$z^n=(z+1)^n=1$, 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!
 A: If $z$ is one of the two nontrivial roots of unity, then $z+1$ is...?
Geometrically: Consider the hexagon inscribed inside the unit circle.
Algebraically: if $z^3=1$ then $z^2+z+1=0$ so $z+1=-z^2$ is a $6$th root.
A: 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 $1$.
A: If you write your two possible values of $z$ as $\frac{-1}{2} \pm \frac{i\sqrt{3}}{2}$, then you are asking to find the argument of $z+1 = \frac{1}{2} \pm \frac{i\sqrt{3}}{2}$, which are standard values.
A: $z^n=1\implies z=\cos\tfrac{2\pi a}n+i\sin\tfrac{2\pi a}n$ where $a\equiv0,1,\cdots,n-1\pmod n$
$(z+1)^n=1\implies z+1=\cos\tfrac{2\pi b}n+i\sin\tfrac{2\pi b}n$ where $b\equiv0,1,\cdots,n-1\pmod n$
Equating the imaginary parts, $\sin\tfrac{2\pi a}n=\sin\tfrac{2\pi b}n$
$\implies\cos\tfrac{2\pi a}n=\pm\cos\tfrac{2\pi b}n$
Equating the real parts, $\cos\tfrac{2\pi a}n=\cos\tfrac{2\pi b}n-1$
Clearly, $\cos\tfrac{2\pi a}n=\cos\tfrac{2\pi b}n$ gives absurdity
$\implies\cos\tfrac{2\pi a}n=-\cos\tfrac{2\pi b}n$
Consequently, $-\cos\tfrac{2\pi b}n=\cos\tfrac{2\pi b}n-1\iff\cos\tfrac{2\pi b}n=\tfrac12=\cos\tfrac\pi3$
$\implies\tfrac{2\pi a}n=2m\pi\pm\tfrac\pi3=\tfrac\pi3(6m\pm1)$
$\iff\dfrac{n(6m\pm1)}6=a$ which is an integer
$\implies6|n(6m\pm1)$
But $(6m\pm1,6)=1$
