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

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 $$x = 0, - 1$$

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

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