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In my complex analysis course, we're asked to calculate the following integrals:

$$\int_{|z|=2}\frac{1}{z^2-1}\,dz\quad\text{and}\quad\int_{|z|=2}z^n(1-z)^m\,dz$$ where $m,n\in\Bbb Z$ for the second integral. I was wondering, for the first one, it seems I can approach it using partial fractions, and then use a keyhole argument, but it seems like this gives zero as an answer; is this correct?

For the second integral, I broke it down into four cases: $(1)$ we have $m,n\geq0$, which is the trivial case and gives zero, $(2)$ we have $m\geq0$ and $n<0$, $(3)$ we have $m<0$ and $n\geq0$, and $(4)$ we have $m,n<0$. Should I approach the latter cases as above; form a keyhole contour and integrate?

I ask these together because it seems like the approaches should be similar, but I'm not sure mine is the easiest/most straightforward. Does anyone have any suggestions about my approach? Thanks!

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For calculating the first one, you might recheck your calculations; it comes out to $4\pi i$. – Clayton Feb 19 '13 at 2:32
@Argon Is it $0$ or $4\pi i$? I know that if it is just one circle around the origin, I get $2\pi i$, so I was guessing this one was zero or $4\pi i$ (since it the circle has two problem points in the interior). – anon271828 Feb 19 '13 at 3:24
@Argon: Why do you think the integral is $0$? If one calculates in the manner that anon describes, I am pretty sure you get $4\pi i$. – Clayton Feb 19 '13 at 4:43
@Argon: You are correct; I went back and reworked it. I had a $+$ where I should have had a minus. Sorry anon271828, the answer is, indeed $0$ for the first integral. – Clayton Feb 20 '13 at 2:38

For the first one:

$$\oint_{|z|=2} \frac{dz}{z^2-1} = \int_{|z|=2} \frac{dz}{(z+1)(z-1)}$$

Considering any closed contour that encloses $z = 1$, $C$:

$$ \oint_{C} \frac{\frac{1}{z+1}}{z-1}dz = 2 \pi i \frac{1}{1+1} = \pi i $$

Considering any closed contour that encloses $z = -1$, $\Gamma$:

$$ \oint_{\Gamma} \frac{\frac{1}{z-1}}{z+1}dz = 2 \pi i \frac{1}{-1-1} = -\pi i $$

by Cauchy's Integral Formula.

Add the values of the integrals to get $0$ that the integral of $|z| = 2$ is zero.

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Note that, for the first integral the poles are at $z=1$ and $z=-1$ and they lie inside the contour $|z|=2$. So the integral equals the sum of residues.

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