1. ## Need help with a quadratic that has complex roots

This is the equation in question:

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

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:

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

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

Can someone give me a hand?

2. ## Re: Need help with a quadratic that has complex roots

Hey Lancet.

Try multiplying both sides by z^2 and consider anything not a z as a constant (if it is a constant). You then get something along the lines of az^2 + bz + c = 0.

3. ## Re: Need help with a quadratic that has complex roots

I would first multiply through by $z^2$ and write in standard form:

$z^2-2r\cos(\theta)z+r^2=0$

$z=\frac{2r\cos(\theta)\pm\sqrt{(-2r\cos(\theta))^2-4(1)(r^2)}}{2(1)}=\frac{2r(\cos(\theta)\pm\sqrt{ \cos^2(\theta)-1})}{2}=$

$r$$\cos(\theta)\pm j\sin(\theta)$$$

Now, apply Euler's formula to obtain:

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

4. ## Re: Need help with a quadratic that has complex roots

Originally Posted by MarkFL2

$z=\frac{2r\cos(\theta)\pm\sqrt{(-2r\cos(\theta))^2-4(1)(r^2)}}{2(1)}=\frac{2r(\cos(\theta)\pm\sqrt{ \cos^2(\theta)-1})}{2}=$
$\sqrt{ \cos^2(\theta)-1}$