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

Solution 1:

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

Solution 2:

Hint 1: $|z-z_0|=r$ is a circle with radius $r$ and center $z_0$;

Hint 2: the unit circle is $z=e^{i\theta}$;

Hint 3: for any $n\in\Bbb{N}$, $(e^{i\theta})^n=e^{in\theta}$;

Hint 4: if $\cos \theta= \pm 1$ and $\sin \theta =0$ then $\theta=k\pi$ for some integer $k$;

Hint 5: if $a,b,c$ are integers such that $a|bc$ and $\gcd(a,b)=1$ then $a|c$.