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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.

Michael Hardy
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Doug
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    You are correct. To show that the first formula is wrong just take $f(z) = 1$. The integral is then zero and not $2\pi$ as the formula suggests. Also your fix seems to work to get the correct result. – Winther Jan 10 '16 at 01:02
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    In view of the residue theorem, I think it a bit more natural to put the $i$ on the other side of the equation. – Ian Jan 10 '16 at 01:47
  • But $|f(z)|^2$ is not analytic! :O – Doug Jan 10 '16 at 01:52
  • I'm not saying it's an instance of the residue theorem, because it is not, but it is in the same spirit. – Ian Jan 10 '16 at 12:27
  • I understood, I was making a jab. I prefer the $i$ where it is because it makes the calculation more apparent. – Doug Jan 10 '16 at 20:32

1 Answers1

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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.

Winther
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Doug
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