TITLE: Another proof of Liouville's theorem.
QUESTION [5 upvotes]: Let $f(z)=\sum_{n} a_n z^n$ has radius of convergence $R>0$ and $0<r<R$. Show:
$$\frac{1}{2\pi} \int_0^{2\pi} |f(re^{it})|^2 dt= \sum_{n}|a_n|^2r^{2n}$$
Use this equality to prove Liouvelle's Theorem. 
Can anyone give me a hint to how to proceed with this?

REPLY [3 votes]: We have $f(z)=\sum_n c_nz^n$.
Substituting $z=re^{it}$ and evaluating,   
$\left|f(re^{it})\right|^2=\sum_n\left|c_n\right|^2r^{2n}+\sum_n\sum_mg(n,m,r)\times e^{i(n-m)t}$
Integrating both sides from $t=0$ to $t=2\pi$ should do it.