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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:
$\alpha$
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
$$
\begin{aligned}
& \alpha^{31}=\left(\omega^{2}\right)^{31}=\omega^{62}=\left(\omega^{3}\right)^{20} \cdot \omega^{2}=\omega^{2}=\alpha \\
\therefore \quad & \alpha^{31}=\alpha
\end{aligned}
$$
$$
\Rightarrow \quad \alpha=\omega \text { or } \omega^{2}
$$
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
$$
\begin{aligned}
& \alpha^{31}=\left(\omega^{2}\right)^{31}=\omega^{62}=\left(\omega^{3}\right)^{20} \cdot \omega^{2}=\omega^{2}=\alpha \\
\therefore \quad & \alpha^{31}=\alpha
\end{aligned}
$$
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