Example 4 (1991 National High School League Question) Let complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=$ $\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3}$, find $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|$.
Solve: Fill in 4000. Reason: Direct calculation is not easy. Seeing $\left|z_{3}\right|, \left|z_{1}+z_{2}\right|, \left|z_{1}-z_{2}\right|$, one can think of a parallelogram, so we can consider the geometric interpretation on the complex plane.

Thus, let $z_{1}, z_{2}, z_{1}+z_{2}$ correspond to points $A, B, C$ on the complex plane, as shown in Figure 5-4. Then $|O A|=\left|z_{1}\right|, |O B|=\left|z_{2}\right|, |O C|=\left|z_{1}+z_{2}\right|, |A B|=\left|z_{2}\right|$. Since quadrilateral $O A C B$ is a parallelogram,

we have $\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}=2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)$.
Thus, $\left|z_{2}\right|=3$, so quadrilateral $O A C B$ is a rhombus, and $\angle A O B=\frac{2 \pi}{3}$,
hence $\frac{z_{2}}{z_{1}}=\cos \frac{2 \pi}{3} \pm \mathrm{i} \sin \frac{2 \pi}{3}$.
Since $\left|z_{1}\right|=3$, we have $\frac{1}{z_{1}}=\frac{\bar{z}_{1}}{9}$,
then $z_{2} \cdot \bar{z}_{1}=9\left(\cos \frac{2 \pi}{3} \pm \mathrm{i} \sin \frac{2 \pi}{3}\right)$,
so $\overline{z_{2} \cdot \bar{z}_{1}}=9\left(\cos \frac{2 \pi}{3} \mp \mathrm{i} \sin \frac{2 \pi}{3}\right)$.
Thus, $\bar{z}_{2} \cdot z_{1}=9\left(\cos \frac{2 \pi}{3} \mp \mathrm{i} \sin \frac{2 \pi}{3}\right)$.

Therefore, $\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}$
$$
=9^{2000}\left(\cos \frac{2000 \times 2 \pi}{3} \pm \mathrm{i} \sin \frac{2000 \times 2 \pi}{3}\right)+9^{2000}\left(\cos \frac{2000 \times 2 \pi}{3} \mp \mathrm{i} \sin \frac{2000 \times 2 \pi}{3}\right)=-9^{2000},
$$

Thus, $\left.\log _{3} \mid z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000} \mid=\log _{3} 3^{4000}=4000$.