# Vertices of Equilateral Triangle in Complex Plane

## Theorem

Let $z_1$, $z_2$ and $z_3$ be complex numbers.

Then:

$z_1$, $z_2$ and $z_3$ represent on the complex plane the vertices of an equilateral triangle
${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$

### Corollary

Let $u, v \in \C$ be complex numbers.

Then:

$0$, $u$ and $v$ represent on the complex plane the vertices of an equilateral triangle.
$u^2 + v^2 = u v$

## Proof

### Sufficient Condition

Let $T$ be the equilateral triangle whose vertices are $z_1$, $z_2$ and $z_3$.

We have that $z_2 - z_1$ and $z_3 - z_1$ are two sides of $T$ which meet at $z_1$.

From the geometry of $T$ it follows that $z_2 - z_1$ is at an angle of $\pi/3$ to $z_3 - z_1$.

Similarly, $z_1 - z_3$ and $z_2 - z_3$ are two sides of $T$ which meet at $z_3$.

From the geometry of $T$ it follows that $z_1 - z_3$ is at an angle of $\pi / 3$ to $z_2 - z_3$.

$(1): \quad z_2 - z_1 = e^{i \pi / 3} \left({z_3 - z_1}\right)$
$(2): \quad z_1 - z_3 = e^{i \pi / 3} \left({z_2 - z_3}\right)$

Then:

 $\ds \dfrac {z_2 - z_1} {z_1 - z_3}$ $=$ $\ds \dfrac {z_3 - z_1} {z_2 - z_3}$ $(1)$ divided by $(2)$ $\ds \leadsto \ \$ $\ds \paren {z_2 - z_1} \paren {z_2 - z_3}$ $=$ $\ds \paren {z_3 - z_1} \paren {z_1 - z_3}$ $\ds \leadsto \ \$ $\ds {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1$ $=$ $\ds - {z_1}^2 - {z_3}^2 + 2 z_3 z_1$ $\ds \leadsto \ \$ $\ds {z_1}^2 + {z_2}^2 + {z_3}^2$ $=$ $\ds z_1 z_2 + z_2 z_3 + z_3 z_1$

$\Box$

### Necessary Condition

Let:

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

Then:

 $\ds {z_1}^2 + {z_2}^2 + {z_3}^2$ $=$ $\ds z_1 z_2 + z_2 z_3 + z_3 z_1$ $\ds \leadsto \ \$ $\ds {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1$ $=$ $\ds - {z_1}^2 - {z_3}^2 + 2 z_3 z_1$ $\ds \leadsto \ \$ $\ds \paren {z_2 - z_1} \paren {z_2 - z_3}$ $=$ $\ds \paren {z_3 - z_1} \paren {z_1 - z_3}$ $\ds \leadsto \ \$ $\ds \dfrac {z_2 - z_1} {z_1 - z_3}$ $=$ $\ds \dfrac {z_3 - z_1} {z_2 - z_3}$

Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_3$ and $z_2 - z_1$.

Similarly:

 $\ds {z_1}^2 + {z_2}^2 + {z_3}^2$ $=$ $\ds z_1 z_2 + z_2 z_3 + z_3 z_1$ $\ds \leadsto \ \$ $\ds - {z_2}^2 - {z_1}^2 + 2 z_1 z_2$ $=$ $\ds {z_3}^2 - z_1 z_3 - z_2 z_3 + z_1 z_2$ $\ds \leadsto \ \$ $\ds \paren {z_2 - z_1} \paren {z_1 - z_2}$ $=$ $\ds \paren {z_3 - z_1} \paren {z_3 - z_2}$ $\ds \leadsto \ \$ $\ds \dfrac {z_2 - z_1} {z_3 - z_2}$ $=$ $\ds \dfrac {z_3 - z_1} {z_1 - z_2}$

Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_2$ and $z_3 - z_2$.

Thus all three angles:

$\angle z_2 z_1 z_3$
$\angle z_1 z_3 z_2$
$\angle z_3 z_2 z_1$

are equal.

By definition, therefore, $\triangle z_1 z_2 z_3$ is equilateral.

$\blacksquare$