Let $f(z)=\sum_{k=0}^na_kz^k$ be a polynomial with coefficients in $\mathbb{C}$. Suppose $\deg f\geq 1$. Prove for any $R>0$,
\begin{equation*} \frac{1}{2\pi i}\int_{|z|=R}z^{n-1}|f(z)|^2dz=a_0\bar{a_n}R^{2n}. \end{equation*}
How to compute out this equation?
Here is how you advance
$$ \frac{1}{2\pi i}\int_{|z|=R}z^{n-1}|f(z)|^2dz = \frac{1}{2\pi i}\sum_{k=0}^n a_k\sum_{k=0}^n \overline a_k \int_{|z|=R}z^{n-1} z^k {(\overline z)}^k \,dz .$$
Now, let $z=Re^{i\theta}$ and try to finish the problem. See related technique.
Note: We used the fact
$$ |f(z)|^2 = f(z) \overline {f(z)}. $$
