# Need help with a quadratic that has complex roots

• Oct 28th 2012, 09:31 PM
Lancet
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?
• Oct 28th 2012, 09:36 PM
chiro
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.
• Oct 28th 2012, 09:45 PM
MarkFL
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:

$z=re^{\pm j\theta}$
• Oct 29th 2012, 07:33 AM
Lancet
Re: Need help with a quadratic that has complex roots
Quote:

Originally Posted by MarkFL2

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