Let $0\leq r_1<r_2$, and $z_0\in \mathbb{C}$, and consider the region $A=\{z\in \mathbb{C}|r_1<|z-z_0|<r_2\}$. Let $f$ analytic in the region $A$. Then we can write, $$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}.$$ Show that if $r_1<r<r_2$, then, $$\int_0^{2\pi} |f(z_0+re^{i\theta})|^2d\theta=2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}+2\pi\sum_{n=1}^{\infty}|b_n|^2r^{-2n}.$$

My attempt:

Note that, $$|f(z_0+re^{i\theta})|^2=f(z_0+re^{i\theta})\overline{f(z_0+re^{i\theta})},$$ then, I arrive to this identity: \begin{align} |f(z_0+re^{i\theta})|^2&=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} a_n\overline{a_m}r^{n+m}e^{i\theta(n-m)}\\ &+\sum_{n=1}^{\infty}\sum_{m=0}^{\infty} b_n\overline{a_m}r^{-n+m}e^{-i\theta(n+m)}\\ &+\sum_{n=0}^{\infty}\sum_{m=1}^{\infty} a_n\overline{b_m}r^{n-m}e^{i\theta(n+m)}\\ &+\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} b_n\overline{b_m}r^{-n-m}e^{i\theta(m-n)}. \end{align} Can I exchange the integral with the series?


1 Answer 1


Yes, you can interchange the integral and the sum. The Laurent series converges uniformly for $0\leq \theta \leq 2\pi$ for fixed $r$. In fact it converges uniformly is any any compact subset of the annulus.

  • $\begingroup$ I see, thanks for the observation. Now, using the first and the last series I obtain the identity. But, What happens with the second and the third series? Thanks. $\endgroup$ Mar 10, 2019 at 23:25
  • 1
    $\begingroup$ @Gerschgorin Note that in the second term $\int_0^{2\pi} e^{i\theta (n+m)}\, d \theta=0$ because $n+m \geq 1$. So all terms vanish after integration. Same thing is true for the third term. $\endgroup$ Mar 10, 2019 at 23:35
  • $\begingroup$ I got it! Thanks a lot for your answer. $\endgroup$ Mar 10, 2019 at 23:38

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