$$\overline z = i{z^2}$$

Let $$z = x + iy$$

$$x - iy = i({x^2} - {y^2} + 2xiy)$$

$$x - iy = i({x^2} - {y^2}) - 2xy$$

$$\therefore$$ $$x = - 2yx$$ or $${x^2} - {y^2} = - y$$

$$x = 0$$ or $$y = - {1 \over 2}$$

Case - I

$$x = 0$$

$$ - {y^2} = - y$$

$$y = 0,\,\,1$$

Case - II

$$y = - {1 \over 2}$$

$$ \Rightarrow {x^2} - {1 \over 4} = {1 \over 2} \Rightarrow x = \pm {{\sqrt 3 } \over 2}$$

$$x = \left\{ {0,i,{{\sqrt 3 } \over 2} - {i \over 2},{{ - \sqrt 3 } \over 2} - {i \over 2}} \right\}$$

Area of polygon $$ = {1 \over 2}\left| {\matrix{ 0 & 1 & 1 \cr {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr {{{ - \sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr } } \right|$$

$$ = {1 \over 2}\left| { - \sqrt 3 \,\,\, - {{\sqrt 3 } \over 2}} \right| = {{3\sqrt 3 } \over 4}$$