Example 5 Given the equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{i}, \overline{r_{i}}(i=1,2,3,4,5)$, where $\overline{r_{i}}$ is the conjugate of $r_{i}$. Find the value of $\sum_{i=1}^{5} \frac{1}{r_{i} \bar{r}_{i}}$.
(12th American Invitational Mathematics Examination)
Explanation: The original equation can be transformed into
$$
\left(\frac{13 x-1}{x}\right)^{10}=-1 \text {. }
$$

Let the 10 complex roots of $y^{10}=-1$ be $\omega_{i}$ and $\overline{\omega_{i}}$ $(i=1,2,3,4,5)$, then we have $\omega_{i}=\frac{13 r_{i}-1}{r_{i}}$, which means
$$
\begin{array}{l}
\frac{1}{r_{i}}=13-\omega_{i} \text {. } \\
\text { Thus, } \frac{1}{r_{i} \overline{r_{i}}}=\left(13-\omega_{i}\right)\left(13-\overline{\omega_{i}}\right) \\
=170-13\left(\omega_{i}+\overline{\omega_{i}}\right) \text {. } \\
\text { Therefore, } \sum_{i=1}^{5} \frac{1}{r_{i} \overline{r_{i}}}=5 \times 170-13 \sum_{i=1}^{5}\left(\omega_{i}+\overline{\omega_{i}}\right) \\
=850 \text {. } \\
\end{array}
$$