Partial fraction decomposition of $\pi\cdot \tan(\pi z)$

Question

Evaluate the partial fraction decomposition of $\pi \tan(\pi z)$

$$2\pi \tan(\pi z)=\cot\left(\frac{\pi}{2}-\pi z\right)-\cot\left(\frac{\pi}{2}+\pi z\right)$$ $$=\frac{2}{1-2z}+\sum_{k=1}^\infty \frac{1-2z}{(\frac{1}{2}-z)^2-k^2}-\frac{2}{1+2z}-\sum_{k=1}^\infty \frac{1+2z}{(\frac{1}{2}-z)^2-k^2}$$ $$=\frac{8z}{1-4z^2}+\sum_{k=1}^\infty \frac{4-8z}{1-2z+4z^2-4k^2}-\sum_{k=1}^\infty \frac{4+8z}{1+2z+4z^2-4k^2}$$ I know that the result should be $$\pi \tan(\pi z)=\sum_{k=1}^\infty \frac{8z}{(2k+1)^2-4z^2}$$ but I dont know how to get there. Any suggestions?

Answer 1

Hint. If you know that, for $0<z<1/2$, $$ \pi \cot (\pi z)=\sum_{k=1}^\infty \left(\frac{1}{z-k}+\frac{1}{z+k-1}\right) $$ then by putting $z \to 1/2-z$ you get $$ \begin{align} \pi \cot (\pi (1/2-z))&=\pi \tan(\pi z) \\\\&=\sum_{k=1}^\infty \left(\frac{1}{1/2-z-k}+\frac{1}{1/2-z+k-1}\right) \\\\&=\sum_{k=1}^\infty\frac{8 z}{(2k-1)^2-4 z^2} \end{align} $$ as wanted.

Answer 2

You may exploit the fact that $\tan(z)=\frac{d}{dz}\log(\cos z)$ and: $$ \cos(z)=\prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)\tag{1}$$ (Weierstrass product) so: $$ \tan(z) = \sum_{n\geq 0}\frac{-8z}{4z^2-(2n+1)^2\pi^2}\tag{2}$$ and: $$\begin{eqnarray*}\pi\tan(\pi z) = \sum_{n\geq 0}\frac{-2z}{z^2-\left(\frac{2n+1}{2}\right)^2}&=&-\sum_{n\geq 0}\left(\frac{1}{z-\frac{2n+1}{2}}+\frac{1}{z+\frac{2n+1}{2}}\right)\\&=&\color{red}{\sum_{m\in\frac{1}{2}+\mathbb{Z}}\frac{-1}{z-m}}.\tag{3}\end{eqnarray*}$$

On the other hand, every singularity of $\pi\tan(\pi z)$ is a simple pole and lie in $\frac{1}{2}+\mathbb{Z}$,
and for every $m$ that belongs to such a set: $$ \text{Res}\left(\pi\tan(\pi z),z=m\right)=\lim_{z\to m}\frac{\pi\sin(\pi z)(z-m)}{\cos(\pi z)}\stackrel{DH}{=}\lim_{z\to m}\frac{\pi\sin(\pi z)}{-\pi \sin(\pi z)}=\color{red}{-1},\tag{4}$$ giving another proof of $(3)$.