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How to prove $\phi(z)=\sum\limits_{m\in Z}(\sum\limits_{n\in Z} \frac{1}{(m-1+nz)(m+nz)})=2-\frac{2\pi i}{z}$ with $Im(z)>0$ and $(m,n)\neq(0,0),(1,0)$?

For $m$ fixed, $a_m=\sum\limits_{n\in Z} \frac{1}{(m-1+nz)(m+nz)}=\sum\limits_{n\in Z} b_n$ converges absolutely since $(m-1+nz)(m+nz)=O(n^2z^2)$, so $a_m$ is independent of the summation order of $n$, but it still needs $\sum\limits_{m}a_m$ to be converged absolutely in order to be independent of the summation order of $m$, and that's hard to show.

It may be useful that $\frac{1}{(m-1+nz)(m+nz)}=\frac{1}{m-1+nz}-\frac{1}{m+nz}$, but I don't know how to proceed further, so how to prove $\phi(z)=2-\frac{2\pi i}{z}$ with $Im(z)>0$?

CYC
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WolframAlpha evaluates the inner sum as $$-\frac{\pi}z\left(\cot\left(\frac{\pi m}z\right)+\cot\left(\frac{\pi-\pi m}z\right)\right)$$
The following answer explains why the cotangent appears. Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$

For the rest of the problem, could you turn it into sums of exponentials?
$$\cot x = \frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}\\ =1+2e^{-2ix}+2e^{-4ix}+...$$

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Empy2
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