jmc

Since $|z{|}^2=z\overline{z},$ we can write $$z^3+z^2-z\overline{z}+2z=0.$$Then $$z(z^2+z-\overline{z}+2)=0.$$So, $z=0$ or $z^2+z-\overline{z}+2=0.$

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Then $$(x+yi{)}^2+(x+yi)-(x-yi)+2=0,$$which expands as $$x^2+2xyi-y^2+2yi+2=0.$$Equating real and imaginary parts, we get $x^2-y^2+2=0$ and $2xy+2y=0.$ Then $2y(x+1)=0,$ so either $x=-1$ or $y=0.$

If $x=-1,$ then $1-y^2+2=0,$ so $y=\pm \sqrt{3}.$ If $y=0,$ then $x^2+2=0,$ which has no solutions.

Therefore, the solutions in $z$ are 0, $-1+i\sqrt{3},$ and $-1-i\sqrt{3},$ and their sum is $\boxed{-2}.$