Prove for an entire function $\displaystyle{f(z)=\sum_{n=0}^\infty a_nz^{n}}$, [duplicate]

Prove that for an entire function $$\displaystyle{f(z)=\sum_{n=0}^\infty a_nz^{n}}$$, $$\displaystyle{\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta=\sum_{n=0}^\infty|a_n|^2r^{2n}}$$ for any $$r>0 \in \mathbb R$$

marked as duplicate by Martin R, RRL, Joshua Mundinger, Lee David Chung Lin, blubApr 21 at 7:51

• If you write out the integral you get $\frac{1}{2\pi} \sum_{n=0}^\infty\sum_{m=0}^\infty a_na_m^* r^{n+m} \int_0^{2\pi}e^{in\theta -im\theta}{\rm d}\theta$. Show that all the integrals are zero except when $n=m$ and conclude from there. – Winther Apr 19 at 12:05
Set $$g(\theta )=f(re^{i\theta })$$. This function is $$2\pi-$$periodic and $$L^2(0,2\pi)$$. Therefore, using Parseval equality, $$\frac{1}{2\pi}\int_0^{2\pi}|g(\theta )|^2=\sum_{n\in \mathbb Z}|c_n|^2,$$
where $$c_n$$ are the Fourier coefficient of $$g$$. To get complex Fourier coefficient of $$g$$, remark that $$g(\theta )=\sum_{n\in\mathbb N}a_nr^ne^{in\theta }.$$