# Thread:

1. ## singularity

$f(z)=\frac{1}{sinh(z)}$
It's ok that z=0 is a singularity, but why
$2ik\pi$, where k is a integer, are singularities too? sinh(z)=0 only when z=0

Regards

2. $\sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$

3. Originally Posted by FernandoRevilla
$\sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$
$sinh(ki\pi)=0$

so, why the argument of the sinh must be an even multiple of $i\pi$ ?

4. Originally Posted by hurz
$sinh(ki\pi)=0$ so, why the argument of the sinh must be an even multiple of $i\pi$ ?

You asked if $2k\pi i$ were singularities of $f(z)$ and the answer is yes. If you want to find all of them:

$\sinh z=\dfrac{e^z-e^{-z}}{2}=0\Leftrightarrow\ldots\Leftrightarrow e^{2z}=1\Leftrightarrow\ldots\Leftrightarrow z=k\pi i\quad (k\in\mathbb{Z})$
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