Question

If ${\left(\frac{1+i}{1-i}\right)}^n=1,$ then n =
(a) 2m + 1
(b) 4m
(c) 2m
(d) 4m + 1 where m ∈ N

Answer 1

Given :- ${\left(\frac{1+i}{1-i}\right)}^n=1$
$\begin{array}{rcl}\mathrm{Since},\frac{1+i}{1-i}& =& \frac{1+i}{1-i}\times \frac{1+i}{1+i}\\ & =& \frac{{\left(1+i\right)}^2}{1-i^2}\\ & =& \frac{1+i^2+2i}{1+1}\end{array}$

$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{1+i}{1-i}=\frac{1-1+2i}{2}=\frac{2i}{2}=2i \\ \Rightarrow {\left(\frac{1+i}{1-i}\right)}^n=1 \\ \Rightarrow {\left(i\right)}^n=1 \\ \Rightarrow n=4\mathrm{m}\mathrm{where}m\in N \end{gathered}$$

Hence, the correct answer is option B.