The area of the triangle whose vertices are $i, \omega$ and $\omega^2$ is (Where $\omega$ is a complex cube…

The area of a triangle with vertices at complex numbers $i$, $\omega$, and $\omega^2$ can be found using their coordinates in the complex plane. Representing $\omega = \cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})$, the points are $A(0,1)$, $B(-\frac{1}{2}, \frac{\sqrt{3}}{2})$, and $C(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$.

Since $B$ and $C$ share the same $x$-coordinate, the base $BC$ is vertical with length $\sqrt{3}$. The height from $A$ to the line $x = -\frac{1}{2}$ is $\frac{1}{2}$.

$\frac{1}{2} \times \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{4}$

The area is $\frac{\sqrt{3}}{4}$ square units.