How special is $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$?

Question

The well-known identity for complex points forming an equilateral triangle reads

$$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$$

I have a doubt concerning the uniqueness of this identity

Is $z_1^2 + z_2^2 + z_3^2 - z_1 z_2 - z_2 z_3 - z_3 z_1$ the only quadratic form in 3 variables that when replacing with the complex points of an equilateral triangle is equal to 0?

Another related question, in case that's not unique,

Is $z_1^2 + z_2^2 + z_3^2 - z_1 z_2 - z_2 z_3 - z_3 z_1$ the smallest quadratic form, in the sense of the smallest absolute determinant of its associated matrix?

A quadratic form in three variables has 6 parameters, and I've recognized 3 equations: 1) the quadratic form is equal to $0$, 2) the translated quadratic form when $z_i \mapsto z_i+\zeta$ for each $i$ is equal to 0 (translation invariance), and 3) the rotated quadratic form when $z_i \mapsto z_ie^{i\pi/3}$ is equal to $0$ (rotational invariance). How can I narrow down the other parameters?

Answer 1

Let $\omega = e^{i\frac{2\pi}{3}}$ be a cubic root of unity. For any $z_1,z_2,z_3 \in \mathbb{C}$, let

$$\begin{align} a &= \frac13 (z_1 + z_2 + z_3)\\ b &= \frac13 (z_1 + z_2\omega + z_3\omega^2)\\ c &= \frac13 (z_1 + z_2\omega^2 + z_3\omega) \end{align}$$ It is easy to check $$\begin{align} z_1 &= a+b+c\\ z_2 &= a+b\omega^2+c\omega\\ z_3 &= a+b\omega+c\omega^2 \end{align}$$ Let $Q(z_1,z_2,z_3)$ be the quadratic form $$z_1^2+z_2^2+z_3^2-z_1z_2 - z_2z_3 -z_3z_1$$ The reason that $Q(z_1,z_2,z_3) = 0$ iff $\triangle z_1z_2z_3$ is equilateral comes down to the fact $$Q(z_1,z_2,z_3) = (z_1 + z_2\omega + z_3\omega^2)(z_1 + z_2\omega^2+z_3\omega) = 9bc$$ So $$Q(z_1,z_2,z_3) = 0 \iff bc = 0 \iff b = 0 \lor c = 0$$ When $b = 0$, $(z_1,z_2,z_3) = (a+c,a+c\omega,a+c\omega^2)$ and we can obtain $z_2,z_3$ by rotating $z_1$ with respect to $a$ counter-clockwisely for angles $120^\circ$ and $240^\circ$ respectively. This implies $\triangle z_1,z_2,z_3$ is equilateral and $z_1,z_2,z_3$ are ordered counter-clockwisely with respect to center $a$.

By a similar argument, if $c = 0$, $\triangle z_1,z_2,z_3$ is again equilateral but $z_1,z_2,z_3$ are ordered clockwisely with respect to $a$.

To show the quadratic form $Q(z_1,z_2,z_3)$ is essentially unique. Let $Q'(z_1,z_2,z_3)$ be another quadratic form which vanishes on and only on equilateral triangles. Expand $Q'(z_1,z_2,z_3)$ in terms of $a,b,c$, there are coefficients $\alpha,\beta_1,\beta_2, \gamma_1,\gamma_2,\gamma_3$ (not al zero) such that

$$Q'(z_1,z_2,z_3) = \alpha a^2 + a (\beta_1 b + \beta_2 c) + (\gamma_1 b^2 + \gamma_2 bc + \gamma_3 c^2)\tag{*1}$$

Pick any $z_1,z_2,z_3$ such that $\triangle z_1z_2z_3$ is equilateral. If one translate all vertices by same amount, $z_k \mapsto z_k + t$, the resulting triangle is also equilateral. However, $a \mapsto a + t$ but $b,c$ remains the same. This means for that specific $b,c$. $(*1) = 0$ for all $a$. This forces

$$\alpha = \beta_1 b + \beta_2 c = \gamma_1 b^2 + \gamma_2 bc + \gamma_3 c^2 = 0$$

It is trivial to find equilateral triangles with $b = 0, c\ne 0$ or triangles with $b \ne 0, c = 0$. This forces $\beta_1 = \beta_2 = \gamma_1 = \gamma_3 = 0$. As a result, $\gamma_2 \ne 0$ and

$$Q'(z_1,z_2,z_3) = \gamma_2 bc = \frac{\gamma_2}{9} Q(z_1,z_2,z_3)$$

From this, we can deduce the quadratic form $Q(z_1,z_2,z_3)$ is unique up to non-zero scaling constants.

Answer 2

I went into full calculation, it's not very elegant:

The vertices of a arbitrary regular triangle, given the parameters $\tilde{z}$, $R$, and $\theta$, are as follows:

$$z_{1} = \tilde{z} + R e^{i\theta}$$ $$z_{2} = \tilde{z} + R e^{i\left(\theta + \frac{2\pi}{3}\right)}$$ $$z_{3} = \tilde{z} + R e^{i\left(\theta + \frac{4\pi}{3}\right)}$$

Replacing these points into

$$az_{1}^2+bz_{2}^2+cz_{3}^2+dz_{1}z_{2}+ez_{1}z_{3}+fz_{2}z_{3}$$

gives

$$e^{2 i \theta} R^2 \left(a - (-1)^{\frac{1}{3}} b + (-1)^{\frac{2}{3}} c + (-1)^{\frac{2}{3}} d - (-1)^{\frac{1}{3}} e + f\right) + \frac{1}{2} e^{i \theta} R \tilde{z} \left(4a + 4(-1)^{\frac{2}{3}} b - 2c - 2i \sqrt{3} c + d + i \sqrt{3} d + e - i \sqrt{3} e - 2f\right) + \left(a + b + c + d + e + f\right) \tilde{z}^2$$

Therefore, we have three equations to satisfy (considering any value of $\tilde{z}$, $R$, or $\theta$) $$ a - (-1)^{\frac{1}{3}} b + (-1)^{\frac{2}{3}} c + (-1)^{\frac{2}{3}} d - (-1)^{\frac{1}{3}} e + f=0$$ $$4a + 4(-1)^{\frac{2}{3}} b - 2c - 2i \sqrt{3} c + d + i \sqrt{3} d + e - i \sqrt{3} e - 2f=0$$ $$ a + b + c + d + e + f=0 $$

Using $(-1)^{\frac{1}{3}}=\frac{1}{2} (1 + i \sqrt3)$ and $(-1)^{\frac{2}{3}}=\frac{1}{2} (-1 + i \sqrt3)$, from the first two equations we get:

$$a=b=c \quad \quad d=e=f$$

And combined with the last equation we get $a=-d$

Therefore, the form should be

$$az_{1}^2+az_{2}^2+az_{3}^2-az_{1}z_{2}-az_{1}z_{3}-az_{2}z_{3}$$

And dividing by $a$, we get the original one, so it's unique (up to non-zero scalar multiples $a$)