The area of the polygon, whose vertices are the non-real roots of the equation z ¯ = i z 2 is

$\overline{z}=i{z}^2\Rightarrow \left(x-iy\right)=i{\left(x+iy\right)}^2$

$\Rightarrow x-iy=\left(x^2-y^2\right)i-2xy$

i.e., $x=-2xy$ and $-y=x^2-y^2$

$\Rightarrow x=0,y=-\frac{1}{2}$ 

When $x=0$; $y=0,1$ 

When $y=-\frac{1}{2}$; $x=\pm \frac{\sqrt{3}}{2}$

$\left(0,0\right)$ will be rejected as vertices would be non-real roots.

So, the vertices will be $\left(0,1\right), \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ and $\left(-\frac{\sqrt{3}}{2},-\frac{-1}{2}\right)$

Hence, area of $∆=\frac{1}{2}\times \sqrt{3}\times \frac{3}{2}=\frac{3\sqrt{3}}{4}$