Contour Integrals: Evaluating I= \oint \, \frac{dz}{(z^2-1)}

[kreil]

Homework Statement

Evaluate the following integral along two different contours, (a) a circle of radius |z|=1/2 centered at the origin, and (b) a circle of radius |z|=3 centered at the origin.

Homework Equations

$$I= \oint \, \frac{dz}{(z^2-1)}$$

The Attempt at a Solution

I'm not sure that I'm doing this right because I keep getting zero for both integrals (maybe the limits are incorrect?)...

(a) Along this circle, |z|=r=1/2 and $z=re^{i \theta} \, \, dz = i r e^{i \theta} d \theta$ and the integral becomes

$$I= \oint \, \frac{dz}{(z^2-1)}= \oint \frac{i r e^{i \theta}d \theta}{r^2-1} = \frac{ir}{r^2-1} \oint_0^{2 \pi}e^{i \theta}d \theta = \frac{ir}{r^2-1} [sin(\theta)-i cos(\theta)]|_0^{2 \pi} = \frac{ir}{r^2-1}(-i+i)=0$$

In part (b) the value of r changes but the integrand stays the same so I get zero again..what am I doing wrong?

 

[Dick]

You should get zero for both integrals. But you are also not doing it right. z^2 isn't r^2, it's r^2*exp(2*i*theta). Aren't you supposed to be doing this with the residue theorem?

 

[ideasrule]

In part (b) the value of r changes but the integrand stays the same so I get zero again..what am I doing wrong?

ANY contour integral will give you exactly the same result, so long as the same poles are surrounded. Put in another way, contour integrals depend only on what poles they enclose (leading to the residue theorem that Dick mentioned), not on the path of integration.

 

[kreil]

You should get zero for both integrals. But you are also not doing it right. z^2 isn't r^2, it's r^2*exp(2*i*theta). Aren't you supposed to be doing this with the residue theorem?

I thought you when you square z you use the complex conjugate (canceling the e's to 1)..?

$$z^2=zz^*=r^2 e^{i \theta}e^{-i \theta}=r^2$$

Also, we haven't gotten to the residue theorem yet in the class so I don't believe I need to use it

 

[Dick]

zz* should be written as |z|^2. If someone is being really sloppy, like a physicist (and I'm one, so don't get me wrong here) they might just write z^2 instead. But I don't think that's what's meant here. If you haven't done the residue theorem, then you might have to do it the hard way. exp(2*i*theta)=cos(2*theta)+isin(2*theta). Try and convert it to real integrals and work each one out.