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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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Thanks I've tried to rewrite the problem as you suggested. –  Raxel Nov 16 '12 at 17:53

1 Answer 1

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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