4. Let the complex numbers be $z_{1}=-3-\sqrt{3} \mathrm{i}, z_{2}=\sqrt{3}+\mathrm{i}$, $z=\sqrt{3} \sin \theta+\mathrm{i}(\sqrt{3} \cos \theta+2)$.
Then the minimum value of $| z-z_{1}|+| z-z_{2} \mid$ is
$\qquad$.
$4.2+2 \sqrt{3}$.
$$
\begin{array}{l}
\left|z-z_{1}\right|+\left|z-z_{2}\right| \geqslant\left|z-z_{1}-\left(z-z_{2}\right)\right| \\
=\left|z_{1}-z_{2}\right|=2+2 \sqrt{3},
\end{array}
$$

The condition for equality to hold is
$$
\arg \left(z-z_{1}\right)=\arg \left(z_{2}-z\right),
$$

which means $\frac{\sqrt{3} \cos \theta+1}{\sqrt{3} \sin \theta-\sqrt{3}}=\frac{\sqrt{3} \cos \theta+2+\sqrt{3}}{\sqrt{3} \sin \theta+3}$. Simplifying, we get $\sin \theta-\sqrt{3} \cos \theta=2$, which is
$$
\sin \left(\theta-60^{\circ}\right)=1, \theta=150^{\circ} .
$$