Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$. Prompt:

Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question:
I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$.
What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane?
Is there any intuition to be gained here?

If $z_1=w_1+x$, $z_2=w_2+x$ and $z_3=w_3+x$, the identity becomes
$$
w_1^2+2w_1x+x^2+
w_2^2+2w_2x+x^2+
w_3^2+2w_3x+x^2=\\
w_1w_2+w_1x+w_2x+x^2+
w_2w_3+w_2x+w_3x+x^2+
w_3w_1+w_3x+w_1x+x^2
$$
that becomes
$$
w_1^2+w_2^2+w_3^2=w_1w_2+w_2w_3+w_3w_1
$$
so, taking $x=-z_3$, we can assume $z_3=0$ and we need to prove that if $z_1$, $z_2$ and $0$ are the vertices of an equilateral triangle, then
$$
z_1^2+z_2^2=z_1z_2.
$$
If $z_1=w_1u$ and $z_2=w_2u$ (with $u\ne0$), the identity above becomes $w_1^2+w_2^2=w_1w_2$, so we can assume $z_1=e^{i\pi/6}$, which means that $z_2=e^{i(\pi/6+\pi/3)}$ or $z_2=e^{i(\pi/6-\pi/3)}$.
If $z_2=e^{i(\pi/6+\pi/3)}=e^{i\pi/2}=i$, then a rotation brings us in the situation with $z_1=e^{-i\pi/6}$ and $z_2=e^{i\pi/6}$. So we are reduced to proving that
$$
e^{i\pi/3}+e^{-i\pi/3}=1
$$
which is obvious.
