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If $\alpha$ is a complex number satisfying the equation $\alpha^{2}+\alpha+1=0$, then $\alpha^{31}$ is equal to
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Verified Answer
The correct answer is:
$\alpha$
Given equation is
$$
\begin{aligned}
\alpha^{2}+\alpha+1 &=0 \\
\therefore \quad \alpha &=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2}
\end{aligned}
$$
Let it be $\quad \alpha=\omega, \omega^{2}$
(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 $\alpha$.
$$
\begin{aligned}
\alpha^{2}+\alpha+1 &=0 \\
\therefore \quad \alpha &=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2}
\end{aligned}
$$
Let it be $\quad \alpha=\omega, \omega^{2}$
(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 $\alpha$.
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