This problem comes from Gamelin, "Complex Analysis", Exercise IX.1.7. Basically it asks us to prove a generalization of the Schwarz Lemma without the assumption $f(0)=0$, by calculating a particular integral.
The statement of the problem is: let $f(z)=\sum_{k=0}^{\infty}a_k z^k$ be analytic in $B(0,1)$, with $|f(z)|\leq M$ for all $z\in B(0,1)$. By integrating $|f(z)|^2$ along $\partial B(0,r)$ for $r<1$, show that $$\sum_{k=0}^{\infty} |a_k|^2\leq M^2.$$ (This will imply that $|f'(0)|=|a_1| \leq M$, which is what Schwarz Lemma states, but here it is not assumed that $f(0)=0$.)
Here is how I attempted to calculate the integral mentioned in the statement of the problem: using the change of variable $z=re^{i\theta}, dz=ire^{i\theta}d\theta$, $$\begin{align} \oint_{\partial B(0,r)} |f(z)|^2 dz & = \int_{0}^{2\pi} (\sum_{k=0}^{\infty}a_k r^k e^{i\theta k})(\sum_{k=0}^{\infty}\overline{{a_k} r^k e^{i\theta k}})ire^{i\theta}d\theta \\ & = \sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\int_{0}^{2\pi}a_{k_1}\overline{a_{k_2}}r^{k_1 + k_2}e^{i\theta (k_1 - k_2)}ire^{i\theta}d\theta \\ & = \sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}a_{k_1}\overline{a_{k_2}}ir^{k_1 + k_2 +1}\int_{0}^{2\pi}e^{i\theta (k_1 - k_2 +1)}d\theta, \end{align}$$
and since $$\int_{0}^{2\pi}e^{i\theta (k_1 - k_2 +1)}d\theta = \begin{cases} 2\pi, & \text{if } k_2 = k_1 + 1 \\ 0, & \text{otherwise} \end{cases}$$
we have $$\oint_{\partial B(0,r)} |f(z)|^2 dz = \sum_{k=0}^{\infty}a_k \overline{a_{k+1}}r^{2k+1}\cdot 2\pi i,$$
But I cannot see how to relate this to the result we wanted to prove. In particular I cannot make this into an expression that involves $\sum_{k=0}^{\infty} |a_k|^2$. However it's quite close to this, since $|a_k|^2 = a_k \overline{a_k}$ and we have terms of the form $a_k \overline{a_{k+1}}$ in the sum on the right hand side. This leads me to suspect that there is some mistake I've made in the computation of the integral; however I cannot see where it is.
If I am mistaken in any step above, can someone help me point it out? If not, can someone give me some hint at this problem? Any help is appreciated!
Let $z=re^{i\theta}$, \begin{align*} \left|f(re^{i\theta})\right|^2 &=f(re^{i\theta})\cdot\overline{f(re^{i\theta})}\\[3pt] &=\sum_{n=0}^{\infty}a_nr^ne^{i n\theta}\cdot \sum_{n=0}^{\infty}\overline{a}_n r^ne^{-i n\theta}\\[3pt] &=\sum_{m,n=0}^{\infty}a_m\overline{\alpha}_nr^{m+n}e^{-i (m-n)\theta}\\ &=\sum_{n=0}^{\infty}|a_n|^2r^{2n} +\sum_{\substack{m,n=0\\ m\ne n}}^{\infty}a_m\overline{a}_nr^{m+n}e^{-i (m-n)\theta}. \end{align*} Note that $$\frac{1}{2\pi}\int_{0}^{2\pi}e^{i n\theta}d\theta=0\ (n\ne 0),$$ we get \begin{align*} \frac{1}{2\pi}\int_{0}^{2\pi}\left|f(re^{i\theta})\right|^2d\theta &=\frac{1}{2\pi}\int_{0}^{2\pi}\left(\sum_{n=0}^{\infty}|a_n|^2r^{2n}\right)d\theta\\[3pt] &=\sum_{n=0}^{\infty}|a_n|^2r^{2n}. \end{align*} So $$\sum_{n=0}^{\infty}|a_n|^2r^{2n} =\frac{1}{2\pi}\int_{0}^{2\pi}\left|f(re^{i\theta})\right|^2d\theta \leq M^2,\ 0<r<1.$$ Let $r\to 1^{-}$, we can get $$\sum_{k=0}^{\infty} |a_k|^2\leq M^2.$$