Q165 / 270 · NDA 2018 (Phase 2) · Mathematics

What is the value of $\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)^{3 \mathrm{n}}+\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right)^{3 \mathrm{n}}$ where $i=\sqrt{-1}$ ?

  • A
    3
  • Correct answer
    B
    2
  • C
    1
  • D
    0

Solution

From cube root of unity $\begin{aligned} & \omega=\frac{-1+\sqrt{3} i}{2} \text { and } \\ & \omega^2=\frac{-1-\sqrt{3} i}{2} \\ & \omega^3=\left(\frac{-1+\sqrt{3} i}{2}\right)\left(\frac{-1-\sqrt{3} i}{2}\right) \end{aligned}$ $\begin{aligned} & =\frac{(-1)^2-(\sqrt{3} i)^2}{4}=\frac{4}{4}=1 \\ & \therefore \quad \omega^3=1 \\ & \text { Now, }(\omega)^{3 \mathrm{n}}+\left(\omega^2\right)^{3 \mathrm{n}}=\left(\omega^3\right)^{\mathrm{n}}+\left(\omega^3\right)^{2 \mathrm{n}} \\ & =1+1=2 \end{aligned}$

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

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