Question

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

Answer 1

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

$\therefore \left(\frac{1+i}{1-i}\right)=1\Rightarrow i^n=1\Rightarrow i^n=i^4\Rightarrow n=4$.