UPSC NDA I 2021 Mathematics Solved Paper

Concept:Simplify the complex fraction to $-i$, then find the smallest positive integer $n$ such that $(-i)^{n^2}=1$.Explanation:First, simplify $\frac{1-i}{1+i}$.Multiply numerator and denominator by $1-i$:$\frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{(1-i)^2}{1-i^2}$.Since $i^2=-1$, denominator becomes $1-(-1)=2$.Expand numerator: $(1-i)^2 = 1 -2i + i^2 = 1 -2i -1 = -2i$.Thus the fraction simplifies to $\frac{-2i}{2} = -i$.The given equation becomes $(-i)^{n^2}=1$.Recall that $(-i)^1=-i$, $(-i)^2=-1$, $(-i)^3=i$, $(-i)^4=1$, and the pattern repeats every $4$.So $(-i)^k=1$ only when $k$ is a multiple of $4$.Therefore we require $n^2$ to be divisible by $4$.The smallest positive integer $n$ satisfying this is $n=2$, because $2^2=4$ is a multiple of $4$.Substituting $n=2$ gives $(-i)^4=1$, which satisfies the equation.Answer:The smallest positive integer $n$ is $2$, which corresponds to option A.