extra_2 - Extra Problems Sheet 2 Stephen Taylor 1 If | a...

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Unformatted text preview: Extra Problems Sheet 2 Stephen Taylor May 11, 2005 1. If | a | < 1 and | b | < 1 show that a- b 1- ¯ ab < 1 Proof : Define a function f ( z ) ≡ a- z 1- ¯ az where a and z are complex valued and | a | < 1. Lemma 1. f ( z ) is analytic on the open unit disk. We compute f ( z ) =- 1(1- ¯ az ) + ¯ a ( a- z ) (1- ¯ az ) 2 = | a | 2- 1 (1- ¯ az ) 2 which is analytic everywhere except at the point z = ¯ a- 1 which is outside the domain of f ( z ). So we find f is an analytic function. We note that a corollary to the maximum modulus principle states Corollary. Suppose that a function f is continuous on a closed bounded region R and is analytic and not constant in the interior of R . Then the maximum value of | f ( z ) | in R , which is always reached, occurs somewhere on the boundary of R and never in the interior. Since the boundary of the unit disk occurs for | z | = 1 from which it is easy to show | f ( z ) | = 1, by the above corollary we conclude that if | z | < 1 then | f ( z ) | < 1. (Any value greater than or equal to one would violate the maximum modulus principle). Fixing z at any complex value b within its domain of definition gives the desired result....
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This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

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extra_2 - Extra Problems Sheet 2 Stephen Taylor 1 If | a...

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