Q265 / 270 · NDA 2010 (Phase 2) · Mathematics

If $\omega$ is a complex cube root of unity, then what is $\omega^{10}+\omega^{-10}$ equal to?

  • A
    2
  • Correct answer
    B
    -1
  • C
    -2
  • D
    1

Solution

$\therefore \omega^3=1 \text { and } 1+\omega+\omega^2=0$ So that $\begin{aligned} & \omega^{10}+\omega^{-10}=\omega^{10}+\frac{1}{\omega^{10}} \\ & =\left(\omega^3\right)^3 \cdot \omega+\frac{1}{\left(\omega^3\right)^3 \cdot \omega} \\ & =\omega+\frac{1}{\omega}=\frac{\omega^2+1}{\omega} \\ & =\frac{-\omega}{\omega}=-1 \end{aligned}$

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

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