My Math Forum Positively oriented circle...

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 November 7th, 2012, 08:23 AM #1 Newbie   Joined: Sep 2012 Posts: 29 Thanks: 0 Positively oriented circle... {z:|z-z0|=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 $\frac{1}{2\pi i}$ times the sum of the residues at each pole of the integrand inside the circle. Here, everything is analytic except for that $z^{-1}$ so the only pole is at z= 0 and that is a pole of order 1. I would expand $e^z+ cos(z)$ 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 $z^{-1}$ 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 $z= e^{i\theta}$ so we can write the integral $i\int_0^{2\pi} e^{e^{i\theta}}- cos(e^{i\theta}) e^{i\theta} d\theta$. 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, $\oint_{|z|\leq 0}\frac{\cos(z)+e{z}}{z}\;\textrm{d}z=2\pi i\left[ \textrm{Residue}\left(\frac{\cos(z)+e^{z}}{z};z=0\ right)\right]=2\pi i \lim_{z\to0}z\frac{\cos(z)+e^{z}}{z}=2\pi i \left(\cos(0)+e^{0}\right)=2\pi i \left(1+1\right)=4\pi i$ .[/color]

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