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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.

2 Answers2

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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?
Pedro
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  • what is the contradiction in (3) can you please tell. I have understood all 3 points but can't conclude. –  Aug 18 '21 at 14:00
  • @Avenger There's no contradiction, rather, you want to show that $h$ must be constant. – Pedro Aug 18 '21 at 15:25
  • How does h(z+1 ) =h(z) implies that h is a constant ? Sorry I am coming back so late but I have been not well –  Sep 10 '21 at 04:08
  • Can you please reply? –  Sep 11 '21 at 05:51
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    Show that $h=f-g$ is bounded on $0 \leq \Re z \leq 1$. Because of $h(z)=h(z+1)$ it is then bounded on the whole of the complex plane. Argue that $h$ is entire. Hence, by Liouville's theorem, $h$ is constant. – Gary Sep 20 '21 at 10:54
  • @Gary I got your argument but I am unable to prove h to be bounded on $0\leq Rz \leq 1$. Can you please help with that? –  Oct 15 '21 at 11:50
  • @Avenger Show that $h(z)$ is holomorphic (has no poles, ect.). Show that $h(z) \to 0$ as $\Im z \to \pm \infty$ in $0\leq \Re z \leq 1$. Conclude from this that $h(z)$ must be bounded in that strip. – Gary Oct 15 '21 at 11:54
  • @Gary I have shown that h(z) is holomorphic but I am unable to understand what you mean by symbol $ \Im z $. You were using Re z symbol in last comment which I think means Real part of z –  Oct 15 '21 at 11:56
  • Imaginary part of $z$. So the function is close to zero in the closed strip $0 \leq \Re z \leq 1$ after some time as you go up or down. So it is bounded after some time. But in the remaining finite part, it is bounded because it is a compact set and $h$ is continuous on it. – Gary Oct 15 '21 at 11:56
  • @Gary Is there an easy or elegant way of proving it that you are aware of , I am only aware of this method: Putting z =x+ iy in the function h(z) and then using the fact that $0\leq x \leq 1 $ and limit y tends to $\infty$. If there is any other method can you please tell me? –  Oct 15 '21 at 12:05
  • @Avenger That is correct. I would just say it is obvious that $h(z)$ tends to zero as $\Im z \to \pm \infty$. – Gary Oct 16 '21 at 05:27
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This is a complex analysis problem.

This is a sum of a series by residues.

$$ \sum_{n = -\infty}^{\infty} f(n) = -\{ sum\, of\, residues\, of\, \pi\cot(\pi z)f(z)\, at\, all\, the\, poles\, of\, f(z) \} \tag{1}$$

Using $x$ to avoid confusion with $z$. $n$ is the variable of summation so replace it with $x$.

$$f(x) = \frac1{(z-x)^2} \tag{2}$$

$f(x)$ has a double pole at $x=z$.

$$ \sum_{n = -\infty}^{\infty} f(n) = - \pi \lim_{x \rightarrow z} \frac{d}{dx} \left[ (x-z)^2 \frac{\cot(\pi x)}{(z-x)^2}\right] \tag{3}$$

$$ \sum_{n = -\infty}^{\infty} f(n) = - \pi \lim_{x \rightarrow z} \frac{d}{dx} {\cot(\pi x)} \tag{4}$$

$$ \sum_{n = -\infty}^{\infty} f(n) = - \pi \lim_{x \rightarrow z} \frac{-\pi}{\sin^2(\pi x)} \tag{5}$$

$$ \sum_{n = -\infty}^{\infty} f(n) = \frac{\pi^2}{\sin^2(\pi z)} \tag{6}$$

  • "$\sum_{n = -\infty}^{\infty} f(n) = -{ sum, of, residues, of, \pi\cot(\pi z)f(z), at, all, the, poles, of, f(z) } $" Can you please write a proof of this statement or give some refence? –  Sep 21 '21 at 16:21
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    @Avenger see for example this. – Evangelopoulos Foivos Sep 21 '21 at 18:42
  • Theorem 3.2. in @Evangelopoulos F reference. My reference is ancient: Schaum's outline Series, Theory and problems of complex variables - Murray R. Spiegel , 4th printing 1987, pg 187 Summation of series. –  Sep 21 '21 at 21:51
  • Take a look at Example 4.3. @Evangelopoulos F reference. Instead of $z \rightarrow \frac1{2}$ just leave it as $z$. –  Sep 21 '21 at 22:01
  • @arthur I am busy. But awarded you bounty. If I have a question I will comment. Thanks a lot. –  Sep 22 '21 at 10:20
  • This can be solved using only real analysis as well. – robjohn Oct 18 '21 at 09:26