# Verify that my contour integral is correct

1. May 15, 2012

### aaaa202

1. The problem statement, all variables and given/known data
Evaluate $\oint$$\frac{1}{a-cos\theta}$ d$\theta$ from [0:2$\pi$]

2. Relevant equations
Method of contour integration.

3. The attempt at a solution
I make the substitution z=exp(i$\theta$) $\Rightarrow$
giving me the integral:

$\oint$1/(a-½(z+z-1)) dz/iz from [0:2$\pi$]
And rearranging:
2/i $\oint$1/(-z2+2az-1) dz from [0:2$\pi$]
The quadratatic equation in the denominator has the roots z0 = a ± √(a2-1). Only the one with negative sign is inside the unit circle for a>1 (actually, how can you know that it will be that for all a>1?) which gives:

Pole at z0 = a - √(a2-1)

So we can write f(z) as:

f(z) = 2i/((z-a + √(a2-1))(z-a - √(a2-1)))

And since it's a simple pole we're dealing with the residue can be found as:

limz->z0{(z-z0)f(z)} = limz->a - √(a2-1){(z-a+√(a2-1) 2i/((z-a + √(a2-1)(z-a - √(a2-1))} = limz->a - √(a2-1){2i/(z-a - √(a2-1)} = 2i/-2√(a2-1) = -i/√(a2-1)

And the residue theorem gives the final value for the integral, I:

I = 2$\pi$/√(a2-1)

Is this correct?

Last edited: May 15, 2012
2. May 15, 2012

### vela

Staff Emeritus
Yes, that's correct.

Try writing $a=1+\delta$ and plug that expression into $a-\sqrt{a^2-1}$. You should be able to see then that the pole will lie inside the unit circle.