# Thread:

1. ## Help needed with algebra

Hi

I've been trying out some exam questions but have become stuck on one and was hoping someone out there might be able to help me.

The question is:
Evaluate using the method of residues
$\int_{0}^{2\pi}\frac{\cos2\theta \: d\theta}{1-2acos\theta + a^2}$

So far I have that
$\int_{0}^{2\pi}\frac{\cos2\theta\:d\theta}{1-2acos\theta+a^2}$ = $\frac{1}{2i}\int\frac{z^4+1}{z^3(a^2-az-az^-1+1)}$

as $cos2\theta= \frac{(z^2+z^-2)}{2}$and $d\theta=\frac{dz}{iz}$

I've tried to multiply all of this out and tidy up but the algebra is causing me problems from this point on and I was hoping you might be able to tell me where I'm going wrong.

Any suggestions would be gratefully appreciated

Thanks

2. What is $a$? What are the restictions on it? Is it $|a|<1$?

3. Originally Posted by michaela-donnelly
$\int_{0}^{2\pi}\frac{\cos2\theta\:d\theta}{1-2acos\theta+a^2}$ = $\frac{1}{2i}\oint\frac{z^4+1}{z^3(a^2-az-az^{-1}+1)}dz$
The denominator factorises as $z^2(a-z)(az-1)$. The double pole at z=0 is obviously inside the contour. The other poles are at z=a and z=a^{-1}. One of these will be inside the contour and the other one outside (unless |a|=1, in which case you are in trouble).

4. ## Help needed with algebra

thank you for your reply. I don't know if there are any restrictions on a as the question doesn't say. I presume it is that $|a|<1$ but I'm not sure....
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