\((\frac{(1+i}{1-i})^m=1\)
\(⇒(\frac{1+i}{1-i}×\frac{1+i}{1+i})^m=1\)
\(⇒(\frac{(1+i)^2}{1^2-1^2})^m=1\)
\(⇒(\frac{1^2+i^2+2i}{2})^m=1\)
\(⇒(\frac{1-1+2i}{2})^m=1\)
\(⇒(\frac{2i}{2})^m=1\)
\(⇒i^m=1\)
\(\text{∴\,m=4k, where k is some integer.}\)
\(\text{Therefore, the least positive integer is 1. }\)
\(\text{Thus, the least positive integral value of m is 4 (= 4 × 1). }\)