12. If the complex numbers $Z_{1}, Z_{2}$ satisfy $\left|Z_{1}\right|=a,\left|Z_{2}\right|=1$, and if $Z_{1} Z_{2}=-a$, then the range of values for $Z_{1}-a Z_{2}$ is $\qquad$
12. $[-2 a, 2 a]$
$$
Z_{1}-a Z_{2}=Z_{2} \bar{Z}_{2} Z_{1}-\frac{Z_{1} \bar{Z}_{1} Z_{2}}{a}=\frac{Z_{1} Z_{2}}{a}\left(a \bar{Z}_{2}-\bar{Z}_{1}\right)=\bar{Z}_{1}-a \bar{Z}_{2}=\overline{Z_{1}-a Z_{2}}
$$

So $Z_{1}=a Z_{2} \in \mathbf{R}$. Also, $0 \leqslant\left|Z_{1}-a Z_{2}\right| \leqslant\left|Z_{1}\right|+a|Z_{2}|=2 a$. When $Z_{1}=a$ and $Z_{2}=-1$, $Z_{1}-a Z_{2}=2 a$. When $Z_{1}=-a$ and $Z_{2}=1$, $Z_{1}-a Z_{2}=-2 a$. Since $Z_{1}-a Z_{2}$ varies continuously, the range of $Z_{1}-a Z_{2}$ is $[-2 a, 2 a]$.