UPSC NDA I 2016 Mathematics Solved Paper

Concept:Given specific complex numbers $z_1 = -i$, $z_2 = \frac{\sqrt{3}}{2} + i\frac{1}{2}$, and $z_3 = -\frac{\sqrt{3}}{2} + i\frac{1}{2}$, we need to verify whether the product $z_1 z_2 z_3$ is purely imaginary and whether the sum $z_1 z_2 + z_2 z_3 + z_3 z_1$ is purely real.Explanation:For statement 1, compute the product:$z_1 z_2 z_3 = (-i) \left( \frac{\sqrt{3}}{2} + i\frac{1}{2} \right) \left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right)$.First multiply the last two terms: $\left( \frac{\sqrt{3}}{2} + i\frac{1}{2} \right) \left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = -\frac{3}{4} + i\frac{\sqrt{3}}{4} - i\frac{\sqrt{3}}{4} + i^2 \frac{1}{4} = -\frac{3}{4} - \frac{1}{4} = -1$.Then multiply by $(-i)$: $(-i) \times (-1) = i$.Since $i$ is purely imaginary, statement 1 is correct.For statement 2, compute the sum $z_1 z_2 + z_2 z_3 + z_3 z_1$:$z_1 z_2 = (-i)\left( \frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = -\frac{\sqrt{3}}{2}i - i^2\frac{1}{2} = -\frac{\sqrt{3}}{2}i + \frac{1}{2}$.$z_2 z_3 = \left( \frac{\sqrt{3}}{2} + i\frac{1}{2} \right) \left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = -1$ (as computed above).$z_3 z_1 = (-i) \left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = \frac{\sqrt{3}}{2}i - i^2\frac{1}{2} = \frac{\sqrt{3}}{2}i + \frac{1}{2}$.Adding these: $\left( -\frac{\sqrt{3}}{2}i + \frac{1}{2} \right) + (-1) + \left( \frac{\sqrt{3}}{2}i + \frac{1}{2} \right) = \left( -\frac{\sqrt{3}}{2}i + \frac{\sqrt{3}}{2}i \right) + \left( \frac{1}{2} + \frac{1}{2} \right) - 1 = 0 + 1 - 1 = 0$.The result is $0$, which is purely real. Hence statement 2 is also correct.Answer:Both statement 1 and statement 2 are correct. Therefore, the correct option is C (Both 1 and 2).