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I'm going through some practice prelim exams, and one of the questions asks to compute $$ \int_{|z|=2}\frac{z^4}{z^5-z-1}dz. $$

The integrand is not quite in the form $f'/f$ to count zeroes and poles inside the circle. I wanted to compute resides, but finding the poles at which to calculate them seems more difficult than usual. What is the more doable approach?

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up vote 4 down vote accepted

Make the substitution $w = \frac{1}{z}$ so that $-\frac{dw}{w^2} = dz$. The integral becomes (the negative sign goes away when we cancel it with the new orientation of the circle) $$ \int_{|w| = \frac{1}{2}} \frac{w^{-6}}{w^{-5} - w^{-1} - 1}dw $$ which is $$ \int_{|w| = \frac{1}{2}} \frac{1}{w - w^5 - w^6}dw $$ The only root of the new denominator inside the new circle is 0, so you can just use residue theory from here.

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Thanks, so it evaluates just to $2\pi i$. – Kotsay Apr 28 '12 at 5:24

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