Reading about Complex Analysis, I came across the following:
Consider first the representation $\frac{\pi ^2}{\sin ^2(\pi z)}=\sum_{n\in \mathbb{Z}}\frac{1} {(z+n)^2}$, which applies for all $z\in \mathbb{C}\backslash \mathbb{Z}$.
For $|z|<\pi$, we can obtain the following expression for the Taylor series of $\frac{z^2}{\sin^2(z)}$:
$\frac{z^2}{\sin^2(z)}=1+2\sum_{m=1}^{\infty}(2m-1)\frac{\zeta (2m)}{\pi^{2m}}z^{2m}$
where $\zeta(z) =\sum \frac{1}{n^z}$. This expression can then be used to obtain the values $\zeta (2)$ and $\zeta (4)$.
My question is, how is the expression for the Taylor series obtained? I am not able to obtain it from the representation for $\frac{\pi ^2}{\sin ^2(\pi z)}$.
First, we can write
$$\begin{align} \sum_{n\in \mathbb{Z}}\frac1{(z+n\pi)^2}&=\frac1{z^2}+\sum_{n=1}^\infty \left(\frac1{(z+n\pi)^2}+\frac1{(z-n\pi)^2}\right)\\\\ &=\frac1{z^2}+2\sum_{n=1}^\infty \frac{z^2+(n\pi)^2}{(z^2-(n\pi)^2)^2}\\\\ &=\frac1{z^2}+\frac2{z^2}\sum_{n=1}^\infty \sum_{m=1}^\infty (2m-1)(z/n\pi)^{2m}\\\\ &=\frac1{z^2}+\frac2{z^2} \sum_{m=1}^\infty (2m-1) \frac{\zeta(2m)}{\pi^{2m}}z^{2m} \end{align}$$
Can you finsish now?