8. 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 $\left|z-z_{1}\right|+\left|z-z_{2}\right|$ is $\qquad$.
Given points $A(-3,-\sqrt{3}), B(\sqrt{3}, 1)$, it is easy to find the line $A B: y=\frac{\sqrt{3}}{3} x$.
Since point $P(\sqrt{3} \sin \theta, \sqrt{3} \cos \theta+2)$ lies on the circle $C: x^{2}+(y-2)^{2}=3$, and the distance $d$ from the center $C$ to the line $A B$ is $d=\frac{2}{\sqrt{1+\left(\frac{\sqrt{3}}{3}\right)^{2}}}=\sqrt{3}$, thus the line $A B$ is tangent to the circle $C$.
Therefore, the minimum value of $\left|z-z_{1}\right|+\left|z-z_{2}\right|$ is $|A B|=\sqrt{(3+\sqrt{3})^{2}+(\sqrt{3}+1)^{2}}=2(\sqrt{3}+1)$.