Sum of the Roots

Using the property $|z|^2=z.\overline { z }$

we can rewrite the equation as ${ z }^{ 3 }+{ z }^{ 2 }-z\overline { z } +2z=0$ $\Longleftrightarrow$ ${ z(z }^{ 2 }+{ z }-\overline { z } +2)=0$ (as we see, zero is a root)

Let $z=a+bi$ ; $i=\sqrt { -1 }$ and $a,b$ real numbers.

Then we can rewrite the equation as:

${ (a+bi) }^{ 2 }+(a+bi)-(a-bi)+2=0$ $\Leftrightarrow$ $a^2 - b^2 +2abi +2bi+2=0$

Hence:

${ a }^{ 2 }{ - }{ b }^{ 2 }+2=0$ (real part) $\wedge$ $2ab + 2b=0$ (imaginary part)

From the imaginary part:

$\rightarrow$ First case: $b=0$ then $a^2+2=0$ and there are no solutions.

$\rightarrow$ Second case: $b\neq0$ $\Longrightarrow$ $a=-1$ . then: $1-b^2+2=0$ and $b= \pm \sqrt { 3 }$ $\therefore$ Solutions: $z=-1\pm i\sqrt { 3 }$

Summing it up: $\boxed { S=-2 }$