Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.