Can anyone prove the identity $\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $ I came across this identity in a paper on elliptic curves, and the proof wasn't provided. It really irked me, and I couldn't find an explanation anywhere else. Can anyone shed some light? 
$$\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $$

Following the suggestion by @Lucian, we use the Euler's infinite product representation of the sine function and write
$$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag 1$$
Next, taking the logarithmic derivative of $(1)$ yields
$$\cot z=\frac1z-2z\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} \tag 2$$
And then, taking the derivative of $(2)$ and multiplying by $-1$ yields
$$\begin{align}
\csc^2z&=\frac{1}{z^2}+2\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} +4z^2\sum_{n=1}^{\infty}\frac{1}{(n^2\pi^2-z^2)^2} \\\\
&=\frac{1}{z^2}+2\sum_{n=1}^{\infty}\frac{(n^2\pi^2+z^2)}{(n^2\pi^2-z^2)^2}\\\\
&=\frac{1}{z^2}+\sum_{n=1}^{\infty}\left(\frac{1}{(n\pi+z)^2}+\frac{1}{(n\pi-z)^2}\right)\\\\
&=\frac{1}{z^2}+\sum_{n=1}^{\infty}\frac{1}{(n\pi+z)^2}+\sum_{n=-\infty}^{-1}\frac{1}{(n\pi+z)^2}\\\\
&=\sum_{n=-\infty}^{\infty}\frac{1}{(n\pi+z)^2}
\end{align}$$
as was to be shown.
