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 November 3rd, 2011, 12:10 AM #11 Member   Joined: Nov 2011 Posts: 40 Thanks: 0 Re: Complex integration using CIF Thanks ZardoZ =D, Hey I just noticed that when I calculate the residue of the simple pole at z=5 i get 5^(-n) as well, but this is irrelavent right because that pole is outside of the circle |z|<2 ? So when your finding the integral using the residue your interested in the sum of the residues within the curve of integration? Sorry if its a silly question
November 3rd, 2011, 03:14 AM   #12
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Re: Complex integration using CIF

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Quote:
 Originally Posted by Grayham1990 Thanks ZardoZ =D, Hey I just noticed that when I calculate the residue of the simple pole at z=5 i get 5^(-n) as well, but this is irrelavent right because that pole is outside of the circle |z|<2 ?
Look at the 4th post from above.

Quote:
 So when your finding the integral using the residue your interested in the sum of the residues within the curve of integration? Sorry if its a silly question
Yes multiplied by 2?i.[/color]

 November 3rd, 2011, 03:35 AM #13 Member   Joined: Nov 2011 Posts: 40 Thanks: 0 Re: Complex integration using CIF So that the Res at z=5 is only that when you take the exterior to the circle, and z=0 is 0 because it isnt within the region your looking at, is that the correct way to see it? So in the actual question the residue of z=5 is zero because its outside the circle where as you can find the residue at z=0 because its inside? Sorry for being a pest, my notes just say: the residue is this given by this formula for 0<|z-a|
 November 3rd, 2011, 03:57 AM #14 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Complex integration using CIF [color=#000000]The Cauchy-Gourstat formula is the following, $\oint_{C}f(z)\;dz=2\pi i \sum_{k}\textrm{Res}\left(f(z);z_{k}\right)$if there are any poles in the interior, the exterior poles are insignificant and we do not use them, this doesn't mean that the residue at these (exterior) poles is zero. I showed you two ways, in the second I inversely used the exterior as the interior, where you had a two poles in the interior z=5 and $z=+\infty$.[/color]
 November 3rd, 2011, 04:37 AM #15 Member   Joined: Nov 2011 Posts: 40 Thanks: 0 Re: Complex integration using CIF Thanks so much for all your help ZardoZ!!

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