# Complex analysis - Power series question

• Jan 4th 2012, 06:17 AM
Speed1991
Complex analysis - Power series question
Suppose that $\displaystyle f(z)=\sum_{n=0}^{\infty} c_nz^n$ for $\displaystyle z\in\mathbb{C}$. Prove that for all R, $\displaystyle \sum_{n=0}^{\infty}|c_n|R^n\leq2M(2R)$ 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.
• Jan 4th 2012, 07:42 AM
FernandoRevilla
Re: Complex analysis - Power series question
Quote:

Originally Posted by Speed1991
$\displaystyle \sum_{n=0}^{\infty}\leq2M(2R)$ where M(r):=sup{|f(z)|:|z|=r}.

What on earth does that inequality mean?
• Jan 4th 2012, 07:45 AM
Speed1991
Re: Complex analysis - Power series question
Sorry, forgot to put in the middle bit
• Jan 4th 2012, 08:31 AM
FernandoRevilla
Re: Complex analysis - Power series question
According to the Cauchy inequalities we have $\displaystyle |c_n|R^n\leq \frac{M(2R)}{(2R)^n}R^n=\frac{M(2R)}{2^n}$ , so

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