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Suppose we have $f(z)=\sum\limits_{n=0}^\infty c_nz^n$ for $|z|<R $

Prove that $\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta=\sum\limits_{n=0}^\infty|c_n|^2r^{2n}$, $r<R$

Got stuck for this proof ... I have no idea what happens to the series when putting $|f(e^{i\theta})|^2$ under the integral sign ....

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A related problem. Using the fact that $|z|^2=z\bar{z}$, we have

$$ \frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta=\frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta}) \overline{f(re^{i\theta})} d\theta$$

$$=\frac{1}{2\pi}\int_0^{2\pi}\sum_{n=0}^{\infty}c_nr^n e^{in\theta} \overline{\sum_{m=0}^{\infty} c_mr^m e^{im\theta}} d\theta $$

$$=\frac{1}{2\pi} \sum_{n=0}^{\infty}c_nr^n \sum_{m=0}^{\infty}\bar{c_m}r^m\int_{0}^{2\pi}e^{i(n-m)\theta} d \theta $$

The other fact you want to use is $\int_{0}^{2\pi}e^{in\theta}e^{-im\theta}=2\pi$ for $n= m$ and $0$ otherwise.

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  • $\begingroup$ @BobaFret: I left it to him to figure this out. $\endgroup$ Oct 26, 2012 at 3:34
  • $\begingroup$ Then as you said, the integrand part becomes $\sum\limits_{n=0}^\infty c_ne^{in\theta}\sum\limits_{m=0}^\infty c_ne^{-im\theta}$ since $c_n=\overline{c_n}$ as a real number. Then the integrand becomes the product of two infinite series and there are many cross-product terms ... $\endgroup$ Oct 26, 2012 at 3:37
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    $\begingroup$ Oh I see. This fact is very helpful actually. Thanks a lot ! $\endgroup$ Oct 26, 2012 at 3:48
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    $\begingroup$ @Cmeteorolite:You are welcome. $\endgroup$ Oct 26, 2012 at 3:57

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