<s>[INST] Complex numbers - Exponential numbers - Proof Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.
For this problem I am stumped...how should I begin?
Also there's a hint for it:
From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z + 1)^n = 1$ tell you? What do the resulting equations tell you about $z$?
Could someone give me a hint on where to begin? thanks in advance
 [/INST] We can say: $$z^n=1 \implies |z|^n=1 \implies |z^n|=1 \implies |z|=1$$
Using similar logic, we can also get: $$(z+1)^n=1 \implies |z+1|^n=1 \implies |(z+1)^n|=1 \implies |z+1|=1$$
Let's set $z=a+bi$: $$|z|=|z+1| \implies a^2 + b^2 = (a+1)^2 + b^2 \implies a^2 + b^2 = a^2 + b^2 + 2a + 1$$ $$\implies 2a + 1 = 0 \implies a = -\frac{1}{2}$$
Pluging this back into the equation, we get: $$\left(-\frac{1}{2}\right)^2 + b^2 = 1 \implies b^2 = \frac{3}{4} \implies b = \pm \frac{\sqrt{3}}{2}$$
With this information, we know that: $$z= -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \implies z = e^{\pm 2\pi i/3}$$
Also: $$z+1= \frac{1}{2} \pm \frac{\sqrt{3}}{2}i \implies z+1 = e^{\pm \pi i/3} \implies (z+1)^n = e^{\pm \pi ni/3}$$
For $(z+1)^n$ to be $1$, $\cos \pm \pi n/3 = 1$ and $\sin \pm \pi n/3 = 0$. This can only happen if $\pm \pi n/3 = 2\pi k$ for some integer $k$
Solving for $n$, we get: $$\pm \pi n/3 = 2\pi k \implies n = \pm 6k \implies \boxed{6 \mid n}.$$
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