Example 3 Let the complex numbers $z_{1}, z_{2}, z_{3}$ satisfy:
$$
\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=\sqrt{3}\left|z_{3}\right|, z_{1}+z_{3}=z_{2} . \\
\text { Find the value of } \frac{z_{1} z_{2}}{z_{3}} \text { . }
\end{array}
$$
【Analysis】Let $\left|z_{1}\right|=\left|z_{2}\right|=r>0$. Then
$$
\frac{z_{3}^{2}}{z_{1} z_{2}}=\frac{\left(z_{2}-z_{1}\right)^{2}}{z_{1} z_{2}}=\left(\frac{z_{2}}{z_{1}}+\frac{z_{1}}{z_{2}}\right)-2 \text {. }
$$

Since $z_{1} \overline{z_{1}}=r^{2}=z_{2} \overline{z_{2}}$, that is
$$
\frac{r^{2}}{z_{1}}=\overline{z_{1}}, \frac{r^{2}}{z_{2}}=\overline{z_{2}},
$$

thus equation (1) can be transformed into finding $\overline{z_{1}} z_{2}+z_{1} \overline{z_{2}}$, which is contained in $\left|z_{1}-z_{2}\right|^{2}$.
Solution From $z_{3}=z_{2}-z_{1}$, we get
$$
\begin{array}{l}
\left|z_{1}\right|^{2}=5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-z_{1} \overline{z_{2}}-\overline{z_{1}} z_{2}\right), \\
\left|z_{2}\right|^{2}=5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-z_{1} \overline{z_{2}}-\overline{z_{1}} z_{2}\right) .
\end{array}
$$
$$
\begin{array}{l}
\text { (2) }+ \text { (3) gives } \\
10\left(z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}\right) \\
=9\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right) \text { (let }\left|z_{1}\right|=\left|z_{2}\right|=r>0 \text { ) } \\
\Rightarrow 10 r^{2}\left(\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}\right)=18 r^{2} \\
\Rightarrow \frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=\frac{9}{5} \\
\Rightarrow \frac{z_{3}^{2}}{z_{1} z_{2}}=\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}-2=-\frac{1}{5} \\
\Rightarrow \frac{z_{1} z_{2}}{z_{3}}=-5 .
\end{array}
$$