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Math Help - Complex analysis - Power series question

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    Complex analysis - Power series question

    Suppose that f(z)=\sum_{n=0}^{\infty} c_nz^n for z\in\mathbb{C}. Prove that for all R, \sum_{n=0}^{\infty}|c_n|R^n\leq2M(2R) 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.
    Last edited by Speed1991; January 4th 2012 at 08:45 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex analysis - Power series question

    Quote Originally Posted by Speed1991 View Post
    \sum_{n=0}^{\infty}\leq2M(2R) where M(r):=sup{|f(z)|:|z|=r}.
    What on earth does that inequality mean?
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    Re: Complex analysis - Power series question

    Sorry, forgot to put in the middle bit
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex analysis - Power series question

    According to the Cauchy inequalities we have |c_n|R^n\leq \frac{M(2R)}{(2R)^n}R^n=\frac{M(2R)}{2^n} , so

    \sum_{n=0}^{\infty}c_nR^n\leq M(2R)\sum_{n=0}^{\infty}\frac{1}{2^n}=2M(2R) .
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