5. Let $P$ be any point in the plane of $\triangle A B C$, and denote $B C=a, C A=b, A B=c, P A=u, P B$ $=v, P C=w$. Then the minimum value of $\frac{u}{a}+\frac{v}{b}+\frac{w}{c}$ is
$5 . \sqrt{3}$.
Given that $P, A, B, C$ correspond to the complex numbers $z, z_{1}, z_{2}, z_{3}$, we have
$$
\begin{array}{l}
\frac{\left(z-z_{1}\right)\left(z-z_{2}\right)}{\left(z_{3}-z_{1}\right)\left(z_{3}-z_{2}\right)}+\frac{\left(z-z_{2}\right)\left(z-z_{3}\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)} \\
+\frac{\left(z-z_{3}\right)\left(z-z_{1}\right)}{\left(z_{2}-z_{3}\right)\left(z_{2}-z_{1}\right)} \equiv 1 .
\end{array}
$$

Therefore, $\frac{u v}{a b}+\frac{v w}{b c}+\frac{w u}{c a} \geqslant 1$,
$$
\left(\frac{u}{a}+\frac{v}{b}+\frac{w}{c}\right)^{2} \geqslant 3\left(\frac{u v}{a b}+\frac{v w}{b c}+\frac{u w}{c a}\right) \geqslant 3 .
$$

Thus, $\frac{u}{a}+\frac{v}{b}+\frac{w}{c} \geqslant \sqrt{3}$.