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I was reading through my book in complex analysis and i encountered this problem. Given, $F=\sum_{n=0}^{\infty} a_nX^n$ is a convergent power series with radius of convergence R. We are asked to show that for every 0$\leq$r$<$R that $$ \int_0^{2\pi}\mid F(re^{it})\mid^2\mathrm{d}t={2\pi}\sum_{n=0}^{\infty}\mid a_n\mid^2r^{2n}$$ This is not my homework but after some effort I have no clue how to solve this one. The assignment was in the chapter about convergent power series and there is nothing about integrating, thus I assume it should be solvable by real analysis calculus. I would appreciate if someone would help me with this one. Regards, Raxel.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Nov 16 '12 at 16:27
Thanks I've tried to rewrite the problem as you suggested. – Raxel Nov 16 '12 at 17:53
up vote 1 down vote accepted

$$|F(re^{it})|^2=\sum_{m=0}^\infty a_mr^me^{imt}\times\sum_{n=0}^\infty \bar{a}_nr^ne^{-int}=\sum_{m,n=0}^\infty a_m\bar{a}_nr^{m+n}e^{i(m-n)t}.$$

Integrating both sides over $[0,2\pi]$ and interchanging the order of integration and summation, and noting that the value of $\int_0^{2\pi}e^{i(m-n)t}dt$ is $2\pi$ when $m=n$ and $0$ when $m\ne n$, then you will obtain the wanted answer. Please check that each step is legal by using the fact $r<R$.

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Of course this is what I was supposed to do. Thank you very much! – Raxel Nov 16 '12 at 18:33
@Raxel: You are welcome! – 23rd Nov 16 '12 at 18:35

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