Example 5 The equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{1}, \bar{r}_{1}, r_{2}, \bar{r}_{2}, r_{3}$, $\bar{r}_{3}, r_{4}, \bar{r}_{4}, r_{5}, \bar{r}_{5}$, , where $\bar{r}_{i}$ is the complex conjugate of $r_{i}$ $(i=1,2,3,4,5)$, find the value of $\frac{1}{r_{1} r_{1}}+$ $\frac{1}{r_{2} r_{2}}+\frac{1}{r_{3} r_{3}}+\frac{1}{r_{4} r_{4}}+\frac{1}{r_{5} r_{5}}$.
Solution: From $x_{10}+(13 x-1)^{10}=0$, we get $\left(\frac{x}{13 x-1}\right)^{10}=-1$.
Let the ten complex roots of $y^{10}=-1$ be $w_{1}, \bar{w}_{1}, w_{2}, \bar{w}_{2}, w_{3}, \bar{w}_{3}, w_{4}, \bar{w}_{4}$, $w_{5}, \bar{w}_{5}$, then $\frac{x}{13 x-1}=w_{i}$ or $\bar{w}_{i}(i=1,2,3,4,5)$, each of which is a linear equation with one root. Therefore, let $\frac{r_{i}}{13 r_{i}-1}=w_{i}$, then $\frac{\bar{r}_{i}}{13 \bar{r}_{i}-1}=\bar{w}_{i}$, from which we deduce $r_{i}=\frac{w_{i}}{13 w_{i}-1}, \bar{r}_{i}=\frac{\bar{w}_{i}}{13 \bar{w}_{i}-1}$. Thus,
$\frac{1}{r_{i} r_{i}}=\frac{\left(13 w_{i}-1\right)\left(13 \bar{w}_{i}-1\right)}{w_{i} w_{i}}=170-13^{3}\left(w_{i}+\bar{w}_{i}\right)$. Therefore, $\sum_{i=1}^{5} \frac{1}{r_{i} r_{i}}=5 \times 170-13 \sum_{i=1}^{5}\left(w_{i}+\bar{w}_{i}\right)=850$.