Question

The least positive integral value of $\textit{m}$ , if $\left(\frac{1 + i}{1 - i}\right)^{\textit{m}}=1$ is: $\left(\right.$ where $i = \sqrt{- 1} \, $

Answer 1

We have $\text{:}$ $\frac{1 + \textit{i}}{1 - \textit{i}}=\frac{1 + \textit{i}}{1 - \textit{i}}\times \frac{1 + \textit{i}}{1 + \textit{i}}=\frac{\left(1 + \textit{i}\right)^{2}}{1 - \left(\textit{i}\right)^{2}}$
$=\frac{1 + 2 \textit{i} + \textit{i}^{2}}{1 - \textit{i}^{2}}$
$=\frac{1 + 2 \textit{i} - 1}{1 + 1}=\frac{2 \textit{i}}{2}=\textit{i}$
$\therefore \left(\frac{1 + \textit{i}}{1 - \textit{i}}\right)^{\textit{m}}=1\Rightarrow \left(\textit{i}\right)^{\textit{m}}=1$
$\Rightarrow $ $\textit{m}$ is a multiple of $4$ .
Hence, the smallest positive value of $\textit{m}$ is $\text{4}$ .