3. If the complex number $z$ satisfies $\left|z^{2}-1\right|=|z|$, then the range of $|z|$ is $\qquad$

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$\begin{array}{l}\text { 3. }\left[\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right] . \\ \text { Let } z=r(\cos \theta+\mathrm{i} \sin \theta)(r>0, \theta \in[0,2 \pi)) \text {. } \\ \text { Then }\left|z^{2}-1\right|=|z| \\ \Leftrightarrow\left|r^{2} \cos 2 \theta-1+\mathrm{i} r^{2} \sin 2 \theta\right|=r \\ \Leftrightarrow\left(r^{2} \cos 2 \theta-1\right)^{2}+\left(r^{2} \sin 2 \theta\right)^{2}=r^{2} \\ \Rightarrow 2 \cos 2 \theta=r^{2}+\frac{1}{r^{2}}-1 \\ \Rightarrow-1 \leqslant r^{2}+\frac{1}{r^{2}} \leqslant 3 \\ \Rightarrow \frac{3-\sqrt{5}}{2} \leqslant r^{2} \leqslant \frac{3+\sqrt{5}}{2} \\ \Rightarrow \frac{\sqrt{5}-1}{2} \leqslant|z| \leqslant \frac{\sqrt{5}+1}{2} .\end{array}$