[Solution] Given $z^{2}-z-5+5 i=0$, let its two roots be

then
$$
\begin{array}{c}
z_{1}=x_{1}+y_{1} i, \quad z_{2}=x_{2}+y_{2} i, \\
z_{1}+z_{2}=1=\left(x_{1}+x_{2}\right)+\left(y_{1}+y_{2}\right) i,
\end{array}
$$
$\therefore \quad\left\{\begin{array}{l}x_{1}+x_{2}=1, \\ y_{1}+y_{2}=0 .\end{array}\right.$
Also, $z_{1} z_{2}=5 i-5=\left(x_{1} x_{2}-y_{1} y_{2}\right)+\left(x_{1} y_{2}+x_{2} y_{1}\right) i$,
$\therefore \quad\left\{\begin{array}{l}x_{1} x_{2}-y_{1} y_{2}=-5, \\ x_{1} y_{2}+x_{2} y_{1}=5 .\end{array}\right.$
From (2) and (4), we have $\left(x_{1}-x_{2}\right) y_{2}=5$,
From (2) and (3), we have $x_{1} x_{2}+y_{2}^{2}=-5$.
Also,
then
$$
\begin{array}{c}
\left(x_{1}+x_{2}\right)^{2}=\left(x_{1}-x_{2}\right)^{2}+4 x_{1} x_{2}, \\
1=\frac{25}{y_{2}^{2}}-20-4 y_{2}^{2},
\end{array}
$$
$\therefore 4 y_{2}^{4}+21 y_{2}^{2}-25=0$, which gives $y_{2}^{2}=1$ (the negative value has been discarded).
From (6), we have $x_{1} x_{2}=-5-y^{2}=-6$.
Therefore, the answer is $(B)$.
12-18 Let $i=\sqrt{-1}$. Then the product of the real parts of the roots of $z^{2}-z=5-5 i$ is
(A) -25 .
(B) -6 .
(C) -5 .
(D) $\frac{1}{4}$.
(E) 25 .
(43rd American High School Mathematics Examination, 1992)