Residue of a function

  • Thread starter DrKareem
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  • #1
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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
 
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Answers and Replies

  • #2
marcusl
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Perfect, except you need z^(-2) in the 3rd term above (i'm sure you made a typo).
 
  • #3
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Oh yes it is a typo. Thanks for the input.
 
  • #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.
 
  • #5
905
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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).
 
  • #6
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Halls is correct. The residue is defined only for a pole of finite order, which you do not have.
 
  • #7
101
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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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