How Does the Value of 'a' Affect the Integral on the Unit Circle Using Residues?

[darthmonkey]

Homework Statement

Calculate the integral ∫dθ/(1+acos(θ)) from 0 to 2∏ using residues.

Homework Equations

Res\underline{zo}(z)=lim\underline{z->zo} (z-z0)f(zo)*2∏i

The Attempt at a Solution

To start I sub cos(θ)=1/2(e^(iθ)+e^(-iθ)) so that de^(iθ)=ie^(iθ)dθ

Re-writing in terms of e^(iθ) and then subbing in z for e^(iθ) I get dz/(iz(1+a(1/2)(z+z*))

This becomes -2idz/(az^2+2z+a). Solving for the pole I get two imaginary roots. The problem is this is all done on the unit circle and the roots I'm getting depend on a so it means the poles are not necessarily in the circle. One should be. 0 is another root but is trivial.

Also I tried solving this integral numerically and it gave me 0 but the answer should be 2∏/√(1-a^2)

 

[haruspex]

I think the answer depends on whether a > 1. If it is, the roots are complex and both lie inside the unit circle, and they cancel. If < 1, only one root lies inside the circle, so you ignore the other.

 

[vela]

Solving for the pole I get two imaginary roots. The problem is this is all done on the unit circle and the roots I'm getting depend on a so it means the poles are not necessarily in the circle.

When |a|>1, you should get two complex roots, namely $z=-\frac{1}{a} \pm i\sqrt{1-\left(\frac{1}{a}\right)^2}$. You should be able to convince yourself that both roots lie on the unit circle.

0 is another root but is trivial.

z=0 isn't a root.

Also I tried solving this integral numerically and it gave me 0 but the answer should be 2∏/√(1-a^2)

From the answer, you should see that you must have |a|<1, so that's the case you're primarily interested in.