Q188 / 270 · NDA 2011 (Phase 2) · Mathematics
The smallest positive integral value of $n$ for which $\left(\frac{1-i}{1+i}\right)^n$ is purely imaginary with positive imaginary part is
- A$1$
- Correct answerB$3$
- C$4$
- D$5$
Solution
$\begin{aligned}\left(\frac{1-i}{1+i}\right)^n & =\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^n=\left[\frac{(1-i)^2}{1-i^2}\right]^n \\ & =\left[\frac{1-2 i+i^2}{1+1}\right]^n \\ & =\left[\frac{1-2 i-1}{2}\right]^n=(-i)^n\end{aligned}$
if $\quad n=3$ then $(-i)^3=i$.
Hence, the correct answer is option: 'B'.