UPSC NDA II 2023 Mathematics Solved Paper

Concept:Find the square root of $-\sqrt{-1}$ by first simplifying it to $-i$, then squaring each option to see which gives $-i$.Explanation:We know $i = \sqrt{-1}$.So $-\sqrt{-1} = -i$.We need $\sqrt{-i}$, meaning a number whose square equals $-i$.Check each option by squaring it.Option A: $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$, not $-i$.Option B: $\left( \frac{1-i}{\sqrt{2}} \right)^2 = \frac{(1-i)^2}{2} = \frac{1 - 2i + i^2}{2} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i$.This matches $-i$, so it is a square root.Option C: $\left( \frac{1+i}{\sqrt{2}} \right)^2 = \frac{1 + 2i + i^2}{2} = \frac{1 + 2i - 1}{2} = i$, not $-i$.Option D: $\left( \frac{1}{\sqrt{2}} i \right)^2 = \frac{i^2}{2} = -\frac{1}{2}$, not $-i$.Thus, only option B satisfies the condition.Answer:B. $\frac{1-i}{\sqrt{2}}$