Equilateral Triangle on a Circle

Necessary and sufficient condition for z₁, z₂, z₃ to form an equilateral triangle. Equivalently: z₁ + z₂ω + z₃ω² = 0 or z₁ + z₂ω² + z₃ω = 0, where ω is a cube root of unity — the vertices are related by 120° rotations. If the triangle is inscribed in |z|=r, its centroid is at the centre iff z₁+z₂+z₃ = 0.

The equilateral triangle condition connects cube roots of unity, rotation, and symmetric polynomials in a single identity.

Setup

Let $z_1$, $z_2$, $z_3$ form an equilateral triangle with centroid $G = (z_1 + z_2 + z_3)/3$.

In an equilateral triangle, each vertex is obtained from another by rotating $120° = 2\pi/3$ about the centroid. Let $\omega = e^{2\pi i/3}$. Then:

$$z_2 - G = \omega(z_1 - G) \quad \text{or} \quad z_2 - G = \omega^2(z_1 - G)$$

Deriving the Condition

Take $z_2 - G = \omega(z_1 - G)$ and $z_3 - G = \omega^2(z_1 - G)$.

Let $w = z_1 - G$. Then:

$$z_1 = G + w, \quad z_2 = G + \omega w, \quad z_3 = G + \omega^2 w$$

Compute $z_1^2 + z_2^2 + z_3^2$:

$$= (G+w)^2 + (G+\omega w)^2 + (G+\omega^2 w)^2$$ $$= 3G^2 + 2Gw(1 + \omega + \omega^2) + w^2(1 + \omega^2 + \omega^4)$$

Since $1 + \omega + \omega^2 = 0$ and $\omega^4 = \omega$:

$$= 3G^2 + 0 + w^2(1 + \omega^2 + \omega) = 3G^2 + 0 = 3G^2$$

Now compute $z_1 z_2 + z_2 z_3 + z_3 z_1$:

$$= (G+w)(G+\omega w) + (G+\omega w)(G+\omega^2 w) + (G+\omega^2 w)(G+w)$$ $$= 3G^2 + Gw(1+\omega + \omega + \omega^2 + \omega^2 + 1) + w^2(\omega + \omega^3 + \omega^2)$$ $$= 3G^2 + 2Gw(1 + \omega + \omega^2) + w^2 \omega(1 + \omega + \omega^2)...$$

Let me compute more carefully:

$$z_1z_2 = G^2 + G(\omega w + w) + \omega w^2$$ $$z_2z_3 = G^2 + G(\omega^2 w + \omega w) + \omega^3 w^2$$ $$z_3z_1 = G^2 + G(w + \omega^2 w) + \omega^2 w^2$$

Sum:

$$= 3G^2 + Gw(1+\omega + \omega + \omega^2 + 1 + \omega^2) + w^2(\omega + 1 + \omega^2)$$ $$= 3G^2 + 2Gw(1 + \omega + \omega^2) + w^2(1 + \omega + \omega^2)$$ $$= 3G^2 + 0 + 0 = 3G^2$$

Therefore:

$$z_1^2 + z_2^2 + z_3^2 = 3G^2 = z_1z_2 + z_2z_3 + z_3z_1$$ $$\boxed{z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1} \qquad \blacksquare$$

Alternative Form

The condition is equivalent to:

$$(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2 = 0$$

Expanding: $2z_1^2 + 2z_2^2 + 2z_3^2 - 2z_1z_2 - 2z_2z_3 - 2z_3z_1 = 0$, which is the same condition. Over $\mathbb{R}$ this would force all differences to zero — but over $\mathbb{C}$, the sum of three complex squares can vanish without each being zero.

When the Centroid Is at the Origin

If $z_1 + z_2 + z_3 = 0$, then $G = 0$ and:

$$z_2 = \omega z_1, \quad z_3 = \omega^2 z_1$$

The three vertices are $z_1$, $\omega z_1$, $\omega^2 z_1$ — the cube roots of $z_1^3$, equally spaced on a circle of radius $|z_1|$.