Question

$If (\frac{(1+i}{1-i})^m=1, \text{then find the least positive integral value of m.}$

Answer 1

$(\frac{(1+i}{1-i})^m=1$

$⇒(\frac{1+i}{1-i}×\frac{1+i}{1+i})^m=1$

$⇒(\frac{(1+i)^2}{1^2-1^2})^m=1$

$⇒(\frac{1^2+i^2+2i}{2})^m=1$

$⇒(\frac{1-1+2i}{2})^m=1$

$⇒(\frac{2i}{2})^m=1$

$⇒i^m=1$

$\text{∴\,m=4k, where k is some integer.}$

$\text{Therefore, the least positive integer is 1. }$

$\text{Thus, the least positive integral value of m is 4 (= 4 × 1). }$