Example 6. Let $a=\sqrt{17}, b=\mathrm{i} \sqrt{19}$. Find the maximum value of $\frac{|a-z|}{|b-z|}$, where $z$ is any complex number with modulus 1.
(2015, Harvard-MIT Mathematics Tournament)
【Analysis】Using the geometric meaning of complex numbers, we know that $z$ is a point on the unit circle centered at the origin in the complex plane, $A(\sqrt{17}, 0), B(0, \sqrt{19})$.
As shown in Figure 1.

Thus, $\frac{|a-z|}{|b-z|}=\frac{|A Z|}{|B Z|}$.
Assume point $Z(\cos \theta, \sin \theta)$. Then
$$
\begin{array}{l}
|A Z|^{2}=18-2 \sqrt{17} \cos \theta, \\
|B Z|^{2}=20-2 \sqrt{19} \sin \theta . \\
\text { If we let } \frac{|a-z|}{|b-z|}=\frac{|A Z|}{|B Z|}=k, \text { then } \\
k^{2}=\left(\frac{|A Z|}{|B Z|}\right)^{2}=\frac{9-\sqrt{17} \cos \theta}{10-\sqrt{19} \sin \theta} .
\end{array}
$$

Simplifying and rearranging, we get
$$
\begin{array}{l}
10 k^{2}-9=\sqrt{19} k^{2} \sin \theta-\sqrt{17} \cos \theta \\
\Rightarrow 110 k^{2}-91 \leqslant \sqrt{19 k^{4}+17} \\
\Rightarrow 19 k^{4}+17 \geqslant\left(10 k^{2}-9\right)^{2} \\
\Rightarrow\left(9 k^{2}-16\right)\left(9 k^{2}-4\right) \leqslant 0 \\
\Rightarrow \frac{2}{3} \leqslant k \leqslant \frac{4}{3} .
\end{array}
$$

Therefore, the maximum value sought is $\frac{4}{3}$.