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I'm stuck on a few of these, but I have most of the details worked out:

(i) $\int_{|r|=1}(z^2-4)^{-1}\,dt=\int_{0}^{2\pi}ie^{i\theta}(e^{2i\theta}-2)^{-1}\,d\theta$

(ii) $\int_{|r|=1}(z^2-2z)^{-1}\,dt=\int_{0}^{2\pi}ie^{i\theta}(e^{2i\theta}-2e^{i\theta})^{-1}\,d\theta$

In both integrals I'm using $z=e^{i\theta}$, and $0 < \theta \leq 2\pi$.

I couldn't think of any integral tricks to make these work out, so I tried putting them into Mathematica. The first yielded something with ArcTanh. I became suspicious because my instructor had us explicity skip Hyperbolic trig functions. The second one I'm not at all sure how to simplify.

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What is $r$, $t$, $z$? How are they related? Have you heard about residue theorem? – TZakrevskiy Feb 3 '14 at 14:34
I fixed it and I do not have the benefit of the residue theorem. – James Feb 3 '14 at 14:43
up vote 0 down vote accepted

The first one do not even bother to compute, as its value is zero. And it is zero because $$ f(z)=\frac{1}{z^2-4} $$ is analytic in the unit disk, and hence its integral on the unit circle is equal to zero.

In the case of the second one, $f(z)=\dfrac{1}{z^2-2z}$ has a single pole at $z=0$. The only singularity inside the unit circle. Then $$ \int_{|z|=1}\frac{dz}{z^2-2z}=2\pi i\,\mathrm{Res}\,(z^2-2z, z=0)=2\pi i\lim_{z\to 0} zf(z)=-\pi i. $$

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I don't have the knowledge of the residue theorem. – James Feb 3 '14 at 14:43

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