12.8 Let $a, b, c$ be non-zero complex numbers, and $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}$, then the value of $\frac{a+b-c}{a-b+c}$ is (where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$ )
(A) 1.
(B) $\pm \omega$.
(C) $1, \omega, \omega^{2}$.
(D) $1,-\omega,-\omega^{2}$.
(China High School Mathematics League, 1991)
[Solution] Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=t$, then $a=b t=c t^{2}=a t^{3}$.
From $a \neq 0$, we know $t^{3}=1$, therefore, $t=1, \omega, \omega^{2}$.
Using the properties of proportions, we have
$$
\begin{aligned}
\text { Original expression } & =\frac{c t^{2}+c t-c}{c t^{2}-c t+c}=\frac{t^{2}+t-1}{t^{2}-t+1} \\
& =\frac{t^{3}+t^{2}-t}{t \cdot\left(t^{2}-t+1\right)}=\frac{t^{2}-t+1}{t\left(t^{2}-t+1\right)}=\frac{1}{t} .
\end{aligned}
$$

Thus, when $t=1, \omega^{2}, \omega$, the original expression takes the values $1, \omega, \omega^{2}$ respectively.
Therefore, the answer is (C).