Problems and Solutions in - Mathematics - University of Hawaii

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2 Complex Analysis 38 2 COMPLEX ANALYSIS
2.1 1989 April 2 COMPLEX ANALYSIS 2.1 1989 April INSTRUCTIONS: Do at least four problems. TIME LIMIT: 1.5 hours 35 1. (a) Let U be the unit disk in the complex plane C, U = {z ∈ C : |z| < 1}, and let f be an analytic function in a neighborhood of the closure of U. Show that if f is real on all the boundary of U, then f must be constant. (b) Let u be a real harmonic function in all the complex plane C. Show that if u(z) ≥ 0 for all z ∈ C, then u must be constant. Solution: (a) The hypotheses imply that Imf(eiθ ) = 0 for all θ ∈ R. Since f is holomorphic in a neighborhood Ω of U, the series ∞ f(z) = anz n n=0 converges uniformly on any compact subset of Ω. The unit circle T = {z : |z| = 1} = {eiθ : θ ∈ R} is one such compact subset, and here the series is f(e iθ ∞ ) = ane inθ . If we write the coefficients as an = cn + ibn, where cn, bn ∈ R, then we have n=0 ane inθ = (cn + ibn)[cos(nθ) + i sin(nθ)] = [cn cos(nθ) − bn sin(nθ)] + i[cn sin(nθ) + bn cos(nθ)]. Thus, by the hypothesis, the series Imf(e iθ ) = ∞ [cn sin(nθ) + bn cos(nθ)] n=0 converges uniformly to zero for all θ ∈ [0, 2π]. Therefore, with the possible exception of c0, we have cn = bn = 0, for all n, so f ≡ c0. ⊓⊔ (b) Since C is simply connected, there is a real-valued harmonic conjugate v(z) such that the function f(z) = u(z) + iv(z) is entire. Now, since u(z) ≥ 0, f(z) maps the complex plane into the right half-plane, {Ref(z) ≥ 0}. It follows immediately from Picard’s theorem that f must be constant. 36 However, an elementary argument using only Liouville’s theorem is probably preferable, so let f be as above, and define g(z) = f(z) + 1. Then g is entire and maps C into {w ∈ C : Rew ≥ 1}. In particular, g(z) is bounded away from zero, so the function h(z) = 1/g(z) is a bounded entire function. (In fact, |h(z)| ≤ 1.) Therefore, by Liouville’s theorem, h is constant, so f is constant, so u = Ref is constant. ⊓⊔ 2. Let f be an analytic function in the region {z : |z| > 1}, and suppose that lim f(z) = 0. z→∞ 35In 1989 there was a single three hour test covering both real and complex analysis. Students were required to do nine problems, with at least four from each part. 36Picard’s theorem states that a non-constant entire function can omit at most one value of C from its range. This is a very powerful theorem, but if you use it for an easy problem like this one, you might be accused of killing a fly with a sledge hammer! 39
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Page 3 and 4: 1 Real Analysis 1.1 1991 November 2
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Page 43 and 44: 2.2 1991 November 21 2 COMPLEX ANAL
Page 49 and 50: 2.3 1995 April 10 2 COMPLEX ANALYSI
Page 69 and 70: 2.9 Some problems of a certain type
Page 71 and 72: A Miscellaneous Definitions and The
Page 73 and 74: A.1 Real Analysis A MISCELLANEOUS D
Page 75 and 76: A.1 Real Analysis A MISCELLANEOUS D
Page 77 and 78: A.2 Complex Analysis A MISCELLANEOU
Page 79 and 80: B List of Symbols F an arbitrary fi
Page 81: INDEX INDEX Riemann mapping theorem

Problems and Solutions in - Mathematics - University of Hawaii

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