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: I conjecture that the strange exponent $1$ should be replaced by a $2$. This means that we are told to compute the quantity
$$Q:={1\over 2\pi i}\int_{\partial D_R}z^{n-1} \bigl|f(z)\bigr|^2\>dz\ .$$
As a hint note that on $\partial D_R$ one has $\bar z={R^2\over z}$ and therefore
$$\bigl|f(z)\bigr|^2=f(z)\overline{f(z)}=f(z)\>\bar f\!\!\left({R^2\over z}\right)\ ,$$
whereby the function $\bar f$ is defined by
$$\bar f(z):=\sum_{k=0}^n \overline{a_k}\,z^k\ .$$
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:
$$ \frac{1}{2 \pi i}\int_{|z|=r} z^{n-1} \cdot |f(z)|^2 dz = R^{2n} a_0 \overline{a_n} $$
A: This is really just a comment, in consideration of the other (I think more useful) two answers.
If the stated equality were correct, one would have that, for $R\not=0$, 
$$ {1\over 2 \pi i R^{2n} }\int_{|z| = R} z^{n-1} | f(z)| \,dz ={1\over 2 \pi}\int_0^{2\pi} e^{i n\theta} \left| f(R e^{i\theta})\over R^n\right| \, d\theta = a_0\overline a_n.$$
In particular, the quantities would not depend on $R$. But (in the second integral), by taking the limit as $R$ tends to infinity, one gets (interchange of limit and integral justified by dominated convergence, say)that 
$$ \lim_{R\to \infty}{1\over 2 \pi }\int_0^{2\pi} e^{i n\theta} \left| f(R e^{i\theta})\over R^n\right| \, d\theta  = {1\over2 \pi}\int_0^{2\pi} e^{i n\theta} |a_n | \, d\theta = 0,$$
for $n\ge 1$. So, assuming the original equality were correct, $a_0\overline a_n = 0$.
