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




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

Proof

 * EquilateralTriangleInComplexPlane.png

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:

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:

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:

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.