Q:

How prove that $\frac{1}{ 2 π i}\int_{ |z|=R} z^{ n−1} |f(z)|^ 2 dz = a_0 \bar{a_n} R^{ 2 n} $.

Let $f(z) = a_0 + a_1z + · · · + a_n z^n$ polynomial of degree $n ≥ 0$. Show that
$$\dfrac{1}{
2 π i} \displaystyle\int_{
|z|=R}
z^{
n−1}
|f(z)|^2
dz = a_0 \bar{a_n} R^{
2 n}
$$
Already tried to use the integral formula of cauchy, induction on $n$, but I can not appear with $\bar{a_n}$ and $R^{
2 n}$. Someone can help.

A:

It must be $ |f(z)|^2 $. Then one can calculate:
$$ \frac{1}{2 \pi i}\int_{|z|=r} z^{n-1} \cdot |f(z)|^2 dz = \frac{1}{2 \pi i}\int_{0}^{2 \pi} R^n e^{int} i (\sum_{k=0}^n R^k a_ke^{ikt}) (\sum_{j=0}^n R^j \overline{a_j}e^{-ijt})dt \\ = \sum_{k,j=0}^n R^{n+k+j}a_k\overline{a_j} \frac{1}{2 \pi i} \int_{0}^{2 \pi} e^{i(n+k-j-1)}e^{it}idt = \sum_{k,j=0}^n R^{n+k+j}a_k\overline{a_j} \frac{1}{2 \pi i}\int_{|z|=1} z^{n+k-j-1} dz $$
Then by Cauchy's Integral Theorem $$ \frac{1}{2 \pi i}\int_{|z|=1} z^{n+k-j-1} dz  = \delta_{n (j-k)}$$ where $ \delta_{n (j-k)} $ denotes the Kronecker Delta. This implies:
$$ \frac{1}{2 \pi i}\int_{|z|=r} z^{n-1} \cdot |f(z)|^2 dz = R^{2n} a_0 \overline{a_n} $$

