3. Given the complex number $z$ satisfies $|z+\sqrt{3} i|+|z-\sqrt{3} i|=4$, then the minimum value of $|z-\mathrm{i}|$ is $\qquad$ .
Reference answer: $z$ corresponds to the curve equation on the complex plane: $\frac{y^{2}}{4}+x^{2}=1$. $z=\cos \theta+2 \sin \theta \mathrm{i}, \quad \theta \in[0,2 \pi)$, then $z-\mathrm{i}=\cos \theta+(2 \sin \theta-1) \mathrm{i}$. Therefore, $|z-i|=\sqrt{\cos ^{2} \theta+(2 \sin \theta-1)^{2}}=\sqrt{3\left(\sin \theta-\frac{2}{3}\right)^{2}+\frac{2}{3}} \geq \sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$, with equality holding if and only if $\sin \theta=\frac{2}{3}$.