singularity

**hurz** · May 12th 2011, 05:08 AM

$f(z)=\frac{1}{sinh(z)}$
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

**FernandoRevilla** · May 12th 2011, 05:24 AM

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

**hurz** · May 12th 2011, 05:29 AM

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 $i\pi$ ?

**FernandoRevilla** · May 12th 2011, 07:51 AM

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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