# Complex Analysis: Integration

1. Mar 6, 2009

### Shay10825

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

Evaluate the following integral for 0<r<1 by writing $$\cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$ reducing the given integral to a complex integral over the unit circle.

$$Evaluate: \displaystyle{\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{1-2r\cos\theta + r^2}\,d\theta}$$

2. Relevant equations

none

3. The attempt at a solution

$$\displaystyle\cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})}$$

$$\displaystyle{z=e^{i\theta}}$$

$$\displaystyle{ \cos t= \frac{1}{2}(z+ \frac{1}{z})}$$

$$\displaystyle{\frac{dz}{iz}= dt}$$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$$\displaystyle{\frac{1}{2\pi}\oint \frac{1}{1-2r[\frac{1}{2}(z+\frac{1}{z})]+r^2}\,\frac{dz}{iz}}$$

$$\displaystyle{\frac{1}{2\pi}\oint \frac{1}{z-2rz[\frac{1}{2}(z+\frac{1}{z})]+r^2}\,\frac{dz}{i}}$$

$$\displaystyle{\frac{-i}{2\pi}\oint \frac{1}{z-2rz[\frac{1}{2}(z+\frac{1}{z})]+r^2}\,dz}$$

$$\displaystyle{\frac{-i}{2\pi}\oint \frac{1}{z-r^2z^2-r+r^2}\,dz}$$

But I get stuck here. What do I do with the "r"? Should I factor it out, and if yes then how?

Thanks

2. Mar 6, 2009

### gabbagabbahey

Leave the "r" where it is, the location of the poles will depend on its value...find those poles by solving the quadratic $z-r^2z^2-r+r^2=0$ for $z$...

Edit: You've also got a couple of algebra errors, double check your last 4 steps

Last edited: Mar 6, 2009