Given the power series $f(z)=\sum^\infty_{n=0}c_n(z-z_0)^n$ I have to prove that $$ \int_0^{2\pi}|f(z_0+re^{it})|^2\;dt =\sum^\infty_{n=0}|c_n|^2r^{2n}$$ Here $ r>0$ is such that $\overline{B_r(z_0)}\subseteq U $ and $ U\subseteq \mathbb{C}$ is the region in which $ f$ is defined. My idea is to prove that the LHS is less or equal than the RHS and vice versa. I already have prooved that $$ \int_0^{2\pi}|f(z_0+re^{it})|^2\;dt \leq\sum^\infty_{n=0}|c_n|^2r^{2n}$$ Replacing $ |f(z_0+re^{it})|^2$ by the series and using the triangular inequality. I also know that $$c_nr^n=\frac{1}{2\pi}\int_0^{2\pi} f(z_0+re^{it})e^{-int}\;dt $$
My next idea was to use Cauchy-Schwartz for the integral like this: $$\frac{1}{2\pi}\int_0^{2\pi} |f(z_0+re^{it})|^2\;dt= $$ $$\frac{1}{2\pi}\int_0^{2\pi} |f(z_0+re^{it})|^2\;dt \underbrace{\left(\frac{1}{2\pi}\int_0^{2\pi}|e^{int}|^2\;dt\right)}_{=1} $$ $$ \geq \frac{1}{4\pi^2}\left|\int_{0}^{2\pi}e^{-int}f(z_0+re^{it})\;dt\right|^2=\frac{1}{4\pi^2}4\pi^2|c_n|^2r^{2n}=|c_n|^2r^{2n} $$
But this does not tell my anything about the series. I think I tried all the standard tools for inequalities with integrals and series and now I'm stuck on this. It would be greate if any of you have any ideas or sugestions (not complete solutions, please). Thank you all in advance