15. Let the 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 $\log _{3}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|$.
15. From the given, $9=\left|z_{1}+z_{2}\right|^{2}=\left(z_{1}+z_{2}\right)\left(\overline{z_{1}}+\overline{z_{2}}\right)=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}, 27=\left|z_{1}-z_{2}\right|^{2}$ $=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-z_{1} \overline{z_{2}}-\overline{z_{1}} z_{2}$, so we have $\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}=18, z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=-9$.

Given $\left|z_{1}\right|=3$, we have $\left|z_{2}\right|=3,\left|z_{1} \overline{z_{2}}\right|=\left|\overline{z_{1}} z_{2}\right|=9$. Let $z_{1} \overline{z_{2}}=9(\cos \theta+\mathrm{i} \sin \theta)$, then $18 \cos \theta=-9$, so $\cos \theta=-\frac{1}{2}, \sin \theta= \pm \frac{\sqrt{3}}{2}$. Thus, we can assume $z_{1} \overline{z_{2}}=9 w$, where $w$ is a cube root of unity, and $\overline{z_{1}} z_{2}=9 w^{2}$, so $\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}=9^{2000}\left(w^{2000}+w^{4000}\right)=9^{2000}\left(w^{2}+w\right)=-9^{2000}$.
Therefore, $\log _{3}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|=\log _{3} 9^{2000}=4000$.