The area of the polygon, whose vertices are the non-real roots of...

\overset{¯}{z} = i{z}^{2} ⇒ \left(x - iy\right) = i{\left(x + iy\right)}^{2}$</p><p>$⇒ x - iy = \left({x}^{2} - {y}^{2}\right)i - 2xy$</p><p>i.e.,$x = - 2xy$and$- y = {x}^{2} - {y}^{2}$</p><p>$⇒ x = 0, y = - \dfrac{1}{2}$</p><p>When$x = 0$;$y = 0, 1$</p><p>When$y = - \dfrac{1}{2}$;$x = \pm \dfrac{\sqrt{3}}{2}$</p><p>\left(0, 0\right)$ will be rejected as vertices would be non-real roots.

So, the vertices will be $\left(0, 1\right), \left(\dfrac{\sqrt{3}}{2}, - \dfrac{1}{2}\right)$ and \left(- \dfrac{\sqrt{3}}{2}, - \dfrac{- 1}{2}\right)$</p><p>Hence, area of$∆ = \dfrac{1}{2} \times \sqrt{3} \times \dfrac{3}{2} = \dfrac{3\sqrt{3}}{4}