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 $\alpha$.