Question

Find the area of triangle whose vertices are cube roots of unity

Answer 1

The vertices of the triangle formed by the cube roots of unity $=1,w,w^2$

The vertices in complex plane is described by -

$\Rightarrow v=(1,0)$ $w=(\frac{-1}{2},\frac{\sqrt{3}}{2})$ $w^2=(\frac{-1}{2},\frac{-\sqrt{3}}{2})$

The side of the equilateral triangle is $\sqrt{3}$

Altitude of the triangle is $=\frac{\sqrt{3}}{2}\times \sqrt{3}=\frac{3}{2}$

So area of the triangle $=\frac{1}{2}\times \frac{3}{2}\times \sqrt{3}=\frac{3}{4}\times \sqrt{3}$

$=\frac{3\sqrt{3}}{4}$

Hence, the answer is $\frac{3\sqrt{3}}{4}.$