Application of Parseval's theorem Let $f(z)$ be holomorphic on the unit disc. Then by Parseval's theorem,
$$\int_{\theta = 0}^{2\pi} |f'(r, \theta)|^2 \; d\theta = 2\pi \sum_{n=1}^\infty n^2 |a_n|^2 r^{2(n-1)}$$
where the $a_n$ are the Fourier coefficients of $f$. I don't see this calculation, can someone explain?
 A: *

*Let $g(r,\theta)$ be a function, continuous and $2\pi$ periodic in $\theta$, then we have that Parseval's theorem gives
$$ \int_0^{2\pi} |g(r,\theta)|^2 \mathrm{d}\theta = 2\pi \sum |b_n|^2 $$
where $b_n$ are given by the Fourier coefficients of $g(r,\cdot)$. 

*So it remains to relate the Fourier coefficients of $f'$ to $f$. Writing $z = r \exp i\theta$ and $\bar{z} = r \exp -i\theta$. We have that $r = \sqrt{z\bar{z}}$ and $\theta = \frac1{2i} \ln z/\bar{z}$. So by the usual change of variables formula
$$ \partial_z = \partial_z r \partial_r + \partial_z \theta \partial_\theta = \frac12 \frac{r}{z} \partial_r + \frac{1}{2i} \frac{1}{z} \partial_\theta $$
Now, since $f$ is holomorphic, we have that $\partial_\bar{z} f = 0$. This implies that $r\partial_r f - \frac{1}{i} \partial_\theta f = 0$. This means that $$ f'(r,\theta) = \frac{1}{i} \frac{e^{-i\theta}}{r} \partial_\theta f(r,\theta) $$ So using the standard properties of the Fourier transform you can related the Fourier coefficients of $f'(r,\theta)$ to that of $f$. 
