Question

If $\alpha$ be a complex number satisfying ${\alpha }^2+\alpha +1=0,$ then ${\alpha }^{32}$ equals

• A. $\alpha$
• B. ${\alpha }^2$
• C. 0
• D. 1

Answer 1

The correct option is B ${\alpha }^2$

${\alpha }^2+\alpha +1=0$
Assuming $\alpha \ne 1$ and multiplying on both sides by $(\alpha -1)$, we get ${\alpha }^3=1$
So, $\alpha$ will be a cube root of unity.
$\Rightarrow {\alpha }^{32}=({\alpha }^3{)}^{10}\cdot {\alpha }^2={\alpha }^2$