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
The most confortable way starts with the Taylor expansion...
$\displaystyle \frac{z} {e^{z}-1} = \sum_{n=0}^{\infty} B_{n}\cdot \frac{z^{n}}{n!}$ (1)
... where the $\displaystyle B_{n}$ are the so called 'Bernoulli numbers'. The first $\displaystyle B_{n}$ are ...
$\displaystyle B_{0}=1$
$\displaystyle B_{1}=-\frac{1}{2}$
$\displaystyle B_{2}=\frac{1}{6}$
$\displaystyle B_{3}=0$
$\displaystyle B_{4}=-\frac{1}{30}$
$\displaystyle \dots$ (2)
Except for $\displaystyle n=1$ all the $\displaystyle B_{n}$ with n odd are null. From (1) and (2) we derive that...
$\displaystyle f(z)= \frac{1}{e^{z}-1} = \frac{1}{z} - \frac{1}{2} + \frac{z}{12} - \frac{z^{3}} {720} + \dots$ (3)
... so that the residue of $\displaystyle f(*)$ in $\displaystyle z=0$ is $\displaystyle 1$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
You can also just use long division:
$\displaystyle \frac{1}{e^z-1}=\frac{1}{z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots}$
I'll do the first one:
$\displaystyle
\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}
$