Suppose $z$ is a complex number and $n\in \mathbf{N}$ be such that $z^n=(1+z{)}^n=1$ , then the least value of $n$ is

$|z{|}^n=|1+z{|}^n=1\Rightarrow \left|z\right|=|z+1|=1$

$\Rightarrow \left|z\right|=1$and $z$ lies on the perpendicular bisector of the segment joining $0+0i$ and $-1+0i,$ that $z$ lies on the line $\mathrm{Re}\left(z\right)=-1/2$

Let     $z=-\frac{1}{2}+iy,$then $\left|z\right|=1$

$\frac{1}{4}+y^2=1\Rightarrow y=\pm \frac{\sqrt{3}}{2}$

$\therefore z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}=\omega ,{\omega }^2$

where $\varpi$ is complex cube root of unit

If $z=\omega ,{\omega }^n=(1+\omega {)}^n=(-1{)}^n{\omega }^{2n}$

$\Rightarrow n$ is even and multiple of 3. Thus, least value of n is 6. Similarly, for $z = {\varpi }^2$