1. ## complex variables]Isolated singularities

Find the Laurent series for the given function about the indicated point and give the residue of the function at the point

1/(e^z-1)
zo=0 (four terms of the Laurent series)

I need help!!!plz

2. The most confortable way starts with the Taylor expansion...

$\frac{z} {e^{z}-1} = \sum_{n=0}^{\infty} B_{n}\cdot \frac{z^{n}}{n!}$ (1)

... where the $B_{n}$ are the so called 'Bernoulli numbers'. The first $B_{n}$ are ...

$B_{0}=1$

$B_{1}=-\frac{1}{2}$

$B_{2}=\frac{1}{6}$

$B_{3}=0$

$B_{4}=-\frac{1}{30}$

$\dots$ (2)

Except for $n=1$ all the $B_{n}$ with n odd are null. From (1) and (2) we derive that...

$f(z)= \frac{1}{e^{z}-1} = \frac{1}{z} - \frac{1}{2} + \frac{z}{12} - \frac{z^{3}} {720} + \dots$ (3)

... so that the residue of $f(*)$ in $z=0$ is $1$ ...

Kind regards

$\chi$ $\sigma$

3. You can also just use long division:

$\frac{1}{e^z-1}=\frac{1}{z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots}$

I'll do the first one:

$
\begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, \frac{1}{z}} \vspace*{0.12cm} \\
\cline{2-3}
\multicolumn{1}{r}{z+\frac{z^2}{2}+\frac{z^3}{6}+\ dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} 1} \\
& \multicolumn{2}{l}{\, \, \, 1+\frac{z}{2}+\frac{z^2}{6}+\dotsb}
\end{array}
$

4. thanks to you all!!!