Help finding value of Contour Integral

[jtleafs33]

Homework Statement

Evaluate the integral:

$\int$$^{2\pi}_{0}$$\frac{d\theta}{(A+Bcos(\theta))^2}$
$a^2>b^2$
$a>0$

The Attempt at a Solution

First, I convert this to contour integration along a full sphere in the complex plane.
I let:

$z=e^(i\theta)$
$dz=ie^(i\theta)$
$d\theta=-idz/z$
$cos(\theta)=(z+z^-1)/2$

Now, substituting back into the integral, I drop the integral sign for now and just work on the integrand:

=$\frac{-idz}{z(a+\frac{bz+bz^-1}{2})}$

=$\frac{-idz}{z(\frac{2a+bz+bz^-1}{2})}$

=$\frac{-4idz}{z(2a+bz+bz^-1)^2}$

=$\frac{-4idz}{b^2z^3+4abz^2+4a^2z+2b^2z+4ab+\frac{b^2}{z}}$


And from here I am stuck on how to find the poles of this function. I want to use the residue theorem to evaluate this integral, but like I said I'm stuck here.

 

[jackmell]

Homework Statement

Evaluate the integral:

$\int$$^{2\pi}_{0}$$\frac{d\theta}{(A+Bcos(\theta))^2}$
$a^2>b^2$
$a>0$

The Attempt at a Solution

First, I convert this to contour integration along a full sphere in the complex plane.
I let:

$z=e^(i\theta)$
$dz=ie^(i\theta)$
$d\theta=-idz/z$
$cos(\theta)=(z+z^-1)/2$

Now, substituting back into the integral, I drop the integral sign for now and just work on the integrand:

=$\frac{-idz}{z(a+\frac{bz+bz^-1}{2})}$

=$\frac{-idz}{z(\frac{2a+bz+bz^-1}{2})}$

=$\frac{-4idz}{z(2a+bz+bz^-1)^2}$

=$\frac{-4idz}{b^2z^3+4abz^2+4a^2z+2b^2z+4ab+\frac{b^2}{z}}$And from here I am stuck on how to find the poles of this function. I want to use the residue theorem to evaluate this integral, but like I said I'm stuck here.

You have to methodically consider all possibilities. I can think of three:

a>0, b>0 and a>b
a>0, b=0
a>0, b<0 and a^2>b^2

Ok, how about the first case. Just for starters, we'll let a=2 and b=1. When you make the subsittutions, you should get:

$$-4i\oint \frac{z dz}{\left(2az+b(z^2+1)\right)^2}$$

so the poles are when $2az+b(z^2+1)=0$. You can do that. Now, just for starters, let a=2 and b=1 and compute the poles for that particular function. Are they both in the unit circle? Check them. What's causing the poles to move in and out of the unit circle? Well, the particular values of a and b by virtue of the expression you get for the zeros of that quadratic expression right? So then those values will in turn determine the value of the integral by means of the Residue Theorem.

Just get a=2 and b=1 working. Analyze that expression for the zeros above based on the particular values of a and b. Get that one, then do the third one. The second just degrades to 2pi/a^2 right?