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Suppose i know the following two formulas:

$\displaystyle \sin(\pi z) = \pi z \prod_{n=1}^\infty \left( 1 - \frac{z^2}{n^2} \right)$ and $\pi \cot(\pi z)= \sum_{n=-\infty}^{n=\infty} \frac {1}{z+n}$ Then i am trying to prove that

$\frac {\pi}{\sin{\pi z}}=\sum_{n=-\infty}^{n=\infty} \frac {(-1)^n}{z-n}$

I am not getting the idea for this,any hints/ideas?

CiaPan
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Math Lover
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  • The proof is almost the same with $1/\sin^2(z)$, using the Liouville theorem for showing the function minus its poles is constant. Then adapt it to $\pi / \sin( \pi z)$, being careful with the convergence of $\sum_n \frac{(-1)^n}{z-n}$ – reuns Jan 01 '17 at 09:31

1 Answers1

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Hint. Use Hadamard Factorization Theorem.