Answer:
Option C
Explanation:
We have,
$\alpha, \beta $ are the roots of equation $x^{2}-2x+4$=0
$x^{2}-2x+4=0$
$x^{2}-2x+1= -3 \Rightarrow (x-1)^{2}=-3$
$x-1= \pm\sqrt{3}i\Rightarrow x=1\pm\sqrt{3}i$
$x=-2\omega,-2\omega^{2}$
$\left[\because \frac{-1+\sqrt{3}i}{2}=\omega , \frac{-1-\sqrt{3}i}{2}=\omega^{2}\right]$
$\therefore$ $\alpha =-2 \omega$ , $\beta = -2 \omega^{2}$
$\alpha^{12}+\beta ^{12}= (-2 \omega) ^{12} +(-2 \omega^{2})^{12}$
= $ 2^{12} (\omega^{12}+ \omega^{24})= 2^{12} (1+1)$ [ $\because \omega^{3}=1$]
=$2^{13}$
