Need help with a quadratic that has complex roots

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?

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.

Re: Need help with a quadratic that has complex roots

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

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

Apply the quadratic formula:

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

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

Now, apply Euler's formula to obtain:

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

Re: Need help with a quadratic that has complex roots

Quote:

Originally Posted by

**MarkFL2**

Apply the quadratic formula:

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

Ah! That's where I was getting stuck. I didn't know what to do with

$\displaystyle \sqrt{ \cos^2(\theta)-1}$

I didn't realize that was a modified trig identity. Thanks!