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Residue theorem and poles of order > 1

 
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PostPosted: Sun, 7 Jan 2007 18:21:26 UTC    Post subject: Residue theorem and poles of order > 1 Reply with quote
http://en.wikipedia.org/wiki/Residue_theorem
It doesn't say anything about uniqueness regarding the set , does this mean that if I have a function like 1/(z-1)^2 then the integral around the circle |z| = 2 will equal or ?

Thank you.
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Kungsman
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PostPosted: Sun, 7 Jan 2007 19:07:28 UTC    Post subject: Re: Residue theorem and poles of order > 1 Reply with quote
Kahsi wrote:
http://en.wikipedia.org/wiki/Residue_theorem
It doesn't say anything about uniqueness regarding the set , does this mean that if I have a function like 1/(z-1)^2 then the integral around the circle |z| = 2 will equal or ?

Thank you.


Clearly the residue at z=1 of 1/(z-1)^2 is 0. Hence the integral is 0 too.
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PostPosted: Sun, 7 Jan 2007 19:25:37 UTC    Post subject: Reply with quote
lol, I took a bad example. But in general, if we have a pole of order greater than 1 will the residue be counted twice in the sum of the residues?
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Opalg
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PostPosted: Sun, 7 Jan 2007 19:39:41 UTC    Post subject: Reply with quote
Kahsi wrote:
lol, I took a bad example. But in general, if we have a pole of order greater than 1 will the residue be counted twice in the sum of the residues?

No. The formula in the residue theorem stays the same regardless of the order of the pole, and the residue of f at a_k is always the coefficient of in the Laurent series of f about a_k.
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