Question

The least positive value of $n$ for which ${\left(\frac{i(i+\sqrt{3})}{1-i^2}\right)}^n$ is a positive integer, is

Answer 1

$={\left[\frac{i(i+\sqrt{3})}{1-i^2}\right]}^2$

$[i^2=-1]$

$={\left[\frac{i^2-i\sqrt{3}}{1-(-1)}\right]}^n$

$={\left[\frac{-1+i\sqrt{3}}{2}\right]}^n$

$\left[\begin{array}{c}\because \omega =\frac{-1+i\sqrt{3}}{2}\\ \because {\omega }^2=\frac{-1-i\sqrt{3}}{2}\end{array}\right]$

$={\omega }^n$

$[{\omega }^{3n}-1]$

Minimum value of $n=3$.