Verify that my contour integral is correct

[aaaa202]

Homework Statement

Evaluate \oint\frac{1}{a-cos\theta} d\theta from [0:2\pi]

Homework Equations

Method of contour integration.

The Attempt at a Solution

I make the substitution z=exp(i\theta) \Rightarrow
giving me the integral:

\oint1/(a-½(z+z^-1)) dz/iz from [0:2\pi]
And rearranging:
2/i \oint1/(-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:

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

 

[vela]

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.