Given: (1−i1+i)2n=1.First simplify 1−i1+i using conjugate (1+i):1−i1+i⋅1+i1+i=(1−i)(1+i)(1+i)2Expand numerator:(1+i)2=1+2i+i2=1+2i−1=2iSimplify denominator:(1−i)(1+i)=12+12=2⇒1−i1+i=22i=iSo the condition becomes:i2n=1i2n=(i2)n=(−1)n(−1)n=1 when n is even.Least positive even n=2.n=2