Question

If $\alpha ,\beta$ are the roots of the equation $x^2-2x+4=0$ prove that ${\alpha }^n+{\beta }^n=2^{n+1}\cos \left(n\pi /3\right)$

Answer 1

$x^2-2x+4=0$

$\therefore x=\frac{2\pm \sqrt{4-16}}{2}=1\pm \sqrt{3}i$

$\therefore \alpha =r{e}^{i\theta },\beta =r{e}^{-i\theta }wheree=2and\theta =\pi /3$

$\therefore {\alpha }^n+{\beta }^n=r^n(e^{in\theta }+e^{-in\theta })$

$=r^n[\cos n\theta +i\sin n\theta +\cos n\theta -i\sin n\theta ]$

$=r^n.2\cos n\theta =2^{n+1}\cos \left(n\pi /3\right)$