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

A problem way too cool and cute to pass up, so check this out:
$(1.) \; z^n = 1 \Rightarrow \vert z \vert^n = 1, \tag{1}$
$(2.) \; \vert z \vert^n = 1 \Rightarrow \vert z \vert = 1 \Rightarrow \exists \theta \in \Bbb R  \;\text{such that} \; z = e^{i\theta}, \tag{2}$
$(3.) \; \vert z \vert = 1 \Rightarrow z \bar z = 1, \tag{3}$
$(4.) \; (z + 1)^n = 1 \Rightarrow \vert z + 1 \vert^n = 1 \Rightarrow \vert z + 1\vert = 1, \tag{4}$
$(5.) \; \vert z + 1 \vert = 1 \Rightarrow (1 + z)(1 + \bar z) = 1, \tag{5}$
$(6.) \;  (1 + z)(1 + \bar z) = 1 \Rightarrow z \bar z + z + \bar z + 1 = 1 \Rightarrow z \bar z + z + \bar z = 0, \tag{6}$
$(7.) \; \text{by (3) and (6),} \; z + \bar z = -1, \tag{7}$
$(8.) \; \text{by (2) and (7),} \; 2 \Re{z} = 2 \cos \theta = -1 \Rightarrow \cos \theta = -\dfrac{1}{2}, \tag{8}$
$(9.) \; \cos \theta = -\dfrac{1}{2} \Rightarrow \Re{(1 + z)} = 1 + \cos \theta = \dfrac{1}{2}, \tag{9}$
$(10.) \; \Re{(1 + z)} = \dfrac{1}{2} \; \text{and} \; \vert 1 + z \vert = 1 \Rightarrow 1 + z = e^{(\pm 2\pi i/ 6) + 2k \pi}, k \in \Bbb Z, \tag{10}$
$(11.) \;  1 + z = e^{(\pm 2\pi i/ 6) + 2k \pi} \Rightarrow 1 = (1 + z)^n = e^{\pm 2n \pi i / 6}, \tag{11}$
$(12.) \; e^{\pm 2n \pi i / 6} = 1 \Rightarrow 6 \mid n. \tag{12}$
QED
