I am trying to prove that $$\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}=\left(\frac{\pi}{\sin \pi z}\right)^2$$ based on some hints.

Let $$f(z)=\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}\hspace{0.5cm}\text{ and }\hspace{0.5cm} g(z)=\left(\frac{\pi}{\sin \pi z}\right)^2$$

The first step is showing that

The Laurent series of both $f$ and $g$ have the same principal part at $z=0$.

I calculated the principal part for $g$ as follows:

$$\frac{\sin \pi z}{\pi}=z\left(1-\frac{\pi^2z^2}{3!}+\frac{\pi^4z^4}{5!}+\dots\right)$$ So near $z=0$ we have $$\frac{\pi}{\sin \pi z}=\frac{1}{z}\left( 1+\frac{\pi^2z^2}{3!}+\dots\right)$$ So $$\frac{\pi}{\sin \pi z}=\frac{1}{z}\left( 1+z^2h_1(z)\right)$$ where $h_1(z)$ is a holomorphic function. Then $$g(z)=\left(\frac{\pi}{\sin \pi z}\right)^2=\frac{1}{z^2}\left(1+z^2h_1(z)\right)^2$$ Hence the principal part of $g(z)$ is $1/z^2$.

According to the calculation above the principal part of $f(z)$ should be $1/z^2$ at $z=0$.

But I don't know how to calculate this principal part. Can you help?

  • 1
    $\begingroup$ For $n\ne 0$, the term $1/(z-n)^2$ is analytic at $0$, so has principal part $0$ there. So you only need to find the principal part of that one last term. $\endgroup$ – GEdgar Sep 19 '17 at 1:01

In a neighborhood of $0$, e.g. $|z|<\frac12$, $\displaystyle\sum\limits_{n\in Z\setminus\{0\}}\dfrac{1}{(z-n)^2}$ is a uniformly convergent series of holomorphic functions, hence is holomorphic. That is $f(z)-\dfrac1{z^2}$.


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