UPSC NDA I 2025 Mathematics Paper

The key concept involves simplifying complex numbers in the polar form. We use the argument (angle) of the complex number and De Moivre's theorem, which states:$(r(\cos \theta + i \sin \theta))^{n} = r^{n} (\cos (n\theta) + i \sin (n\theta))$Calculation:$\Rightarrow \left( \frac{\sqrt{3}+i}{\sqrt{3}-i} \right)^3$Multiply the numerator and the denominator by the conjugate of the denominator$\Rightarrow \frac{\sqrt{3}+i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i}$$\Rightarrow \frac{(\sqrt{3}+i)^2}{(\sqrt{3}-i)(\sqrt{3}+i)}$$\frac{2+2\sqrt{3}i}{4} = \frac{1+\sqrt{3}i}{2}$Convert to polar form. The modulus $r$ is$r = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 } = \sqrt{ \frac{1}{4} + \frac{3}{4} } = \sqrt{1} = 1$The argument $\theta$ is$\theta = \tan^{-1} \left( \frac{ \frac{\sqrt{3}}{2} }{ \frac{1}{2} } \right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$So the polar form of the complex number is$1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$To cube the complex number, we use De Moivre's TheoremIn our case, $r = 0$ so,$\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)^3 = \cos \pi + i \sin \pi = -1 + 0i = -1$