QJEEM22Ma8782: JEE Main 2022 Mathematics - Class 11 JEEM22Ma8782 Solution: Let . Then is equal to ______________.

$\because$ ${z^2} + \overline z = 0$

Let $z = x + iy$

$\therefore$ ${x^2} - {y^2} + 2ixy + x - iy = 0$

$({x^2} - {y^2} + x) + i(2xy - y) = 0$

$\therefore$ ${x^2} + {y^2} = 0$ and $(2x - 1)y = 0$

if $x = \, + \,{1 \over 2}$ then $y = \, \pm \,{{\sqrt 3 } \over 2}$

And if $y = 0$ then $x = 0, - 1$

$\therefore$ $z = 0 + 0i, - 1 + 0i,{1 \over 2} + {{\sqrt 3 } \over 2}i,{1 \over 2} - {{\sqrt 3 } \over 2}i$

$\therefore$ $\sum {\left( {{R_e}(z) + m(z)} \right) = 0}$