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

Find $\alpha,\beta$ as a complex-conjugate pair, write them in modulus-argument (polar) form, then apply De Moivre's theorem to each power and add.

For $x^2-2x+4=0$: sum of roots $\alpha+\beta=2$, product $\alpha\beta=4$. The discriminant is $4-16=-12<0$, so the roots are a complex-conjugate pair:

$\alpha,\beta = \frac{2\pm\sqrt{-12}}{2} = 1\pm i\sqrt3.$

Write $\alpha=1+i\sqrt3$ in polar form. Modulus: $r=\sqrt{1^2+(\sqrt3)^2}=\sqrt{4}=2$. Argument: $\tan\theta=\dfrac{\sqrt3}{1}=\sqrt3 \Rightarrow \theta=\dfrac{\pi}{3}$ (first quadrant, since both real and imaginary parts are positive). So

$\alpha = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right), \qquad \beta = 2\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right)$

(since $\beta$ is the conjugate of $\alpha$).

By De Moivre's theorem, $\left(\cos\phi+i\sin\phi\right)^n=\cos n\phi+i\sin n\phi$, so

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