Complex Analysis: Integration

[Shay10825]

Homework Statement

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.

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

Homework Equations

none

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

 

[gabbagabbahey]

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

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

Thanks

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