$12 \cdot 4$ When $n$ takes $1,2, \cdots, 100$, the number of different values that $\left[\left(\frac{\sqrt{3}+i}{2}\right)^{8}+1\right]^{n}$ can take is
(A) 8.
(B) 6.
(C) 4.
(D) 2.
. (High School Mathematics Competition in Shanghai, China, 1985)
[Solution] $\left[\left(\frac{\sqrt{3}+i}{2}\right)^{8}+1\right]^{n}=\left\{\left[-i\left(\frac{-1+\sqrt{3} i}{2}\right)\right]^{8}+1\right\}^{n}$
$$
\begin{array}{l}
=\left[(-i \omega)^{8}+1\right]^{n}=\left(\omega^{2}+1\right)^{n} \\
=(-\omega)^{n},
\end{array}
$$

where $\omega$ is a root of $1+x+x^{2}=0$, $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$,
thus $(-\omega)^{n}$ for $n=1,2, \cdots, 100$ can take $\pm 1, \pm \omega, \pm \omega^{2}$, a total of 6 different values.
Therefore, the answer is $(B)$.