Q87 / 180 · COMEDK 2019 · Mathematics

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

  • Correct answer
    A
    $\alpha$
  • B
    $\alpha^{2}$
  • C
    1
  • D
    $i$

Solution

We have, $\alpha^{2}+\alpha+1=0$ $$ \Rightarrow \quad \alpha=\omega \text { or } \omega^{2} $$ Now, if $\alpha=\omega$, then $$ \left.\alpha^{31}=\omega^{31}=\left(\omega^{3}\right)^{10} \cdot \omega=\omega\right)=\alpha $$ and if $\alpha=\omega^{2}$, then $$ \begin{aligned} & \alpha^{31}=\left(\omega^{2}\right)^{31}=\omega^{62}=\left(\omega^{3}\right)^{20} \cdot \omega^{2}=\omega^{2}=\alpha \\ \therefore \quad & \alpha^{31}=\alpha \end{aligned} $$

Hence, the correct answer is option: 'A'.

Relevant Questions