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This question already has an answer here:

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 $r>0 \in \mathbb R$

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

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  • $\begingroup$ 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. $\endgroup$ – Winther Apr 19 at 12:05
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    $\begingroup$ If you did not understand the answer to your previous question (math.stackexchange.com/questions/3193399/… ) you could have asked for details. No point in posting another question. $\endgroup$ – Kavi Rama Murthy Apr 19 at 12:09
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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 $$g(\theta )=\sum_{n\in\mathbb N}a_nr^ne^{in\theta }.$$

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