2. Let $Z$ be a complex number, $M=\left\{Z\left|(Z-1)^{2}=\right| Z-\left.1\right|^{2}\right\}$, then
(A) $M=\{$ pure imaginary numbers $\}$;
(B) $M=\{$ real numbers $\}$;
(C) $\{$ real numbers $\} \subset M \subset\{$ complex numbers $\}$;
(D) $M=\{$ complex numbers $\}$.
2. (B)
$$
\begin{array}{lc}
\because & (Z-1)^{2}=|Z-1|^{2}, \\
\text { i.e., } & (Z-1)^{2}=(Z-1)(\bar{Z}-1), \\
\therefore & (Z-1)(Z-\bar{Z})=0, \\
\text { therefore } & Z=1 \text { or } Z=\bar{Z} .
\end{array}
$$

i.e., $Z$ is a real number, the answer is $(\mathrm{B})$.