Problem 1060

Analysis Note that $x \neq 0$, so the original equation can be transformed into $\left(13-\frac{1}{x}\right)^{10}=-1$. Let $y=13-\frac{1}{x}$, converting the problem into studying the complex roots of the equation $y^{10}=-1$.

Solution Clearly, $x \neq 0$, so the original equation can be transformed into $\left(13-\frac{1}{x}\right)^{10}=-1$. Let $y=13-\frac{1}{x}$, then $y^{10}=-1$. Suppose the 10 complex roots of the equation $y^{10}=-1$ are $\varepsilon_{k}, \overline{\varepsilon_{k}}(k=1,2,3,4,5)$, where $\varepsilon_{k}=\cos \frac{(2 k-1) \pi}{10} + \mathrm{i} \sin \frac{(2 k-1) \pi}{10}, k=1,2,3,4,5$. Without loss of generality, let $13-\frac{1}{r_{k}}=\varepsilon_{k}$, then $\frac{1}{r_{k}}=13-\varepsilon_{k}, k=1,2,3,4,5$. Therefore, $\sum_{k=1}^{5} \frac{1}{r_{k} \overline{r_{k}}}=\sum_{k=1}^{5}\left(13-\varepsilon_{k}\right)\left(13-\overline{\varepsilon_{k}}\right)=\sum_{k=1}^{5}\left[170-13\left(\varepsilon_{k}+\overline{\varepsilon_{k}}\right)\right]$ $=850-26\left(\cos \frac{\pi}{10}+\cos \frac{3 \pi}{10}+\cos \frac{5 \pi}{10}+\cos \frac{7 \pi}{10}+\cos \frac{9 \pi}{10}\right)=850$. Thus, the value of the required algebraic expression is 850.