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In the complex plane, let C be the circle |z| = 2 with positive (counterclockwise) orientation. Show that:
[tex]\int _C \frac{dz}{(z-1)(z+3)^2} = \frac{\pi i}{8}[/tex]
This isn't homework, it was a problem in one of the practice GREs. It looks like a straightforward application of the residue theorem, the only problem is that I never understood the second half of complex analysis? The only pole of the integrand contained in the interior of C is 1, so I have to find the residue of the integrand at 1? What's a residue, and how do I find it? Mathworld gives a definition in terms of the Laurent series, but I'm sure there's a simpler way when it comes to basic rational functions like these. And once I have the residue, what do I do, multiply by [itex]2\pi i[/itex]? Thanks.
[tex]\int _C \frac{dz}{(z-1)(z+3)^2} = \frac{\pi i}{8}[/tex]
This isn't homework, it was a problem in one of the practice GREs. It looks like a straightforward application of the residue theorem, the only problem is that I never understood the second half of complex analysis? The only pole of the integrand contained in the interior of C is 1, so I have to find the residue of the integrand at 1? What's a residue, and how do I find it? Mathworld gives a definition in terms of the Laurent series, but I'm sure there's a simpler way when it comes to basic rational functions like these. And once I have the residue, what do I do, multiply by [itex]2\pi i[/itex]? Thanks.