# Why is the ratio of any two sides of an equilateral triangle on a complex plane equal to a complex cubic root of unity?

Why is the ratio of any two sides of an equilateral triangle on a complex plane equal to a complex root of unity?

This question is derived from an excerpt from the following note, which states that:

"If $$a_1, a_2, a_3$$ are vertices of an equilateral triangle, then triple of complex numbers $$a_1 − a_2, a_2 −a_3$$ and $$a_3 −a_1$$ have the same length and the ratio of any consecutive pair is the same (complex) cube root of unity."

I am not sure why the case is true. I know how to represent a ratio of two complex numbers on a complex plane but I am lost on how to prove my question.

Intuitive AND rigorous answer would be greatly appreciated, though either is fine

If $$a_1, a_2, a_3$$ are the vertices of the triangle then $$a_1-a_2, a_2 -a_3, a_3-a_1$$ are `vectors' representing the edges of the triangle.
You can turn one edge of an equilateral triangle into the next edge by rotating it by angle $$2\pi/3$$. This corresponds to multiplication by $$e^{2\pi i 3}$$, aka the complex cube root of unity.
In other words, $$a_1-a_2 = e^{2\pi i /3} (a_2-a_3)$$ and so on.