UPSC NDA I 2025 Mathematics Paper

Concept:To simplify a complex fraction raised to a power, first rationalize the denominator or convert to polar form. Use De Moivre's theorem: $(r(\cos\theta + i\sin\theta))^n = r^n (\cos(n\theta) + i\sin(n\theta))$.Explanation:Step 1: Multiply numerator and denominator by the conjugate of the denominator ( $\sqrt{3}+i$ ):$\frac{\sqrt{3}+i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} = \frac{(\sqrt{3}+i)^2}{(\sqrt{3})^2 - i^2}$.Step 2: Simplify: numerator $(\sqrt{3}+i)^2 = 3 + 2\sqrt{3}i + i^2 = 2 + 2\sqrt{3}i$; denominator $3 - (-1) = 4$.So the fraction becomes $\frac{2+2\sqrt{3}i}{4} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$.Step 3: Convert to polar form. Modulus $r = \sqrt{\left(\frac12\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac14 + \frac34} = 1$.Argument $\theta = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Thus the number is $\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$.Step 4: Cube using De Moivre's theorem: $(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})^3 = \cos\pi + i\sin\pi = -1$ .So the result is $-1$.Answer:Option A: $-1$