complex number to a power divisible by 6

Question

I actually have a follow-up question to this post -- given that n is a positive integer such that $z^n = (z+1)^n = 1$, I need to show that n is divisible by 6. I can now show that $z$ and $z+1$ both lie on the unit circle, but can this be of any help to me?

Answer 1

As you have noticed, $|z|=1$ and $|z+1|=1$ and so $z$ is in the intersection of the unit circle centered at the origin with the unit circle centered at $-1$.

The two intersection points of these circles are the primitive cubic roots of unit: $\omega$ and $\omega^2$.

Now $\omega^2+\omega+1=0$ and so $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$.

The result follows because $-\omega$ and $-\omega^2$ are primitive $6$-th roots of unit.

Answer 2

Note that $$z^n=1 \implies z = e^{i2\pi k/n}, \tag 1$$ where $k \in \mathbb{Z}$. Likewise, $$(1+z)^n = 1 \implies 1+z = e^{i2\pi r/n} \implies 1=e^{i2\pi r/n}-e^{i2\pi k/n},\tag 2$$ where $r \in \mathbb{Z}$.

Now $$e^{i2\pi r/n}-e^{i2\pi k/n} = 2e^{i\pi (r+k)/n}\sin\left(\frac{\pi(r-k)}{n}\right)=1 \implies \sin\left(\frac{\pi(r-k)}{n}\right)=\pm \frac{1}{2}.\tag 3$$ Consequently, $$\frac{\pi(r-k)}{n}=\pm\frac{\pi}{6}+t\pi, \tag 4$$ where $t \in \mathbb{Z}$. Simplify $(6)$ to get $$n = \pm\frac{6(r-k)}{1+6t}.$$ Observe that $n$ is an integer, and $\gcd(6,1+6t)=1$. Therefore, $n=6p$, where $p \in \mathbb{Z}$.