13. 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 the value of $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|$.
13. From the problem, we know that $\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=9$, and $\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right)=27$. Therefore, $\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}=18$, and $\left|z_{1}\right|^{2}=9$, so $\left|z_{2}\right|=3$, and $z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=-9$, $\left|z_{1} \bar{z}_{2}\right|=\left|\bar{z}_{1} z_{2}\right|=9$. Let $z_{1} \bar{z}_{2}=9(\cos \theta+i \sin \theta), \bar{z}_{1} z_{2}=9(\cos \theta-i \sin \theta)$, then $\cos \theta=\frac{1}{2}$. Thus, $z_{1} \bar{z}_{2}=9 \omega$ or $9 \omega^{2}\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)$,

When $z_{1} \bar{z}_{2}=9 \omega$, $\bar{z}_{1} z_{2}=9 \omega^{2}$, from which we get $\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}=-9^{2000}$; when $z_{1} \bar{z}_{2}=9 \omega^{2}$, $\bar{z}_{1} z_{2}=9 \omega$, similarly we get $\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}=-9^{2000}$. Therefore, $\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|=9^{2000}$, so the required value is 4000.