QJEEM22Mab92c: JEE Main 2022 Mathematics - Class 11 JEEM22Mab92c Solution: The area of the polygon, whose vertices are the non-real roots…

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

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 $$=\frac{1}{2}\left|\begin{array}{ccc}0& 1& 1\\ \frac{\sqrt{3}}{2}& \frac{-1}{2}& 1\\ \frac{-\sqrt{3}}{2}& \frac{-1}{2}& 1\end{array}\right|$$

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