Let \[ f(z)=a_0+a_1z+\cdots+a_nz^n \] be a complex polynomial of degree \(n>0\). Prove that \[ \frac1{2\pi i}\int_{|z|=R} z^{n-1}|f(z)|^2\,dz =a_0\overline{a_n}R^{2n}. \]
Solution
1. On the circle \(|z|=R\),\[ \overline{f(z)} = \sum_{k=0}^n \overline{a_k}R^{2k}z^{-k}. \]
If \(|z|=R\), then \[ z\overline z=R^2, \] so \[ \overline z=\frac{R^2}{z}. \] Therefore \[ \overline{f(z)} = \sum_{k=0}^n \overline{a_k}\,\overline z^{\,k} = \sum_{k=0}^n \overline{a_k}R^{2k}z^{-k}. \]
2. The coefficient of \(z^{-1}\) in the Laurent polynomial\(z^{n-1}|f(z)|^2\) is\[ a_0\overline{a_n}R^{2n}. \]
By step 1, \[ z^{n-1}|f(z)|^2 = \sum_{j=0}^n\sum_{k=0}^n a_j\overline{a_k}R^{2k}z^{n-1+j-k}. \] A term has exponent \(-1\) precisely when \[ n-1+j-k=-1, \] equivalently when \[ k=n+j. \] Since \(0\leq j,k\leq n\), this forces \(j=0\) and \(k=n\). Hence the unique \(z^{-1}\) term is \[ a_0\overline{a_n}R^{2n}z^{-1}. \]
3. The required integral is\[ \boxed{ \frac1{2\pi i}\int_{|z|=R} z^{n-1}|f(z)|^2\,dz = a_0\overline{a_n}R^{2n} }. \]
For every integer \(m\), \[ \frac1{2\pi i}\int_{|z|=R} z^m\,dz = \begin{cases} 1,&m=-1,\\ 0,&m\neq-1. \end{cases} \] Thus termwise integration of the finite Laurent expansion in step 2 extracts exactly its \(z^{-1}\) coefficient. Step 2 identifies that coefficient as \(a_0\overline{a_n}R^{2n}\).
4. Q.E.D.
Step 3 is the asserted identity.