Given $x,y,z\in \mathbb C$, $x+y+z=0$ and $|x|=|y|=|z|$, prove that $x^3=y^3=z^3$.

Question

Given $z_1,z_2,z_3\in \mathbb C$ such that $$z_1+z_2+z_3=0 \tag {C1}$$ $$|z_1|=|z_2|=|z_3|, \tag{C2}$$ prove that $$z_1^3=z_2^3=z_3^3.\tag{C3}$$

Source: List of problems for math-contest training.

My attempt: By (C2), let $\rho$ be the common modulus for $|z_j|,j=1,2,3.$ Therefore $\forall j$, $$z_j=\rho e^{i\theta_j}\ \ \text{and} \ \ z_j^3=\rho^3 e^{i 3\theta_j}$$ And, by (C1) it holds that $$e^{i \theta_1}+e^{i \theta_2}+e^{i \theta_3}=0.$$ But (C3) implies $$e^{i 3\theta_1}=e^{i 3\theta_2}=e^{i 3\theta_3}.$$ and that implies $\theta_1=\theta_2= \theta_3$ (for $\theta_j \in [0,2\pi[$). But that appears to be inconsistent with (C1).

Is there a problem with the problem statement? hints and answers are welcomed.

Possible duplicate of showing certain vertices form an equilateral triangle — Martin R, Dec 22 '18 at 13:52
https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question — lab bhattacharjee, Dec 22 '18 at 14:10

score 2 · Answer 1 · answered Dec 22 '18 at 13:48

(C3) doesn't imply that $\theta_1=\theta_2=\theta_3$.

Take for example $\theta_1 = 0, \theta_2 = \frac{2\pi}{3}, \theta_3 = \frac{4\pi}{3}$

Then $$e^{i3\theta_1} = e^0 = 1$$ $$e^{i3\theta_2} = e^{2\pi i} = 1$$ $$e^{i3\theta_3} = e^{4\pi i} = 1$$

But clearly $\theta_1\not = \theta_2, \theta_1 \not = \theta_3,\theta_2\not = \theta_3$.

Given $x,y,z\in \mathbb C$, $x+y+z=0$ and $|x|=|y|=|z|$, prove that $x^3=y^3=z^3$.

1 Answers1