If $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1and{z}_1+z_2+z_3=0$ then area of the triangle whose vertices are $z_1,z_2,z_3$ is 

$\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1\Rightarrow$circum centre of triangle is origin Also  $z_1+z_2+z_3=0\text{, }$ which coincide with centriod since radius is 1, length of side is $a=\sqrt{3}$

Then area of triangle  $\frac{\sqrt{3}}{4}{a}^2=\frac{3\sqrt{3}}{4}$