# singularity

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• May 12th 2011, 05:08 AM
hurz
singularity
$\displaystyle f(z)=\frac{1}{sinh(z)}$
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
• May 12th 2011, 05:24 AM
FernandoRevilla
$\displaystyle \sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$
• May 12th 2011, 05:29 AM
hurz
Quote:

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

$\displaystyle sinh(ki\pi)=0$

so, why the argument of the sinh must be an even multiple of $\displaystyle i\pi$ ?
• May 12th 2011, 07:51 AM
FernandoRevilla
Quote:

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

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:

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