Question

Find the least positive integer $n$ such that ${\left(\frac{1+i}{1-i}\right)}^n=1$

Answer 1

Given, ${\left(\frac{1+i}{1-i}\right)}^n=1$

$\Rightarrow {(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^n=1$

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

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

$\Rightarrow (i{)}^n=1$
This is possible for $n=4,8,12,.....$
The least positive integer $n$ for this to be true $=4$.