Q.If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$, then for any $n \in N$, show that $\alpha^n + \beta^n = 2^{n+1} \cdot \cos\left(\dfrac{n\pi}{3}\right)$.

The roots are complex conjugates of modulus 2 and argument $\pm\pi/3$; De Moivre's theorem then gives $\alpha^n+\beta^n=2^{n+1}\cos(n\pi/3)$ for every natural number $n$.

Step 1 — Find the roots.

$x^2-2x+4=0 \implies x=\dfrac{2\pm\sqrt{4-16}}{2}=\dfrac{2\pm\sqrt{-12}}{2}=1\pm i\sqrt3$.

So $\alpha=1+i\sqrt3,\ \beta=1-i\sqrt3$.

Step 2 — Write in polar (modulus-argument) form.

$|\alpha|=\sqrt{1^2+(\sqrt3)^2}=\sqrt4=2$, $\arg(\alpha)=\tan^{-1}\left(\dfrac{\sqrt3}{1}\right)=\dfrac{\pi}{3}$.

So $\alpha=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$ and, since $\beta=\bar\alpha$, $\beta=2\left(\cos\dfrac{\pi}{3}-i\sin\dfrac{\pi}{3}\right)$.

Step 3 — Apply De Moivre's theorem to raise to the $n^{\text{th}}$ power. …