3. Given the complex number $z$ satisfies
$$
|z+\sqrt{3} \mathrm{i}|+|z-\sqrt{3} \mathrm{i}|=4 \text {. }
$$

Then the minimum value of $|z-\mathrm{i}|$ is $\qquad$
3. $\frac{\sqrt{6}}{3}$.

Note that, the curve equation corresponding to $z$ on the complex plane is: $\frac{y^{2}}{4}+x^{2}=1$.
$$
\begin{array}{l}
\text { Let } z=\cos \theta+2 \mathrm{i} \sin \theta(\theta \in[0,2 \pi)) \text {. Then } \\
z-\mathrm{i}=\cos \theta+(2 \sin \theta-1) \mathrm{i} . \\
\text { Therefore, }|z-\mathrm{i}|=\sqrt{\cos ^{2} \theta+(2 \sin \theta-1)^{2}} \\
=\sqrt{3\left(\sin \theta-\frac{2}{3}\right)^{2}+\frac{2}{3}} \\
\geqslant \sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3},
\end{array}
$$

The equality holds if and only if $\sin \theta=\frac{2}{3}$.