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 Post subject: Contour IntegrationPosted: Mon, 4 Apr 2011 15:39:51 UTC
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Joined: Mon, 21 Mar 2011 21:36:12 UTC
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C is the circle |z|=2, oriented counterclockwise in the complex plane.

What I did: There is a pole at z=1 (order 1) and a pole at z=-3 (order 2). Using the residue theorem, I calculated the residues as 1/16 and -1/16, respectively, so I was expecting the value of the integral to be zero, but that's incorrect. Any ideas what I did wrong?

Edit: I see the problem now. z=-3 is not in C. Now I need to figure out what to do in a case like this...

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 Post subject: Re: Contour IntegrationPosted: Mon, 4 Apr 2011 15:46:50 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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StarKid wrote:
C is the circle |z|=2, oriented counterclockwise in the complex plane.

What I did: There is a pole at z=1 (order 1) and a pole at z=-3 (order 2). Using the residue theorem, I calculated the residues as 1/16 and -1/16, respectively, so I was expecting the value of the integral to be zero, but that's incorrect. Any ideas what I did wrong?

Edit: I see the problem now. z=-3 is not in C. Now I need to figure out what to do in a case like this...

Cauchy representation formula...

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 Post subject: Re: Contour IntegrationPosted: Mon, 4 Apr 2011 15:47:25 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
StarKid wrote:
C is the circle |z|=2, oriented counterclockwise in the complex plane.

What I did: There is a pole at z=1 (order 1) and a pole at z=-3 (order 2). Using the residue theorem, I calculated the residues as 1/16 and -1/16, respectively, so I was expecting the value of the integral to be zero, but that's incorrect. Any ideas what I did wrong?

Edit: I see the problem now. z=-3 is not in C. Now I need to figure out what to do in a case like this...

You just use the residue theorem as usual. There's no problem. You sum the residues inside the curve, that's the rule.

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