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Which type of singularity the function
$ f(z)= \frac{e^z}{z(1- e^{-z})} $ at $ z=0 $

I Know, we find the singularity of a function by expanding the given function by using Laurent series expansion.

I tried like this. $ f(z)= \frac{e^z}{z(1- e^{-z})} $
$ f(z)= \frac{1}{z}{e^z}{(1- e^{-z})^{-1}} $
$ f(z)= \frac{1}{z}({e^z+1+e^{-z}+e^{-2z}+\dots)} $
from this i concluded that this function has essential singularity at $z=0$.
Is my conclusion correct??

Please help me. thank you.

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$z=0$ is a zero both for $z$ and for $1-e^{-z}$, while the numerator is never zero. Hence it is a pole of order $2$. – Federica Maggioni Apr 26 '13 at 6:34
$e^{-z}$ has infinite terms, how pole of order 2? can you explain.... @FedericaMaggioni – prasad Apr 26 '13 at 6:40
up vote 3 down vote accepted

1. Function $f(z)$ can be rewritten as $$f(z)= \dfrac{e^z}{z(1- e^{-z})}=\dfrac{e^{2z}}{z({e^{z}-1})}=\dfrac{ze^{2z}}{z^2({e^{z}-1})}=\dfrac{e^{2z}}{z^2}h(z)$$ 2. Function $h(z)=\dfrac{z}{e^{z}-1}$ has a removable singularity at $z=0.$

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later function has a pole of order 2. so, what is the conclusion? @M.Strochyk – prasad Apr 26 '13 at 6:46
Thus $f(z)$ has a pole of order $2$ at $z=0.$ – M. Strochyk Apr 26 '13 at 6:56

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