The area of the polygon, whose vertices are the non-real roots of the equation $\overline z = i{z^2}$ is :

<p>$\overline z = i{z^2}$</p> <p>Let $z = x + iy$</p> <p>$x - iy = i({x^2} - {y^2} + 2xiy)$</p> <p>$x - iy = i({x^2} - {y^2}) - 2xy$</p> <p>$\therefore$ $x = - 2yx$ or ${x^2} - {y^2} = - y$</p> <p>$x = 0$ or $y = - {1 \over 2}$</p> <p>Case - I</p> <p>$x = 0$</p> <p>$- {y^2} = - y$</p> <p>$y = 0,\,\,1$</p> <p>Case - II</p> <p>$y = - {1 \over 2}$</p> <p>$$ \Rightarrow {x^2} - {1 \over 4} = {1 \over 2} \Rightarrow x = \pm {{\sqrt 3 } \over 2}$$</p> <p>$$x = \left\{ {0,i,{{\sqrt 3 } \over 2} - {i \over 2},{{ - \sqrt 3 } \over 2} - {i \over 2}} \right\}$$</p> <p>Area of polygon $$ = {1 \over 2}\left| {\matrix{ 0 & 1 & 1 \cr {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr {{{ - \sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr } } \right|$$</p> <p>$$ = {1 \over 2}\left| { - \sqrt 3 \,\,\, - {{\sqrt 3 } \over 2}} \right| = {{3\sqrt 3 } \over 4}$$</p>