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# HW3s - Key to Complex Analysis Homework 3 Chapter 1(in...

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Key to Complex Analysis Homework 3 September 27, 2011 Chapter 1 (in Greene & Krantz’s book) 29. Suppose that f : D (0 , 1) C is holomorphic and that | f | ≤ 2. Derive an estimate for 3 ∂z 3 f i 3 . Proof. Choose 0 < r < 2 / 3 so that D ( i 3 , r ) D (0 , 1). By Cauchy’s esti- mate, 3 ∂z 3 f i 3 2 * 3! r 3 . Letting r 2 / 3, we get 3 ∂z 3 f i 3 2 * 3! ( 2 3 ) 3 = 81 2 . 2 32. Suppose that f is bounded and holomorphic on C \ { 0 } . Prove that f is constant. [ Hint: Consider the function g ( z ) = z 2 · f ( z ) and endeavor to apply Theorem 3.4.4] Proof. We consider the following auxiliary function, g ( z ) = z 2 f ( z ) , if z 6 = 0; 0 , if z = 0 . Since f is holomorphic on C \{ 0 } , so is g on C \{ 0 } . However, at z = 0, we need to check whether g 0 (0) exists. By the condition that f is bounded on C \ { 0 } , we obtain g 0 (0) = lim z 0 g ( z ) - g (0) z - 0 = lim z 0 z 2 f ( z ) z = lim z 0 zf ( z ) = 0 . 1

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Hence, we know that g is holomorphic on entire C . Choose M such that | f | ≤ M , on C \ { 0 } , we have | g ( z ) | ≤ M | z | 2 , for all z C this implies, by Theorem 3.4.4, g is a complex polynomial of at most 2 degree. Write as z 2 f ( z ) = g ( z ) = a 0 + a 1 z + a 2 z 2 on C \ { 0 } . Let letting z 0 and noticing that f is bounded, we get a 0 = 0. Hence z 2 f ( z ) = a 1 z + a 2 z 2 on C \{ 0 } or zf ( z ) = a 1 + a 2 z . Again, by letting z
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HW3s - Key to Complex Analysis Homework 3 Chapter 1(in...

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