2. 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
(A) 1
(B) $\pm \omega$
(C) $1, \omega, \omega^{2}$
(D) $1,-\omega,-\omega^{2}$, where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$
(C)
2.【Analysis and Solution】Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=t$, then $a=b t=c t^{2}=a t^{3}$.
Since $a \neq 0$, we know $t^{3}=1$,
Therefore, $t=1, \omega, \omega^{2}$.
Using the properties of proportions, we know the original expression equals $\frac{1}{t}$,
Thus, when $t=1, \omega^{2}, \omega$, the original expression takes the values $1, \omega, \omega^{2}$ respectively.