Question

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

• A. 4

Answer 1

The correct option is A 4
$\Rightarrow {(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^n={\left(\frac{(1+i{)}^2}{1^2-i^2}\right)}^n$
$\Rightarrow {\left(\frac{1+i^2+2i}{1-(-1)}\right)}^n={\left(\frac{1-1+2i}{2}\right)}^n$
$=(i{)}^n$
$i^2=-1$ and $i^4=1$
$\therefore$ The least value of $n=4$