The area of Delta whose vertices are z, omega | vedclass

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Question

(B) Let the vertices be $A = z$,$B = \omega z$,and $C = z + \omega z$.The side lengths are:$|AB| = |z - \omega z| = |z(1 - \omega)| = |z| |1 - \omega|

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{4} |z|^2$

Vedclass · Answer

(B) Let the vertices be $A = z$,$B = \omega z$,and $C = z + \omega z$.The side lengths are:$|AB| = |z - \omega z| = |z(1 - \omega)| = |z| |1 - \omega| = |z| \sqrt{1^2 + 1^2 - 2 \cos(120^{\circ})} = |z| \sqrt{3}$.$|BC| = |(z + \omega z) - \omega z| = |z|$.$|AC| = |(z + \omega z) - z| = |\omega z| = |z|$.Since $|BC| = |AC| = |z|$ and $|AB| = |z|\sqrt{3}$,the triangle is an isosceles triangle with sides $|z|, |z|, |z|\sqrt{3}$.The angle between sides $BC$ and $AC$ is $120^{\circ}$.Area $= \frac{1}{2} 	imes |BC| 	imes |AC| 	imes \sin(120^{\circ})$Area $= \frac{1}{2} 	imes |z| 	imes |z| 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} |z|^2$.

The area of $\Delta$ whose vertices are $z, \omega z, z + \omega z$ is (where $\omega$ is a complex cube root of unity):

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