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.

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.

share|improve this question
1  
What is $r$, $t$, $z$? How are they related? Have you heard about residue theorem? –  TZakrevskiy Feb 3 at 14:34
    
I fixed it and I do not have the benefit of the residue theorem. –  James Feb 3 at 14:43

1 Answer 1

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 $$ \int_{|z|=1}\frac{dz}{z^2-2z}=2\pi i\,\mathrm{Res}\,(z^2-2z, z=0)=2\pi i\lim_{z\to 0} zf(z)=-\pi i. $$

share|improve this answer
    
I don't have the knowledge of the residue theorem. –  James Feb 3 at 14:43

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.