MATH502_HW6 - Serge Ballif MATH 502 Homework 6(1 The...

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Serge Ballif MATH 502 Homework 6 February 29, 2008 (1) The function f ( z ) is entire, and it is known that | f ( z ) | ≤ 1 whenever | z | = 1 and that | f ( z ) | ≤ 10 whenever | z | = 10 . Show that | f ( z ) | ≤ 5 whenever | z | = 5 . What can be said about | f (0) | ? Claim 1. For all 1 ≤ | z | ≤ 10 we have | f ( z ) | ≤ | z | . Proof. Consider the function f ( z ) /z defined on A = { 1 < | z | < 10 } . This region is a connected open set on which f ( z ) /z is holomorphic (since no poles lie in the domain). Hence, we can invoke the maximum principle to see that | f ( z ) /z | attains its maximum on the the boundary ∂A = {| z | = 1 , 10 } . By assumption we know that | f ( z ) /z | ≤ 1 on ∂A . Thus, for every z A , we must have | f ( z ) /z | ≤ 1 or equivalently | f ( z ) | ≤ | z | . In particular, this shows that | f ( z ) | < 5 for | z | = 5. The only thing that we can conclude about f (0) is that it lies somewhere in the unit disc, because for any u the constant function f ( z ) = u satisfies the requirements of the problem. (2) Let denote the open unit disk. Suppose that f : is a holomorphic map with two distinct fixed points.
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  • Summer '08
  • JOHNRIDENER
  • Math, TA, Holomorphic function, Types of functions, Functions and mappings, Conformal map

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