How to use Liouville's theorem to prove the formula of πsin⁡πz

The RHS h(z)=limN→∞∑n=−NN(−1)nz−n=1z+∑n≥1(−1)n2zz2−n2 converges locally uniformly away from Z so it is meromorphic with simple poles, it is 2-periodic because h(z+2)−h(z)=limN→∞(−1)N+1z−N−1+(−1)N+2z−N−2−(−1)−Nz−N−(−1)N+1z−N+1 Also the second expression for h means h(z)=o(|z|) as Im(z)→±∞,R(z)∈[0,2] Together with the decay of πsin⁡πz it means f(z)=o(|z|) which implies from Liouville's theorem that f is constant. Since f is odd then f(0)=0.

Answered: "How to use Liouville's theorem to prove the..."

Tori Hines

Answered question

2022-01-23

How to use Liouville's theorem to prove the formula of

\frac{π}{\sin π z}

Answer & Explanation

Anabelle Miller

Beginner2022-01-24Added 12 answers

The RHS
$h (z) = lim_{N \to \infty} \sum_{n = - N}^{N} \frac{{(- 1)}^{n}}{z - n} = \frac{1}{z} + \sum_{n \geq 1} {(- 1)}^{n} \frac{2 z}{z^{2} - n^{2}}$
converges locally uniformly away from Z so it is meromorphic with simple poles,
it is 2-periodic because $h (z + 2) - h (z) = lim_{N \to \infty} \frac{(- 1)^{N + 1}}{z - N - 1} + \frac{(- 1)^{N + 2}}{z - N - 2} - \frac{(- 1)^{- N}}{z - N} - \frac{(- 1)^{N + 1}}{z - N + 1}$
Also the second expression for h means $h (z) = o (| z |)$ as $I m (z) \to \pm \infty, R (z) \in [0, 2]$
Together with the decay of $\frac{π}{\sin π z}$ it means f(z)=o(|z|) which implies from Liouville's theorem that f is constant.
Since f is odd then f(0)=0.

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