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NDA 1 2025 Maths, question 13

Topic: Algebra

Q13

What is (3+i3−i)3\left( \frac{\sqrt{3} + i}{\sqrt{3} - i} \right)^{3} equal to?

  1. (a) -1Correct answer
  2. (b) 0
  3. (c) 1
  4. (d) 3

Solution.

Answer: (a) -1

  1. Rationalising the denominator gives 3+i3−i=(3+i)24=12+i32=cos⁡60∘+isin⁡60∘\frac{\sqrt{3} + i}{\sqrt{3} - i} = \frac{\left( \sqrt{3} + i \right)^{2}}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = \cos 60^{\circ} + i\sin 60^{\circ}.
  2. By De Moivre’s theorem, its cube is cos⁡180∘+isin⁡180∘=−1\cos 180^{\circ} + i\sin 180^{\circ} = - 1.