4. Given complex numbers $z, z_{1}, z_{2}\left(z_{1} \neq z_{2}\right)$ satisfy $z_{1}^{2}=z_{2}^{2}=-2-2 \sqrt{3} \mathrm{i}$, and $\left|z-z_{1}\right|=\left|z-z_{2}\right|=4$, then $|z|=$ $\qquad$ .
Solve $z_{1}^{2}=z_{2}^{2}=-2-2 \sqrt{3} \mathrm{i} \Rightarrow z_{1}=1-\sqrt{3} \mathrm{i}, z_{2}=-1+\sqrt{3} \mathrm{i}$, the points corresponding to the complex numbers $z_{1}, z_{2}$ on the complex plane are symmetric with respect to the origin, and $\left|z_{1}\right|=\left|z_{2}\right|=2$. Therefore, $|z|=\sqrt{4^{2}-2^{2}}=2 \sqrt{3}$.