Question

If $\alpha$ is a complex number satisfying the equation $\alpha^2 + \alpha +1 = 0$ then $\alpha^{31}$ is equal to

• A. $\alpha$
• B. $\alpha^2$
• C. $1$
• D. $i$

Answer 1

Answer: (A)

Given equation is
$\alpha^{2}+\alpha+1=0$
$\therefore \alpha=\frac{-1 \pm \sqrt{1-4}}{2}$
$=\frac{-1 \pm \sqrt{3} i}{2}$
Let it be $\,\,\,\,\,\alpha=\omega, \omega^{2}$
(1) If $\,\,\,\,\,\,\,\alpha=\omega$, then
$\alpha^{31}=(\omega)^{31}=\omega =\alpha$
(2) If $\,\,\,\,\,\,\,\,\alpha=\omega^{2}$, then
$\alpha^{31}=\left(\omega^{2}\right)^{31}$
$=\omega^{62}=\omega^{2}=\alpha$
Hence, $\alpha^{31}$ is equal to $\alpha$.