How to compute this series
$$\sum_{n=-\infty}^\infty \frac1{a^2+(z+2n\pi)^2}~~~~~~~a, z\in\mathbb R$$
Wolfram gives the following result:
$$\sum_{n=-\infty}^\infty \frac1{a^2+(z+2n\pi)^2}=\frac1{4a}\coth\frac{a-iz}{2}+\frac1{4a}\coth\frac{a+iz}{2}+\frac1{2a}\lfloor \frac{\arg(z+ia)}{2\pi}\rfloor$$
I know how to compute the series
$$\sum_{n=-\infty}^\infty \frac1{a^2+(2n\pi)^2}$$
which is the Mittag-Leffler's series. But for this problem, there is a shift $(z+2n\pi)$, and how to deal with this? Another question is about the term
$$\frac1{2a}\lfloor \frac{\arg(z+ia)}{2\pi}\rfloor$$
which is appearing in Wolfram's output. I don't know why there is this term, especially, if both $z$ and $a$ are real. Any hint will be appreciated.