KCET · Maths · Complex Number
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 \(\mathrm{i}\)
Answer & Solution
Correct Answer
(A) \(\alpha\)
Step-by-step Solution
Detailed explanation
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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