Suppose z is a complex number and n∈N be such that zn=(1+z)n=1 , then the least value of n is

|z|n=|1+z|n=1 ⇒ |z|=|z+1|=1⇒|z|=1 and z lies on the perpendicular bisector of the segment joining 0+0i and −1+0i, that z lies on the line Re⁡(z)=−1/2Let     z=−12+iy,then |z|=114+y2=1 ⇒ y=±32∴ z=−12±i32=ω,ω2where ϖ is complex cube root of unitIf z=ω,ωn=(1+ω)n=(−1)nω2n⇒n is even and multiple of 3. Thus, least value of n is 6. Similarly, for z = ϖ2