Need help with a quadratic that has complex roots

**Lancet** · October 28th 2012, 09:31 PM

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?

**chiro** · October 28th 2012, 09:36 PM

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.

**MarkFL** · October 28th 2012, 09:45 PM

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

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

Apply the quadratic formula:

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

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

Now, apply Euler's formula to obtain:

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

**Lancet** · October 29th 2012, 07:33 AM

Originally Posted by MarkFL2

Apply the quadratic formula:

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

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

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

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

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