How can I show that
$$\left( {\frac{1}{\sin z}} \right)^{2}=\sum_{-\infty}^{+\infty}\frac{1}{\left( {z-\pi n} \right)^2}.$$
I don't know how to do this.
Any help will be appreciated.
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Vincenzo Tibullo
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unicornki
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I think the easiest way is proving that $\frac{1}{\sin(z)^2}$ minus the RHS is a bounded entire function, then proving that the constant is $0$ by some way. this is also the easiest way for proving that $\pi \cot(\pi z) = \frac{1}{z}+\sum_k \frac{1}{z-k}+\frac{1}{z+k}$ – reuns Apr 24 '16 at 15:02