Let $a, b, c$ be real numbers where $0\leq c\leq b\leq a\leq 1$.
If $\omega$ is one of the complex roots of polynomial $z^3+az^2+bz+c$ and $|\omega| \geq 1$.
Show that $\omega^4 = 1$.
My work :
$\omega^3+a\omega^2+b\omega+c = 0$
$(\omega-1)(\omega^3+a\omega^2+b\omega+c) = 0$
$\omega^4 + \omega^3(a-1) + \omega^2(b-a) + \omega(c-b) = c$
$|(1-a)\omega^3 + (a-b)\omega^2 + (b-c)\omega + c| = |\omega^4|$
$|(1-a)\omega^3| + |(a-b)\omega^2| + |(b-c)\omega| + |c| \geq |\omega^4|$
As $|\omega| \geq 1$, so
$ |\omega^4| \leq |\omega|^3[(1-a)+(a-b)+(b-c)+c] = |\omega|^3$
Thus $|\omega| \leq 1$ so $|\omega| = 1$
Please suggest how to proceed to get $\omega^4 = 1$.
