The smallest positive integer for which $\left(\frac{1-i}{1+i}\right)^{2 n}=-1$ where $i=\sqrt{-1}$

Question

The smallest positive integer for which $\left(\frac{1-i}{1+i}\right)^{2 n}=-1$ where $i=\sqrt{-1}$ is

Professor PC Thomas Classes · Answer

Simplify $(1-i)/(1+i) = -i$.