Originally Posted by

**lpd** Hi. I have this problem.

a) Use the Residue Theorem to calculate

$\displaystyle \int_C{\frac{dz}{z^2-2az+1}$

where $\displaystyle C$ is the unit circle {$\displaystyle z:|z|=1$} and $\displaystyle a$ is real with $\displaystyle a>1$.

Okay, this is my attempt. I found the roots of $\displaystyle z^2-2az+1$ and came up with $\displaystyle a \pm \sqrt{a^2-1}$, and noticed that we can only use $\displaystyle a-\sqrt{a^2-1}$, and i denote this as $\displaystyle \beta=a-\sqrt{a^2-1}$ (and $\displaystyle \alpha =a+\sqrt{a^2-1}$)

Thus by the residue theorem

$\displaystyle \int_C{\frac{dz}{z^2-2az+1}=2\pi i Res[\frac{1}{z^2-2az+1}, a-\sqrt{a^2-1}]$

$\displaystyle =lim_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim_{z\to\beta}\frac{1}{z-\alpha}$

$\displaystyle =\frac{1}{\beta-\alpha}=\frac{1}{-2\sqrt{a^2-1}}$

$\displaystyle \int_C{\frac{dz}{z^2-2az+1}=2\pi i \frac{1}{-2\sqrt{a^2-1}} = -\frac{\pi i}{2\sqrt{a^2-1}}$

b) Use this to calculate

$\displaystyle \int_0^{2\pi}\frac{dx}{a-cos(x)}$

So I write the integral by letting $\displaystyle z=e^{ix}$ and $\displaystyle cos(x)=\frac{1}{2}(z+z^{-1})...$

and yields $\displaystyle \frac{-2}{i}\int_C \frac{dz}{z^2-2az+1}$

and i use the previous answer, and I get $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$

The main problem is...

c) **What happens to your argument (and the values of the integrals) when **$\displaystyle a<-1$?

What did you just write here above?? Is it $\displaystyle a<-1$ ? If so then you'll have to take $\displaystyle a+\sqrt{a^2-1}$ , since the other

root is out of the zone bounded by the path of integration...

Tonio

I'm not entirely sure about this one. My guess is $\displaystyle \alpha$ will now be the simple pole, but when i rework it all out again, it comes to the same conclusion of $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$.. so i'm not sure please help.