The least positive integer $n$ such that ${\left(\frac{1-i}{1+i}\right)}^{2n}=1$ is .....

The correct option is C $2$
$z=\left(\frac{1-i}{1+i}\right)$
$=\frac{(1-i)(1-i)}{1+1}$
$=\frac{(1-i{)}^2}{2}$
$=\frac{1-1-2i}{2}$
$=-i$...(i)
Hence
$(-i{)}^{2n}=1$

$i^{2n}=1$

$(i^2{)}^n=1$

$(-1{)}^n=1$

Hence
$n=0,2,\mathrm{4...}$
Thus the least positive integer is 2.