Example 2 The equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{1}, \overline{r_{1}}, \overline{r_{2}}, \overline{r_{2}}, \overline{r_{3}}, \overline{r_{3}}, \overline{r_{4}}, \overline{r_{4}}, \overline{r_{5}}, \overline{r_{5}}$, where $\overline{r_{i}}$ is the conjugate of $r_{i}$ $(i=1,2,3,4,5)$, find the value of $\frac{1}{r_{1} \overline{r_{1}}}+\frac{1}{r_{2} \overline{r_{2}}}+\frac{1}{r_{3}} \overline{r_{3}}+\frac{1}{r_{4} \overline{r_{4}}}+\frac{1}{r_{5} r_{5}}$.
Solve: From $x^{10}+(13 x-1)^{10}=0$, we get
$$
\left(\frac{x}{13 x-1}\right)^{10}=-1 \text {. }
$$

Let the 10 complex roots of $y^{10}=-1$ be $\omega_{1}, \overline{\omega_{1}}, \omega_{2}, \overline{\omega_{2}}, \omega_{3}, \overline{\omega_{3}}, \omega_{4}, \overline{\omega_{4}}, \omega_{5}, \overline{\omega_{5}}$, then
$$
\frac{x}{13 x-1}=\omega_{i} \text { or } \overline{\omega_{i}}(i=1,2,3,4,5) .
$$

They are all linear equations, each with one root, so let $\frac{r_{i}}{13 r_{i}-1}=\omega_{i}$, then $\frac{\overline{r_{i}}}{13 \overline{r_{i}}-1}=\overline{\omega_{i}}$, from which we deduce
$$
r_{i}=\frac{\omega_{i}}{13 \omega_{i}-1}, \overline{r_{i}}=\frac{\overline{\omega_{i}}}{13 \overline{\omega_{i}}-1} .
$$

Thus,
$$
\frac{1}{r_{i} \cdot \overline{r_{i}}}=\frac{\left(13 \omega_{i}-1\right)\left(13 \overline{\omega_{i}}-1\right)}{\omega_{i} \cdot \overline{\omega_{i}}}=170-13\left(\omega_{i}+\overline{\omega_{i}}\right) .
$$

Therefore, by Vieta's formulas, we have
$$
\sum_{i=1}^{5} \frac{1}{r_{i} \bar{r}_{i}}=5 \times 170-13 \sum_{i=1}^{5}\left(\omega_{i}+\overline{\omega_{i}}\right)=850 .
$$