Complexified!!zzz

Since $z=0$ is not a solution of our equation, we multiply by $z^2$ and simplify to get $z^5+3\lvert z \rvert^3=0$ . Substituting $z=r e^{i\theta}$ , we get $r=\sqrt{3}$ and $e^{i5\theta}=-1$ which has $\boxed{5}$ solutions.