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Residue of a function

  1. Feb 19, 2007 #1
    Hi. I'm trying to find the residue of

    [tex]\exp{\frac{1}{z}}[/tex]

    at z=0 since it is a pole, so I can integrate the function over the unit circle counterclockwise. I expanded this function in Laurent Series to get

    [tex]\exp{\frac{1}{z}} = 1 + \frac{1}{1!z} + \frac{1}{2!z^2}+ ...[/tex]

    So in this case the residue is the coefficient of 1/z which is 1. Is this method correct? There is no answer to it in the book...

    EDIT: If you guys can find the residue using another method, please teach me.
    EDIT 2: Fixed the typo :p
     
    Last edited: Feb 19, 2007
  2. jcsd
  3. Feb 19, 2007 #2

    marcusl

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    Perfect, except you need z^(-2) in the 3rd term above (i'm sure you made a typo).
     
  4. Feb 19, 2007 #3
    Oh yes it is a typo. Thanks for the input.
     
  5. Feb 19, 2007 #4

    HallsofIvy

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    Except that z= 0 is NOT a pole for e1/z. A point is a "pole of order n" for function f(z) is znf(z) is analytic but no zkf(z) is analytic for k< n. In particular, the Laurent series for f(z) has no power of z less than -n. Since the Laurent series for e1/z has all negative integers as powers, z= 0 is an essential singularity, not a pole.
     
  6. Feb 19, 2007 #5
    It's been a couple years since I took complex analysis, but yes, this seems like a perfectly accurate way to find the Laurent Series (and by consequence the residue as well).
     
  7. Feb 19, 2007 #6
    Halls is correct. The residue is defined only for a pole of finite order, which you do not have.
     
  8. Feb 20, 2007 #7
    Thank you HallsofIvy for clarifying this matter. This must be the reason why the other methods for finding the residue are not applicable, just like what Deadwolfe suggested. That explains a lot. And thanks for the rest for your input.
     
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