Thread: Integral
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Old January 1st, 2009, 03:46 AM
asi123 asi123 is offline
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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.
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