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Thread: 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...

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

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    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}
    $
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  4. #4
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    thanks to you all!!!
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