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This question already has an answer here:

We need to show that for an entire function $f$ on $\mathbb{C}$, there are constants $a_1,...,a_n$ such that $\int^{2 \pi}_{0}|f(re^{i \theta})|^2 d\theta=2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}$.

Thoughts so far are that we can find a power series expansion about $0$ by Taylor's Theorem. Given the result it seems natural to use $f(z)^2$ somewhere but I can't see how.

Any help appreciated.

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marked as duplicate by 23rd, Amzoti, TMM, Lord_Farin, Martin May 25 '13 at 14:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $f(z)=\sum\limits_{n=0}^\infty a_nz^n$. For $z=re^{i\theta} $ we have $$f(re^{i\theta })=\sum\limits_{n=0}^\infty a_nr^ne^{in\theta }$$ We have that $$|f(re^{i\theta})|^2=f(re^{i\theta})\cdot \overline{f(re^{i\theta})}=$$ $$\sum\limits_{n=0}^\infty a_nr^ne^{in\theta }\cdot \sum\limits_{m=0}^\infty \overline {a_m}r^me^{-im\theta }=$$ $$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty a_n \overline {a_m}r^{n+m}e^{i(n-m)\theta }$$ Therefore $$\int_0^{2\pi }|f(re^{i\theta})|^2d\theta =$$ $$\int_0^{2\pi}\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty a_n \overline {a_m}r^{n+m}e^{i(n-m)\theta }d\theta =(*)$$ $$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \int_0^{2\pi} a_n \overline {a_m}r^{n+m}e^{i(n-m)\theta }d\theta=(*)$$ $$\sum_{n=0}^\infty a_n\overline{a_n}r^{n+n}\cdot 2\pi= $$ $$2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}$$ $(*)$Where we can use uniform convergence to switch integral and sums, and also note that $$\int_0^{2\pi}e^{i(n-m)\theta}d\theta=\begin{cases}0&n\neq m\\2\pi&n=m\end{cases}$$

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Let $f(z)=\sum_{n\geq 0}a_nz^n$ the power series expansion of $f$ with infinite radius of convergence. Then, for $r\geq 0$ fixed, the function $$ g:\theta\longmapsto f(re^{i\theta})=\sum_{n\geq 0}a_nr^ne^{in\theta} $$ is a Fourier series which converges normally as $\sum_{n\geq 0}|a_n|r^n$ converges. In particular, $g$ is $2\pi$-periodic and continuous, whence in $L^2(0,2\pi)$. So the equality you want is Parseval.

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Using the power series representation is a good idea. Take $r$ (strictly) smaller than the radius of convergence of the Taylor series at $0$, say $\sum_na_nz^n$. Use uniform convergence to switch the series and the integral.

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