For $a,b,c\in\mathbb{C}$, prove that $a^2+b^2+c^2=ab+bc+ca$ iff $a$, $b$, and $c$ form an equilateral triangle.

Question

Let $a$, $b$, $c$ are different complex numbers. Prove that the points $a$, $b$, and $c$ form an equilateral triangle in the complex plane if and only if $$a^2 + b^2 + c^2 = ab + bc + ca.$$

I found the same question few times here and there, but they all use $e^i$ which I am not familiar yet with. So I need a proof that is as simple as possible using only "common" algebra knowledge.

Answer 1

Hint:

By equating sides and angles, $$\frac{a-b}{b-c}=\frac{b-c}{c-a}=\frac{c-a}{a-b}=\omega$$

where $\omega=e^{\pm 2i\pi/3}$

Observe that $a=b+(b-c)\omega$ which means rotating $c$ about $b$ by an exterior angle of $\pm \dfrac{2i\pi}{3}$ to get $a$.

Answer 2

We have $a^2+b^2+c^2-ab-ac-bc=(a+\xi b+\xi^2 c)(a+\xi^2 b+\xi c)$, where $\xi$ is the primitive $3$rd root of unity $\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)$. Observe that $a+\xi b+\xi^2 c=0$ iff $a$, $b$, $c$ form a positively oriented (counterclockwise) equilateral triangle. Similarly, $a+\xi^2b+\xi c=0$ iff $a$, $b$, and $c$ form a negatively oriented (clockwise) equilateral triangle.