Question

Let $\alpha$ be a complex number such that ${\alpha }^2+\alpha +1=0$, then ${\alpha }^{31}$ is equal to

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

Answer 1

Answer: (A)

$\alpha$

Given $\alpha$ satisfies the equation ${\alpha }^2+\alpha +1=0$.

Then $\alpha =\frac{-1+i\sqrt{3}}{2}$ and $\frac{-1-i\sqrt{3}}{2}$.

Among which one is $\omega$ and the other is ${\omega }^2$ where $\omega$ is the cube root of unity such that ${\omega }^3=1$.

Now if $\alpha =\omega$ then ${\alpha }^{31}={\omega }^{31}=({\omega }^3{)}^{10}.\omega =\omega =\alpha$ [Using ${\omega }^3=1$]

Now if $\alpha ={\omega }^2$ then${\alpha }^{31}={\omega }^{62}=({\omega }^3{)}^{20}.{\omega }^2={\omega }^2=\alpha$ [Using ${\omega }^3=1$].

So, ${\alpha }^{31}=\alpha$.