Assume that $f(z)$ is holomorphic on $\{|z|<R\}$ and has Taylor expansion $f(z)=\sum\limits_{n=0}^{\infty}c_nz^n$. Prove that $$\dfrac{1}{2\pi}\int\limits_{0}^{2\pi}|f(re^{i\varphi})|^2d\varphi=\sum\limits_{n=0}^{\infty}|c_n|^2r^{2n},\quad (0<r<R).$$
We have $c_n=\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{n+1}}dz$. Moreover, if we set $z=re^{i\varphi}$, we obtain $$\dfrac{1}{2\pi}\int\limits_{0}^{2\pi}|f(re^{i\varphi})|^2d\varphi=\frac{1}{2\pi i}\int\limits_{|z|=r}\dfrac{|f(z)|^2}{z}dz.$$ I don't have any idea to go on. Could someone help me? Thanks in advance!
