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How to find this Laurent series?

  1. May 7, 2007 #1
    I understand perfectly well how to do Taylor series, but I am foggy on these Laurent series. Say, we have something like,

    [tex]f\left( z \right)\; =\; \frac{1}{z^{2}\cdot \sin \left( z \right)}[/tex]

    I think I need to use the taylor series expressions for sin(z) but otherwise, I am not sure what to do about that z^2. If I use that in a taylor series with z=0, then I get a singularity.

    Since its 1/sin(z), do I just inverse the taylor series for sin(z) Yes, you can see I am very unknowledgeable about this. I turn here because the explanations I found in my book are totally lacking.
  2. jcsd
  3. May 7, 2007 #2
    It's clear that z=0 is a pole for your function, thats why the Taylor series is not well defined there. The Laurent series will tell you how this singularity behaves (how it grows to infinity).

    A Laurent series has the form

    [tex]\sum_{n=1}^\infty \frac{b_n}{z^n}+\sum_{n=0}^\infty a_nz^n,[/tex]

    so you are right, in order to calculate the series, you just have write down the series for [itex]\csc z[/itex] and multiply it by [itex]z^{-2}[/itex].
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