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Math Help - complex variables]Isolated singularities

  1. #1
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    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
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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  3. #3
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    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:

    <br />
\begin{array}{rc@{}c}<br />
& \multicolumn{2}{l}{\, \, \, \frac{1}{z}} \vspace*{0.12cm} \\<br />
\cline{2-3}<br />
\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} \\<br />
& \multicolumn{2}{l}{\, \, \, 1+\frac{z}{2}+\frac{z^2}{6}+\dotsb}<br />
\end{array}<br />
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  4. #4
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    thanks to you all!!!
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