[Solution] From $(z-1)^{2}=|z-1|^{2}=(z-1)(\bar{z}-1)$ we know that
if $z-1=0$, then $z=1$;
if $z-1 \neq 0$, then $z-1=\bar{z}-1$, which implies $z=\bar{z}$, indicating that $z$ is a real number. Therefore, the answer is $(B)$.
12・25 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 $\}$.
(China High School Mathematics League, 1986)