7. The complex number $z$, such that $z^{3}+z=2|z|^{2}$, then all possible values of $z$ are
7. $0,1,-1 \pm 2 i$

Analysis: Clearly, $z=0$ satisfies. Suppose $z=r(\cos \theta+i \sin \theta)(r>0)$, substituting into the original equation, we get $\left\{\begin{array}{l}r^{2} \cos 3 \theta+\cos \theta=2 r, \quad (1) \\ r^{2} \sin 3 \theta+\sin \theta=0, \quad (2)\end{array}\right.$ Using the triple angle formula $\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta$ and substituting into (2), we get $\sin \theta\left(3 r^{2}-4 r^{2} \sin ^{2} \theta+1\right)=0$. If $\sin \theta=0$, then $z \in \mathbb{R}$, solving gives $z=0$ or 1; If $\sin \theta \neq 0$, then $\sin ^{2} \theta=\frac{3 r^{2}+1}{4 r^{2}}$. Then $(1)^{2}+(2)^{2}=r^{4}+1+2 r^{2} \cos 2 \theta=4 r^{2}$, and $\cos 2 \theta=1-2 \sin ^{2} \theta=-\frac{r^{2}+1}{2 r^{2}}$, substituting in gives $r^{2}=5$, hence $\sin ^{2} \theta=\frac{4}{5}, \cos ^{2} \theta=\frac{1}{5}$, yielding $z= \pm 1 \pm 2 i$. Verification shows that $z=1 \pm 2 i$ satisfies the condition. Therefore, all possible values of $z$ are $0,1,-1 \pm 2 i$.