Vertices of equilateral triangle on complex plane

Question

I come cross a problem when I read a book of complex analysis:

If three complex number $a,b,c$ satisfy the relation of

$a^2+b^2+c^2=ab+ac+bc$.

Prove that: these numbers must be three vertices of an equilateral triangle on the complex plane.

if $a,b,c$ are real numbers, we have $a=b=c$. but I’m not sure how to prove it with complex number. The hint I got is:

Calculate $((b-a)\omega+(b-c))\cdot((b-a)\omega^2+(b-c))$, where $\omega$ is nonreal cube root of unity.

Answer 1

What is the geometric meaning of

$((b-a)\omega + (c-a))((b-a)\omega^2 + (c-a))$

If we subtract $a$ from each point we translate them such that $a$ is on the origin.

$(b-a)\omega$ rotates $(b-a)$ 120 degrees clockwise $(b-a)\omega^2$rotates $(b-a)$ 120 degrees counter-clockwise.

If $a,b,c$ form an equilateral triangle, one of these will be exactly the negative of $(c-a)$

In which case one of the factors of $((b-a)\omega + (c-a))((b-a)\omega^2 + (c-a))$ equals $0.$

Answer 2

The hint is one way to write it. Another is $$ (x + y \omega + z \omega^2)(x + y \omega^2 + z \omega) = x^2 + y^2 + z^2 - yz-zx-xy $$