7. Suppose that $x, y$, and $z$ are complex numbers of equal magnitude that satisfy
$$
x+y+z=-\frac{\sqrt{3}}{2}-i \sqrt{5}
$$
and
$$
x y z=\sqrt{3}+i \sqrt{5} \text {. }
$$

If $x=x_{1}+i x_{2}, y=y_{1}+i y_{2}$, and $z=z_{1}+i z_{2}$ for real $x_{1}, x_{2}, y_{1}, y_{2}, z_{1}$, and $z_{2}$, then
$$
\left(x_{1} x_{2}+y_{1} y_{2}+z_{1} z_{2}\right)^{2}
$$
can be written as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.
Answer:
1516
Solution: From the conditions, it is clear that $a, b, c$ all have magnitude $\sqrt{2}$. Conjugating the first equation gives $2\left(\frac{a b+b c+c a}{a b c}\right)=-\frac{\sqrt{3}}{2}+i \sqrt{5}$, which means $a b+b c+c a=\left(-\frac{\sqrt{3}}{4}+i \frac{\sqrt{5}}{2}\right)(\sqrt{3}+i \sqrt{5})=$ $\frac{-13+i \sqrt{15}}{4}$. Then,
$$
\begin{aligned}
a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} & =\frac{1}{2} \operatorname{Im}\left(a^{2}+b^{2}+c^{2}\right) \\
& =\frac{1}{2} \operatorname{Im}\left((a+b+c)^{2}\right)-\operatorname{Im}(a b+b c+c a) \\
& =\frac{\sqrt{15}}{4}
\end{aligned}
$$
so the answer is 1516 .
Remark:
$$
\{a, b, c\}=\left\{\frac{-\sqrt{3}-i \sqrt{5}}{2}, \frac{-3 \sqrt{3}-i \sqrt{5}}{4}, \frac{3 \sqrt{3}-i \sqrt{5}}{4}\right\}
$$