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Thread: Help with Laurent Expansions

  1. #1
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    Help with Laurent Expansions

    Basically I've been working through the problem below, and when I went to check my answers they differed, so I was wondering are the answers below the question both the same? Thanks.

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  2. #2
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    Quote Originally Posted by Monkens View Post
    Basically I've been working through the problem below, and when I went to check my answers they differed, so I was wondering are the answers below the question both the same? Thanks.



    $\displaystyle z^2-3iz-2=(z-i)(z-2i)\Longrightarrow $ the function $\displaystyle f(z)=\frac{1}{z^2-3iz-2}$ is analytic in the given

    annulus and thus its Laurent expansion there is just a power series...

    Tonio
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  3. #3
    MHF Contributor chisigma's Avatar
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    First we separate the two terms of f(*)...

    $\displaystyle \displaystyle f(z)= \frac{1}{z^{2} -3\ i\ z -2} = \frac{i}{z-i} - \frac{i}{z-2\ i}$ (1)

    ... and then we consider that is...

    $\displaystyle \displaystyle \frac{-i}{z-2\ i} = \frac{-i}{-i + (z-i)}= \frac{1}{1+i (z-i)} = $

    $\displaystyle \displaystyle = \sum_{n=0}^{\infty} (-i)^{n} (z-i)^{n}$ (2)

    ... so that for $\displaystyle 0<|z-i|<1$ is...

    $\displaystyle \displaystyle \frac{1}{z^{2} -3\ i\ z -2} = \frac{i}{z-i} + \sum_{n=0}^{\infty} (-i)^{n} (z-i)^{n}$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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