Series representation of hyperbolic cotangent

Question

I recently encountered this series representation of hyperbolic cotangent function. How this equation can be derived? $$\coth(z) =\sum_{k=-\infty}^{\infty}\frac{z}{z^{2}+k^{2}\pi^{2}}, \quad \frac{iz}{\pi} \not \in \mathbb{Z}$$

Answer 1

First note that the sum can be written as $$\sum_{k=-\infty}^{\infty}\frac{z}{z^{2}+k^{2}\pi^{2}}=\frac{1}{z}+2z\sum_{k=1}^{\infty}\frac{1}{z^{2}+k^{2}\pi^{2}}$$ From Euler's formula for the trigonometric cotangent (see e.g. Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$ or http://mathworld.wolfram.com/Cotangent.html, formula (18)) $$\pi \cot \pi z=\frac{1}{z}+2z\sum_{k=1}^{\infty}\frac{1}{z^{2}-k^{2}}$$ you find $$\cot z=\frac{1}{z}+2z\sum_{k=1}^{\infty}\frac{1}{z^{2}-k^{2}\pi^{2}}$$ Now use $\cot z = i \coth(iz)$ to get

$\begin{align*} \coth(z) &= -i\cot(-iz)\\ &= -i\left(\frac{1}{-iz}+2(-iz)\sum_{k=1}^{\infty}\frac{1}{(-zi)^{2}-k^{2}\pi^{2}}\right)\\ &=\frac{1}{z}-2z\sum_{k=1}^{\infty}\frac{1}{-z^{2}-k^{2}\pi^{2}}\\ &=\frac{1}{z}+2\sum_{k=1}^{\infty}\frac{1}{z^{2}+k^{2}\pi^{2}} \end{align*}$