSIS QUALS 17 Problem 8: L
Page 19 and 20: ANALYSIS QUALS 19 Consider ˆxn ∈
Page 21 and 22: ANALYSIS QUALS 21 b) Suppose S −
Page 23 and 24: ANALYSIS QUALS 23 Problem 12: Let f
Page 25 and 26: ANALYSIS QUALS 25 b) I present the
Page 27 and 28: ANALYSIS QUALS 27 Solution. Picard
Page 29 and 30: ANALYSIS QUALS 29 Problem 2: Is eve
Page 31 and 32: δ 0 Then f(ξ) δf(w) = δ dξ
Page 33 and 34: ANALYSIS QUALS 33 But as D is compa
Page 35 and 36: so y ∈ B(x, ε), so y ∈ B(x (n1
Page 37 and 38: Problem 10: Let the power series f(
Page 39 and 40: ANALYSIS QUALS 39 where the inequal
Page 41 and 42: ANALYSIS QUALS 41 Problem 8: Let f
Page 43 and 44: ANALYSIS QUALS 43 Winter 2007 Probl
Page 45 and 46: Then for N sufficiently large, we h
Page 47: ANALYSIS QUALS 47 see that on ( √

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