What does the group of the geometric images of z, such that $z+\overline{z}+|z^2|=0$ define? 
What does the group of the geometric images of z, such that
  $z+\overline{z}+|z^2|=0$ define?
A) A circumference of center (0,-1) and radius 1.
B) Two lines, of equations $y=x+2$ and $y=x$.
C)A circumference of center (1,0) and radius 1.
D)A circumference of center (-1,0) and radius 1.

I tried to simplify the expression:
$$z+\overline{z}+|z^2|=0 \Leftrightarrow \\
x+yi+x-yi+|(x+yi)^2|=0\Leftrightarrow \\
2x+\sqrt{(x^2+y^2)^2+(2xy)^2}=0 \Leftrightarrow \\
\sqrt{(x^2+y^2)^2+(2xy)^2}=-2x \Leftrightarrow \\
(x^2+y^2)^2+(2xy)^2=4x^2 \Leftrightarrow \\
x^4 - 2 x^2 y^2 + y^4+4x^2y^2 = 4x^2 \Leftrightarrow \\
x^4+y^4+2x^2y^2 = 4x^2 \Leftrightarrow \\
???$$
How do I continue from here?
I have also been thinking that if the sum of those three numbers is zero then they could be the vertices of a triangle. I rewrote the expression:
$$\rho \cdot cis(\theta)+\rho cis(-\theta)+|\rho^2 \cdot cis(2\theta)|=0 \Leftrightarrow \\
\rho \cdot cis(\theta)+\rho cis(-\theta)+\rho^2 =0 \Leftrightarrow \\
\rho(cis(\theta)+cis(-\theta)+\rho) = 0  \Leftrightarrow \\
\rho = 0 \lor cis(\theta)+cis(-\theta)+\rho = 0 \Leftrightarrow \\
cis(\theta)+cis(-\theta) = -\rho \Leftrightarrow \\
\cos(\theta)+\sin(\theta)i+\cos(-\theta)+\sin(-\theta)i = -\rho \Leftrightarrow \\
\cos(\theta)+\cos(\theta) = -\rho  \Leftrightarrow \\
2\cos(\theta) = -\rho \Leftrightarrow \\
\rho = -2\cos(\theta)$$
This means that $\rho$ will be between 0 and 2. If $|z^2|=\rho^2 = \rho^2 cis(0)$, then one of the vertices is $4$.
But what do I do next? How do I solve this?

Hint:  using $|z^2|=|z|^2= z \bar z\,$:
$$
0 = z+\bar{z}+|z^2| = z + \bar z + z \bar z \color{red}{+1-1}=(z+1)(\bar z + 1) -1 = |z+1|^2 - 1 \iff |z+1|^2 = 1
$$
