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Question: Answered & Verified by Expert
If $\alpha$ is a complex number such that $\alpha^{2}+\alpha+1=0$, then what is $\alpha^{31}$ equal to?
MathematicsComplex NumberNDANDA 2009 (Phase 2)
Options:
  • A $\alpha$
  • B $\alpha^{2}$
  • C 0
  • D 1
Solution:
2548 Upvotes Verified Answer
The correct answer is: $\alpha$
Since, $\alpha$ is a complex root therefore $\alpha^{2}+\alpha+1=0 \Rightarrow \alpha=\omega$ or $\omega^{2}$
consider $\alpha^{31}=(\omega)^{31}$
$=\left(\omega^{3}\right)^{10} \cdot \omega$
$=\omega \quad\left(\because \omega^{3}=1\right)$
$=\alpha$

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