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Math Help - Laurent series

  1. #1
    Senior Member Twig's Avatar
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    Laurent series

    Hi

    Complex Analysis....

    Problem:
    Find the Laurent series for  f(z)=\frac{z^{2}}{z^{2}-1} about the point z_{0}=1

    I got that f(z)=1+\frac{1}{2}\cdot \frac{1}{z-1}-\frac{1}{2}\cdot \frac{1}{z+1} , which I wrote as 1+\frac{1}{2}\cdot \frac{1}{z-1}+\frac{1}{4}\cdot \sum_{k=0}^{\infty}\left(\frac{z-1}{4}\right)^{k}

    Need some input.

    I guess all the terms need to be expressed as powers of (z-1) .

    Thx
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  2. #2
    Super Member
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    \frac{1}{z+1}=\frac{1}{2}\left(\frac{1}{1-(-\frac{z-1}{2})}\right)=\frac{1}{2}\sum_{k=0}^{\infty}\frac  {(-1)^k (z-1)^k}{2^k}

    so:

    \frac{z^2}{z^2-1}=1+\frac{1}{2}(z-1)^{-1}-\frac{1}{4}\sum_{k=0}^{\infty}\frac{(-1)^k (z-1)^k}{2^k}

    Try checking your work in Mathematica. I'm not saying let the software do it for you but rather let the software check your work. Enter:

    Code:
    Series[z^2/(z^2-1),{z,1,5}]
    and then compare that to what you obtain manually.
    Last edited by shawsend; September 23rd 2009 at 01:52 PM. Reason: corrected code
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  3. #3
    Senior Member Twig's Avatar
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    thanks, IŽll do that
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