The least positive integer n such that ( 2i ) ^ n ( 1 - i ) ^ n - 2...

$\dfrac{( 2i{)}^{n}}{( 1 - i{)}^{n - 2}}$

$= \dfrac{( 2i{)}^{n}}{( - 2i) \dfrac{n - 2}{2}}$

$= \dfrac{( 2i{)}^{\dfrac{n + 2}{2}}}{( - 1{)}^{\dfrac{n - 2}{2}}}$

$= \dfrac{\left(2\right)\dfrac{n + 2}{2}i\dfrac{n + 2}{2}}{( - 1) \dfrac{n - 2}{2}}$
Clearly $n$ must be even $n = 2, 4$ rejected.

So the least positive integer possible is $n = 6$