7. If $z$ is a complex number, and $|z|=1$, then the maximum value of $\left|\frac{\sqrt{3} \mathrm{i}-z}{\sqrt{2}-z}\right|$ is

保留源文本的换行和格式，翻译结果如下：

7. If $z$ is a complex number, and $|z|=1$, then the maximum value of $\left|\frac{\sqrt{3} \mathrm{i}-z}{\sqrt{2}-z}\right|$ is
Let $z=\cos \theta+\mathrm{i} \sin \theta$, then $\left|\frac{\sqrt{3} \mathrm{i}-z}{\sqrt{2}-z}\right|^{2}=\frac{\cos ^{2} \theta+(\sin \theta-\sqrt{3})^{2}}{(\cos \theta-\sqrt{2})^{2}+\sin ^{2} \theta}=\frac{4-2 \sqrt{3} \sin \theta}{3-2 \sqrt{2} \cos \theta}$.
Let $k=\frac{4-2 \sqrt{3} \sin \theta}{3-2 \sqrt{2} \cos \theta}$, then $2 \sqrt{2} k \cos \theta-2 \sqrt{3} \sin \theta=3 k-4$ $\Rightarrow(2 \sqrt{2} k)^{2}+(2 \sqrt{3})^{2} \geqslant(3 k-4)^{2} \Rightarrow k^{2}-24 k+4 \leqslant 0 \Rightarrow 12-2 \sqrt{35} \leqslant k \leqslant 12+2 \sqrt{35}$. Therefore, the maximum value of the original expression is $\sqrt{12+2 \sqrt{35}}=\sqrt{7}+\sqrt{5}$.