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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  
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] Quote:
Quote:
 
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<za<some r just trying to understand it properly 
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 CauchyGourstat formula is the following, 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 .[/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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