Question

Find the least positive integar n such that ${\left(\frac{2i}{1+i}\right)}^n$ is a positive integer.

Answer 1

${\left(\frac{2i}{i+1}\right)}^n$ is an integer

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

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

$=(1+i{)}^n$

Put $n=1$

$=1+i$ this is imaginary

Put $n=2$

$=(1+i{)}^2=2i$

Put $n=4$

$\left(2i\right)\left(2i\right)=-4$

Put $n=8$

$=(-4)(-4)=16$ (positive integer)

Least positive integer $n=8$