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