# Double integral over the unit circle and Cauchy's integral formula

Let $$f$$ be analytic on the unit disc $$D$$ and assume $$\int \int _{D} |f|^{2} dxdy$$ exists. Let $$\begin{equation*} f(z)=\sum_{n=0}^{\infty} a_{n}z^{n} \end{equation*}$$ Prove that $$\begin{equation*} \frac{1}{2\pi}\int \int _{D} |f(z)|^{2} dxdy=\sum_{n=0}^{\infty} \frac{|a_{n}|^{2}}{2n+2} \end{equation*}$$

I'm struggling to start this problem. This could be because my double integral knowledge is a bit rusty but I've reviewed Green's theorem and how to do double integral over a circle but nothing really seems to be working out. I've also tried applying the Cauchy Integral Formula without much success though I'm sure it will come in handy.

Additionally, this problem appears in a section where only one theorem is given, so I'm wondering if it is useful. Here's the theorem:

Let $$\{f_{n}\}$$ be a sequence of analytic functions on an open set $$U$$, converging uniformly on every compact subset $$K\subseteq U$$ to a function $$f$$. Then $$f$$ is holomorphic. Furthermore, the sequence of derivatives $$\{f'_{n}\}$$ converges uniformly on every compact subset $$K$$ to $$f'$$.

Any hints would be appreciated. Thanks so much in advance.

(I posted another question about the application of this same theorem so I'm sorry if this seems repetitive, I'm just really struggling with my homework and the fact that I can't see my professor to ask questions due to the COVID-19 outbreak)

I think there is factor of $$2 \pi$$ missing in the equation and there cannot be a $$z$$ appearing on RHS after integration.
In polar coordinates the given integral becomes $$\int_0^{1} \int_0^{2\pi} |\sum\limits_{k=0}^{\infty} a_kr^{k}e^{ik\theta}|^{2}rd\theta dr$$. Just expand the square and note that the cross terms vanish: $$\int_0^{2\pi} e^{ik\theta}e^{-ij\theta}d\theta=0$$ if $$j \neq k$$. Hence we get $$2\pi \int_0^{1} \sum\limits_{k=0}^{\infty} |a_k|^{2} r^{2k+1} dr$$. Now compute the integral.