Example 3 Let the complex numbers $z_{1}$ and $z_{2}$ satisfy
$$
\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3} \text {. }
$$

Find the value of $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|$.
(1991, National High School Mathematics Competition)
Solution: From the given, we have
$$
\begin{aligned}
9 & =\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2} \\
& =\left(z_{1}+z_{2}\right)\left(\overline{z_{1}+z_{2}}\right) \\
& =\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right), \\
27 & =\left|z_{1}-z_{2}\right|^{2}=\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-z_{2}\right) \\
& =\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right) .
\end{aligned}
$$

Given $\left|z_{1}\right|=3$, from the above two equations, we get $\left|z_{2}\right|=3, z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=-9$, and $\left|z_{1} \bar{z}_{2}\right|=\left|\bar{z}_{1} z_{2}\right|=9$.

Let $z_{1} \bar{z}_{2}=9(\cos \theta+i \sin \theta)$, then $\bar{z}_{1} z_{2}=$ $9(\cos \theta-i \sin \theta)$. From $-9=z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=$ $18 \cos \theta$, we get $\cos \theta=-\frac{1}{2}, \theta=\frac{2 \pi}{3}$. Therefore, $\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}$ $=9^{2000} \cos 2000 \theta=-9^{2000}$.
Thus, the original expression $=4000$.