TITLE: Error on a complex analysis qualifying exam?
QUESTION [5 upvotes]: Let $f(z) = \sum_{n = 0}^\infty c_n z^n$ for $\left| z \right| < R$.  
The problem as stated. 

For all $r < R$, $$\int_{\left\{ |z| = r \right\}} \left| f(z) \right|^2 \, dz = 2\pi \sum_{n=0}^\infty \left| c_n \right|^2 r^{2n}. $$  

What I think the statement should be. 

For all $r < R$, $$\int_{\left\{ |z| = r \right\}} \frac{\left| f(z) \right|^2}{iz} \, dz = 2\pi \sum_{n=0}^\infty \left| c_n \right|^2 r^{2n}. $$

Basically, I think the professor forgot to account for the derivative $\gamma^\prime(t) = ire^{it}$ that enters into the integrand when we calculate the integral over $[0, 2\pi]$.  
Question.  Am I correct?
Thanks for your help.

REPLY [3 votes]: As pointed out in the comments the first formula is wrong since when $f(z)=1$ the integral is zero and not $2\pi$ as the formula suggests. 
In addition, here is a proof that the second formula I suggested gives the correct result.
Let $\gamma(t) = re^{it}$.  Then
\begin{align} \int_\left\{ \left| z \right| = r \right\} \frac{\left| f(z) \right|^2}{iz} dz &= \int_0^{2\pi} \left( \sum_{n=0}^\infty c_n r^n e^{i n t} \right) \left( \sum_{m = 0}^\infty \overline{c_m} r^m e^{-i m t} \right) dt \\ &= \sum_{i = 0}^\infty \sum_{n + m = i} c_n \overline{c_m} r^i \int_0^{2\pi}e^{i(n-m)t} dt \\ &= 2\pi \sum_{n = 0}^\infty \left| c_n \right|^2 r^{2n} \end{align} as claimed.  The equalities are justified essentially by Cauchy's theorem together with the fact that the power series $f(z)$ converges absolutely within its radius of convergence.