The smallest positive integral value of n for which (1+√3i)n/2 is real is
Given, (1+√3i)n2
=2n2[1+√3i2]n2
=2n2(eiπ3n2)
=2n2.einπ6
Hence we get a real value for nϵ6k where k=0,1,2...
Hence the smallest integral value of n is 6.
Find the smallest positive integer value of n for which (1+i)n(1−n)n−2 is a real number.