7. Given that $z_{1}, z_{2}, \ldots, z_{7}$ are the seventh roots of $2021+i$, and their principal arguments in ascending order are $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{7}$, find the value of $\tan \alpha_{1} \tan \alpha_{3}+\tan \alpha_{2} \tan \alpha_{4}+\cdots+\tan \alpha_{7} \tan \alpha_{2}$ is $\qquad$
Answer: -7
Solution: It is evident that $\left\{a_{n}\right\}$ forms an arithmetic sequence with a common difference of $\frac{2 \pi}{7}$. Moreover, we have $1+\tan a_{1} \tan a_{3}=\frac{\tan a_{3}-\tan a_{1}}{\tan \frac{4 \pi}{7}}, 1+$ $\tan \alpha_{2} \tan \alpha_{4}=\frac{\tan \alpha_{4}-\tan a_{2}}{\tan \frac{4 \pi}{7}}, \cdots, 1+\tan \alpha_{7} \tan \alpha_{2}=\frac{\tan a_{2}-\tan a_{7}}{\tan \frac{4 \pi}{7}}$. Adding these seven equations, we get $\tan \alpha_{1} \tan \alpha_{3}+\tan \alpha_{2} \tan \alpha_{4}+\cdots+\tan \alpha_{7} \tan \alpha_{2}=-7$.