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)

$\alpha$

Cube root of unity is $x^3=1$

$⟹x^3-1=0$

$⟹(x-1)(x^2+x+1)=0$

Either $x=1$ or $x^2+x+1=0$

But $x\ne 1$ $⟹x^2+x+1=0$

If $\alpha$ is a number satisfying ${\alpha }^2+\alpha +1=0$

Roots will be $\omega ,{\omega }^2$ and ${\omega }^3=1$

$\therefore {\alpha }^{31}={\omega }^{31}={\left({\omega }^3\right)}^{10}\omega =\omega =\alpha$