Suppose that for . Prove that for all R, where M(r):=sup{|f(z)|:|z|=r}.

I have absolutely no idea where to start with this question. Help starting would be much appreciated.

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- Jan 4th 2012, 06:17 AMSpeed1991Complex analysis - Power series question
Suppose that for . Prove that for all R, where M(r):=sup{|f(z)|:|z|=r}.

I have absolutely no idea where to start with this question. Help starting would be much appreciated. - Jan 4th 2012, 07:42 AMFernandoRevillaRe: Complex analysis - Power series question
- Jan 4th 2012, 07:45 AMSpeed1991Re: Complex analysis - Power series question
Sorry, forgot to put in the middle bit

- Jan 4th 2012, 08:31 AMFernandoRevillaRe: Complex analysis - Power series question
According to the Cauchy inequalities we have , so

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