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Finding the residue of e^(-1/z^2) at z=0

  1. Apr 1, 2009 #1
    This may be because im tired but i cant find the residue of

    [itex]e^{-\frac{1}{z^2}}[/itex] at [itex]z=0[/itex]

  2. jcsd
  3. Apr 2, 2009 #2


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    Re: Residue

    You know the Taylor's series for ez about z= 0 don't you? Replace z in that by -1/z2 to get a Laurent series for this function.
  4. Apr 2, 2009 #3
    Re: Residue



    so the residue is 0?

    i was going to do this but i got confused - why is this an acceptable form of the laurent series - we derived it from the taylor series which doesn't admit negative n values. so surely there could be some other terms that we're not taking into account?
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