This is the equation in question:

$\displaystyle 1-2r cos(\Theta) z^{-1}+r^2 z^{-2}=0$

I'm out of practice for these and don't remember how best to handle this. If I plug it into the quadratic equation, I get a square root that I'm not sure how to simplify. I'm guessing I need to apply Euler's Formula, but again, I'm out of practice.

I know the answer to this is:

$\displaystyle z=re^{\pm j\theta}$

...I'm just not sure how to get there.

Can someone give me a hand?