Prove that $\frac {\pi^2}{ \sin(\pi z)^2 } = \sum_{n=-\infty , n\neq 0 }^{n=\infty} 1/ (z-n)^2 $

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The following question was part of my complex analysis assignment and I am not able to prove it.

How to prove that $$\frac {\pi^2 } { \sin(\pi z)^2 } = \sum_{n=-\infty , n \neq0 }^{n=\infty} \frac{1}{(z-n)^2 }$$

I tried by using the identity $$\sin(\pi z) = \pi z \prod_{n=1}^{\infty} \left(1- \left(\frac zn \right)^2 \right) $$ But that is in product so can't be used here.

I don't have any other ideas except these. Actually the prof is known in department to be not good in teaching and online classes added more to it. In particularly in these type of questions where I have to prove some identity I am having very much difficulty.

It is my very humble request to you to help me.

Best Answer

A standard argument is as follows:

  1. Let $f$ denote your infinite sum, and let $g$ denote your function built out of the sine.
  2. Show that $f$ and $g$ are meromorphic functions with double poles only at the integers except zero. They also have no roots.
  3. This means that the difference $h=f-g$ is a holomorphic function, but it is also a function that satisfies $h(z)=h(z+1)$. Can you conclude?
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