5. Let the complex numbers $Z_{1}, Z_{2}$ satisfy $\left|Z_{1}\right|=\left|Z_{1}+Z_{2}\right|=$ $3,\left|Z_{1}-Z_{2}\right|=3 \sqrt{3}$, then $\log _{3} \mid\left(Z \cdot \overline{Z_{2}}\right)^{2009}+$ $\left(\overline{Z_{1}} Z_{2}\right)^{2009} \mid=$ $\qquad$
5. 4018 .

Solution: From $\left\{\begin{array}{l}\left|Z_{1}+Z_{2}\right|=3 \\ \left|Z_{1}-Z_{2}\right|=3 \sqrt{2}\end{array}\right.$,
we get $\left\{\begin{array}{l}\left|Z_{2}\right|=3 \\ Z_{1} \overline{Z_{2}}+\overline{Z_{1}} Z_{2}=-9\end{array}\right.$, and $\left|Z_{1} \overline{Z_{2}}\right|+\left|\overline{Z_{1}} Z_{2}\right|=9$,
thus let $Z_{1} \overline{Z_{2}}=9(\cos \theta+i \sin \theta) \quad \overline{Z_{1}} Z_{2}=9(\cos \theta-$ $i \sin \theta)$
From $-9=Z_{1} \overline{Z_{2}}+\overline{Z_{1}} Z_{2}=18 \cos \theta$
we get $\cos \theta=-\frac{1}{2}$,
hence $Z_{1} \overline{Z_{2}}=9 \omega$ or $Z_{1} \overline{Z_{2}}=9 \omega^{2}$,
When $Z_{1} \overline{Z_{2}}=9 \omega$, $\left(Z_{1} \overline{Z_{2}}\right)^{2009}+\left(\overline{Z_{1}} Z_{2}\right)^{2009}=$ $-9^{2009}$.
Therefore, $\log _{3}\left|\left(Z_{1} \overline{Z_{2}}\right)^{2009}+\left(\overline{Z_{1}} Z_{2}\right)^{2009}\right|=4018$
Similarly, when $\overline{Z_{1}} Z_{2}=9 \omega^{2}$, the result is still 4018.