# 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 $z_2 - z_3$.

From Complex Multiplication as Geometrical Transformation/Corollary:

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

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

Then:

\(\displaystyle \dfrac {z_2 - z_1} {z_1 - z_3}\) | \(=\) | \(\displaystyle \dfrac {z_3 - z_1} {z_2 - z_3}\) | $(1)$ divided by $(2)$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {z_2 - z_1} \paren {z_2 - z_3}\) | \(=\) | \(\displaystyle \paren {z_3 - z_1} \paren {z_1 - z_3}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1\) | \(=\) | \(\displaystyle - {z_1}^2 - {z_3}^2 + 2 z_3 z_1\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle {z_1}^2 + {z_2}^2 + {z_3}^2\) | \(=\) | \(\displaystyle z_1 z_2 + z_2 z_3 + z_3 z_1\) |

$\Box$

### 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:

\(\displaystyle {z_1}^2 + {z_2}^2 + {z_3}^2\) | \(=\) | \(\displaystyle z_1 z_2 + z_2 z_3 + z_3 z_1\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1\) | \(=\) | \(\displaystyle - {z_1}^2 - {z_3}^2 + 2 z_3 z_1\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {z_2 - z_1} \paren {z_2 - z_3}\) | \(=\) | \(\displaystyle \paren {z_3 - z_1} \paren {z_1 - z_3}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac {z_2 - z_1} {z_1 - z_3}\) | \(=\) | \(\displaystyle \dfrac {z_3 - z_1} {z_2 - z_3}\) |

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:

\(\displaystyle {z_1}^2 + {z_2}^2 + {z_3}^2\) | \(=\) | \(\displaystyle z_1 z_2 + z_2 z_3 + z_3 z_1\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle - {z_2}^2 - {z_1}^2 + 2 z_1 z_2\) | \(=\) | \(\displaystyle {z_3}^2 - z_1 z_3 - z_2 z_3 + z_1 z_2\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {z_2 - z_1} \paren {z_1 - z_2}\) | \(=\) | \(\displaystyle \paren {z_3 - z_1} \paren {z_3 - z_2}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac {z_2 - z_1} {z_3 - z_2}\) | \(=\) | \(\displaystyle \dfrac {z_3 - z_1} {z_1 - z_2}\) |

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.

$\blacksquare$

## Sources

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $7$