Question

What is the area of the triangle whose vertices are the cube roots of unity?

Answer 1

The vertices of the triangle formed by the cube roots of unity = 1 , ω, ω² .

the vertices in the complex plane are described by :

1 = (1, 0)        ω = (-1/2 , √3/2 )   ω² = ( -1/2 , -√3/2)

The side of this equilateral  triangle is = √3

Altitude of the triangle is =  √3/2 * √3 = 3/2

So area of triangle = 1/2 * 3/2 * √3  = 3√3 / 4