# Unit circle and the roots

1. Dec 20, 2012

### darthmonkey

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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)

Last edited by a moderator: Feb 4, 2013
2. Dec 20, 2012

### haruspex

Re: Residues

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.

3. Dec 21, 2012

### vela

Staff Emeritus
Re: Residues

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.

z=0 isn't a root.

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