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. $1,-\omega,-\omega^{2}$
When $q \neq 1$, $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=q \Rightarrow a=b q, b=c q, c=a q$
$\Rightarrow c=a q, b=a q^{2}$ and $q^{3}=1 \Rightarrow q^{2}+q+1=0 \Rightarrow \frac{a+b-c}{a-b+c}=\frac{1+q^{2}-q}{1-q^{2}+q}$ $=\frac{-2 q}{-2 q^{2}}=\frac{1}{q}=\omega^{2}, \omega$; When $q \neq 1$, $\frac{a+b-c}{a-b+c}=1$. Therefore, the answer is $C$.