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Math Help - laurent series

  1. #1
    Junior Member
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    laurent series

    I need a small bit of help finding the laurent series of

    (e^Z)/(Z-Z^2) that converge for 0 < |Z| < R. I also need to find R.

    i took z out of the bottom line leaving me with (1/z)(e^z/1-z). As far as i know i should only be interested in the case of z=1 as it's the first non-analytic point after z = 0. Correct?.

    Im kinda lost on how to continue this problem.. should i be using the maclaurin series somewhere?

    thanks
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  2. #2
    MHF Contributor

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    Use the geometric series to argue that \frac{1}{1- z}= \sum_{n=0}^\infty z^n. For what values of z does that converge? Dividing each term by z gives \frac{1}{z(1- z}= \sum_{n=-1}^\infty, a "Laurent" series since it now includes a negative power.

    Of course e^z= \sum_{n=0}^\infty \frac{z^n}{n!}. Multiply those two series term by term.

    Alternatively find the MacLaurin series for the function \frac{e^z}{1- z} which is analytic at z= 0. Then divide each term by z to get the Laurent series for \frac{e^z}{z(1- z)}
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