# Prove the equation won't be satisfied for any complex number Z?

• Apr 30th 2009, 09:23 PM
fardeen_gen
Prove the equation won't be satisfied for any complex number Z?
If $|Z| < \frac{1}{3}$, then prove that the equation $1 + 2\cos^2 \theta_{1}Z + 2\cos^2\theta_{2}Z^2 + ... + 2\cos^2\theta_{n}Z^n = 0$, will not be satisfied by any complex number z.
• May 1st 2009, 02:22 AM
Showcase_22
I haven't solved it yet, but here's what I have so far:

$

1 + 2\cos^2 \theta_{1}Z + 2\cos^2\theta_{2}Z^2 + ... + 2\cos^2\theta_{n}Z^n = 0
$

$1+(1+ \cos 2 \theta_1)Z+(1+ \cos 2 \theta_2)Z^2+....+ (1+\cos 2 \theta_n)Z^n=0$

$(1+Z+Z^2+.......+Z^n)+(Z \cos 2\theta_1+ Z^2 \cos 2\theta_2+.....+ Z^n \cos 2 \theta_n)=0$

$0=(1+Z+Z^2+.......+Z^n)+(Z \cos \theta_2+ Z^2 \cos \theta_2+.....+ Z^n \cos 2 \theta_n)$ $< \left(1+\frac{1}{3}+\frac{1}{9}+......+\frac{1}{3^ n} \right)+ \left( \frac{1}{3} \cos 2\theta_1+\frac{1}{9} \cos 2\theta_2+.....+ \frac{1}{3^n} \cos 2 \theta_n \right)$

$\left( 1+\frac{1}{3}+\frac{1}{3^2}+...+ \frac{1}{3^n} \right)= \frac{1-\left( \frac{1}{3} \right)^n}{1-\frac{1}{3}}=\frac{3}{2} \left( 1- \left( \frac{1}{3} \right)^n \right)$

I'm not really sure where to go next (Worried)
• May 1st 2009, 03:34 AM
CaptainBlack
Quote:

Originally Posted by fardeen_gen
If $|Z| < \frac{1}{3}$, then prove that the equation $1 + 2\cos^2 \theta_{1}Z + 2\cos^2\theta_{2}Z^2 + ... + 2\cos^2\theta_{n}Z^n = 0$, will not be satisfied by any complex number z.

Because the Cauchy lower bound on the roots of this polynomial is greater than or equal to 1/3.

(see here)

CB
• May 1st 2009, 04:09 AM
Showcase_22
aww, shoot!

I was way off!!