<s>[INST] clarification for the solution of $(z^3=|z|^2 +4)$!? $(z-2)(z^2+z+2)=0$
so the solutions are 
$$z=2;$$$$z^2+z+2=0;$$
$$z=-(1+-\sqrt{-7})/2$$
I removed the absolute value for imposing a solution in the field of real, but only because I saw it on wolfram alpha, now I ask you! I can always impose a solution in the real numbers to remove the absolute value? and if yes I have to discard all imaginary values​​?
 [/INST] Let us proceed in the conventional way:
Let $z=r(\cos\theta+i \sin\theta)$ where $r$ is real
$\implies |z|=r$ and $z^3=\{r(\cos\theta+i \sin\theta)\}^3=r^3\cos3\theta+ir^3\sin3\theta$ (using de Moivre's formula)
$\implies r^3\cos3\theta+ir^3\sin3\theta=r^2+4$
Clearly $r\ne0$
Equating the imaginary terms $r^3\sin3\theta=0\implies \sin3\theta=0\implies 3\theta=n\pi$ where $n$ is any integer
$\theta=\frac{n\pi}3$ where $n=0,1,2$ (A more general set of values of $n$ has been discussed here)
If $n=0,2,\cos3\theta=1$ and $r^3-r^2-4=0\implies r=2$ (as you have found)
$\implies z=2(\cos0+i \sin0)=2$ and $z=2(\cos\frac{2\pi}3+i\sin\frac{2\pi}3)=-1+\sqrt3i$
If $n=1,\cos3\theta=-1\implies  z=-2(\cos\frac\pi3+i \sin\frac\pi3)=-1-\sqrt3i$
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