# 4U Roots of Unity/Polynomials Question (1 Viewer)

#### nerrma

##### New Member
Part (a) and (b) are straightforward enough. Any help for part (c)?

#### jathu123

##### Active Member
$\bg_white \noindent Since |z|=1, we can use the identity: z^n+z^{-n}=\cos n\theta (you can prove this by subbing in z= cis\theta and using DMT. \\ \\ Dividing the LHS by z^3, we obtain : \\ z^3+z^{-3} = 2\cos 3\theta \qquad Using the above identity \\ \\ On the RHS, we divide each factor by z, this is equivalent to dividing the whole side by z^3: \\ \left ( \frac{z^2 +1}{z} \right )\left ( \frac{z^2-\sqrt{3}z+1}{z} \right )\left ( \frac{z^2+\sqrt{3}z+1}{z} \right )=\left ( z+z^{-1} \right )\left ( z+z^{-1}-\sqrt{3} \right )\left ( z+z^{-1}+\sqrt{3} \right ) \\ =(2\cos \theta)\left ( 2\cos \theta-2\cos \frac{\pi}{6} \right )\left ( 2\cos \theta-2\cos \frac{5\pi}{6} \right ) \\ =8 \cos \theta \left ( \cos \theta - \cos\frac{\pi}{6} \right )\left ( \cos \theta - \cos\frac{5\pi}{6} \right ) \\ \\ \therefore \cos 3\theta =4\cos \theta \left ( \cos \theta - \cos\frac{\pi}{6} \right )\left ( \cos \theta - \cos\frac{5\pi}{6} \right )$