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November 7th, 2012, 08:23 AM  #1 
Newbie Joined: Sep 2012 Posts: 29 Thanks: 0  Positively oriented circle...
{z:zz0=p}. Find ?_c_1(0) of (e^z + cosz)z^1 dz I know it should be 4pi(i) 
November 7th, 2012, 08:25 AM  #2 
Newbie Joined: Sep 2012 Posts: 29 Thanks: 0  Re: Positively oriented circle...
actually is Find ?_c_1(0) of (e(z) + cos(z))z^1 dz

December 4th, 2012, 08:14 AM  #3 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Positively oriented circle...
The integral around a closed path (in this case a circle), is equal to times the sum of the residues at each pole of the integrand inside the circle. Here, everything is analytic except for that so the only pole is at z= 0 and that is a pole of order 1. I would expand as a Taylor's series around z= 0 and multiply each term by 1/z to get the Laurent series for the entire integrand. The residue is the coefficient of in the Laurent series. Or you can do it directly. I presume that by "C_1(0)" is the unit circle with center at z= 0. A standard parameterization for the circle is so we can write the integral . That looks difficult to me so I would use the "residue" method! 
December 6th, 2012, 12:14 AM  #4 
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: Positively oriented circle...
[color=#000000]I guess you mean, .[/color] 

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