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$(z-2)(z^2+z+2)=0$ 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​​?

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  • $\begingroup$ You can't. Surely $2$ is a solution. But this does not imply you can “divide by $z-2$”: it's not a polynomial function. $\endgroup$
    – egreg
    Oct 12, 2013 at 13:28

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To solve $z^3=|z|^2+4$, one may first look for $r=|z|$. Obviously, $r^3=r^2+4$ hence $(r-2)(r^2+r+2)=0$ Since $r^2+r+2\ne0$ for every real number $r$, one knows that $r=2$. Now, if $z=2\mathrm e^{\mathrm i t}$, the equation to solve becomes $8\mathrm e^{3\mathrm i t}=2^2+4=8$, that is, $\mathrm e^{3\mathrm i t}=1$.

Finally, the equation $z^3=|z|^2+4$ has exactly three solutions $z$, corresponding to the third roots of unity, namely, $$ -1+\mathrm i\sqrt3,\qquad-1-\mathrm i\sqrt3,\qquad2. $$

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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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As can be seen easily that $z^3$ is real. Now as it is real hence, $z^3=\bar{z^3}$ $\implies$ $z=\bar{z}$.
The equation is $$ z^3=|z|^2+4$$ $\implies$ $$ z^3=z\bar{z}+4 $$ $\implies$ $$z^3=z^2+4$$ $\implies$ $$(z-2)(z^2+z+2) $$ And then as you proceeded.

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