Prove that the area of the triangle with vertices $z, wz, z+wz$ is $\frac{\sqrt{3}}{4}|z|^2$
My Attempt $$ A=\frac{i}{4}\begin{vmatrix} 1&1&1\\ z&wz&z+wz\\ \bar{z}&\bar{w}\bar{z}&\bar{z}+\bar{w}\bar{z}\\ \end{vmatrix} \\ =\frac{i}{4}\bigg[wz(\bar{z}+\bar{w}\bar{z})-\bar{w}\bar{z}(z+wz)-z(\bar{z}+\bar{w}\bar{z})+\bar{z}(z+wz)-z(\bar{w}\bar{z})-\bar{z}(wz)\bigg]\\ =\frac{i}{4}\bigg[ w|z|^2+|w|^2|z|^2-\bar{w}|z|^2-|w|^2|z|^2-|z|^2-\bar{w}|z|^2+|z|^2-{w}|z|^2-\bar{w}|z|^2+w|z|^2 \bigg]\\ =\frac{i|z|^2}{4}\bigg[ \color{red}w+\color{blue}{|w|^2}-\color{green}{\bar{w}}-\color{blue}{|w|^2}-\color{brown}1-\color{orange}{\bar{w}}+\color{brown}1-\color{red}{w}-\bar{w}+w \bigg] =\frac{i|z|^2}{4}(w-\bar{w}) $$ How can it be same as $\frac{\sqrt{3}}{4}|z|^2$ ?
HINT...It is easier just to draw a picture. Assuming that you mean $\omega$ is a cube root of unity, the three points form an isosceles triangle whose two equal sides have length $|z|$, since $|\omega z|=|z|$, and the angle between these sides is $120^o$
The answer you obtained is equivalent to the answer given because $\omega=\cos120^o+i\sin120^o$