$\displaystyle f(z)=\frac{1}{sinh(z)}$
It's ok that z=0 is a singularity, but why
$\displaystyle 2ik\pi$, where k is a integer, are singularities too? sinh(z)=0 only when z=0
Regards
You asked if $\displaystyle 2k\pi i$ were singularities of $\displaystyle f(z)$ and the answer is yes. If you want to find all of them:
$\displaystyle \sinh z=\dfrac{e^z-e^{-z}}{2}=0\Leftrightarrow\ldots\Leftrightarrow e^{2z}=1\Leftrightarrow\ldots\Leftrightarrow z=k\pi i\quad (k\in\mathbb{Z})$