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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Complex Number

NDA 2021 · Mathematics · Complex Number

The smallest positive integer n for which ((1-i)/(1+i))^ n^2 where...

The smallest positive integer nn for which (1−i1+i)n2\left(\frac{1-i}{1+i}\right)^{n^2} where i=−1i=\sqrt{-1}, is

Options

  1. A22
  2. B44
  3. C66
  4. D88

Correct answer

A. 22

Step-by-step solution

(1−i1+i)n2=1⇒((1−i)(1−i)(1+i)(1−i))n2=1\begin{aligned} & \left(\frac{1-i}{1+i}\right)^{n^2}=1 \\ & \Rightarrow\left(\frac{(1-i)(1-i)}{(1+i)(1-i)}\right)^{n^2}=1\end{aligned}⇒((1)2+(i)2−2i1−(i)2)n2=1⇒(−2i2)n2=1⇒(−i)n2=1\begin{aligned} & \Rightarrow\left(\frac{(1)^2+(i)^2-2 i}{1-(i)^2}\right)^{n^2}=1 \\ & \Rightarrow\left(\frac{-2 i}{2}\right)^{n^2}=1 \Rightarrow(-i)^{n^2}=1\end{aligned}ForForn=2,(-i)^{2^2}=(-i)^4=1Hence,Hence,n=2$.
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