Thread: Integral
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Old January 1st, 2009, 03:53 AM
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Originally Posted by asi123 View Post
Well, I think you are right with the second idea because we exactly learning how to work with Cauchy's integral formula so it's got to be it
My question due is how can you break it into f(z)=\frac{i}{(z-a)(az-1)}?

Thanks a lot.
Substitute z = e^{it}:

1. dt = \frac{dz}{iz} = \frac{-i \, dz}{z}.

2. \cos t = \frac{e^{it} + e^{-it}}{2} = \frac{z + \frac{1}{z}}{2} = \frac{z^2 + 1}{2z}.

Simplify. The integrand becomes \frac{i}{az^2 - z - a^2 z + a} which factorises in the way Mathstud gave.
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