Question

If $1,\omega ,{\omega }^2$ are cube roots of unity, prove that $1,\omega ,{\omega }^2$ are vertices of an equilateral triangle.

Answer 1

Let $z_1,z_2,z_3$ be the 3 vertices.

$z_1=e^{i{\theta }_1},z_2=e^{i{\theta }_2},z_3=e^{i{\theta }_3}$

$\therefore z_1+z_2+z_3=0\Rightarrow e^{i{\theta }_1}+e^{i{\theta }_2}+e^{i{\theta }_3}=0\Rightarrow 1+e^{i({\theta }_2-{\theta }_1)}+e^{i({\theta }_3-{\theta }_1)}=0$

$\Rightarrow e^{i({\theta }_2-{\theta }_1)}+e^{i({\theta }_3-{\theta }_1)}=-1$

Using Euler's formula, we get,

$\sin ({\theta }_2-{\theta }_1)+\sin ({\theta }_3-{\theta }_1)=0$ and

$\cos ({\theta }_2-{\theta }_1)+\cos ({\theta }_3-{\theta }_1)=-1$

Let $({\theta }_2-{\theta }_1)=x,({\theta }_3-{\theta }_1)=y$

$\Rightarrow \sin x+\sin y=0\Rightarrow \sin x=-\sin y=\sin (-y)\iff x=\{2k\pi -y,2k\pi +\pi +y$

$\cos x+\cos y=-1\Rightarrow 2\cos y=-1\iff y=2k\pi \pm \frac{2\pi }{3},x=2kp\mp \frac{2\pi }{3}$

So, ${\theta }_2,{\theta }_1,and{\theta }_3,{\theta }_1$ differ $\frac{2\pi }{3}$

$\Rightarrow$ triangle being equivalent.