This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Extra Problems Sheet 2 Stephen Taylor May 11, 2005 1. If  a  < 1 and  b  < 1 show that a b 1 Â¯ ab < 1 Proof : Define a function f ( z ) â‰¡ a z 1 Â¯ 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 Â¯ az ) + Â¯ a ( a z ) (1 Â¯ az ) 2 =  a  2 1 (1 Â¯ az ) 2 which is analytic everywhere except at the point z = Â¯ 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....
View
Full
Document
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.
 Three '09
 Cong
 Math

Click to edit the document details