$z^n=(z+1)^n=1$, show that $n$ is divisible by $6$. we are given $z^n=(z+1)^n=1$, $z$ complex number. we want to prove that $n$ is divisible by $6$. I showed that $|z|=|z+1|=1$. Hence $z$ is on the intersection of two unit circles, one centered at $(0,0)$ and the other one centered at $(-1,0)$. Then $z$ can take two values $z=\operatorname{cis}(2 \pi /3)$ or $z=\operatorname{cis}(4 \pi/3)$. Then from $z^n=1=\operatorname{cis}(2 \pi k)$, $k$ integer I showed that $n$ should be divisible by $3$. I am left to show that $n$ should be divisible by $2$. I think if I figure out a way to show that $\operatorname{arg}(z+1)=\pi/3$ or $-\pi/3$ then I would be done, by using $(z+1)^n=\operatorname{cis}(2\pi k)$. but how can I do this? Any ideas are welcome! Thanks in advance!

If $z=\operatorname{cis}\left(\dfrac{2\pi}3\right)$ then what is $z+1$?
It is
\begin{align}
& \left(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\right)+1 \\[6pt]
= {} & \left(-\frac 1 2 + i\frac{\sqrt 3}2\right)+1 \\[6pt]
= {} & \frac 1 2 + i \frac{\sqrt 3} 2 \\[6pt]
= {} & \operatorname{cis}\left( \frac \pi 3 \right).
\end{align}
It's not hard to see that if you raise that to the $6$th power you get $1$ but to any lower power you get something other than $1$.
