Question

If ${\left(\frac{(1+i)}{(1-i)}\right)}^m=1$, then least integral value of $m$ is

Answer 1

Finding the least integral value of $\mathit{m}$ :

Given that $[\frac{(1+i)}{(1-i)}{]}^m=1$

Multiply numerator and denominator with $(1+i)$

$\Rightarrow$${\left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]}^m=1$

$\Rightarrow$ ${\left[\frac{{(1+i)}^2}{(1+1)}\right]}^m=1$

$\Rightarrow$ ${\left[\frac{(1+2i-1)}{2}\right]}^m=1$

$\Rightarrow$ ${\left(\frac{2i}{2}\right)}^m=1$

$\Rightarrow$ $$\begin{gathered} i^m=1 \\ m=4 \end{gathered}$$

Hence. the value of $m$ is $\mathbf{4}\mathbf{.}$