Question

Determine least positive value of n for which ${\left(\frac{1+i}{1-i}\right)}^n=1$

Answer 1

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

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

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

Hence least positive value of $n$ is 4. $(i^4=1)$