12・30 The set $S=\left\{z^{-2} \mid \arg z=\alpha, \alpha\right.$ is a constant $\}$ in the complex plane is represented by
(A) Ray $\arg z=2 \alpha$.
(B) Ray $\arg z=-2 \alpha$.
(C) Ray $\arg z=-\alpha$.
(D) None of the above answers is correct.
(China High School Mathematics League, 1984)
[Solution] $\because \arg z=\alpha, \alpha$ is a constant, and from the principal value range of the argument of a complex number, we have $0 \leqslant \alpha < 2 \pi$. Since $-\alpha, -2 \alpha$ are not in $[0, 2 \pi)$, options $(B)$ and $(C)$ are incorrect.
Given this, we can set $z=r(\cos \alpha+i \sin \alpha)$,

thus $\bar{z}=r(\cos \alpha-i \sin \alpha)$,

and
$$
\begin{aligned}
z^{-2} & =[r(\cos \alpha-i \sin \alpha)]^{2}=r^{2}(\cos 2 \alpha-i \sin 2 \alpha) \\
& =r^{2}[\cos (-2 \alpha)+i \sin (-2 \alpha)],
\end{aligned}
$$

which means the argument of $\bar{z}^{2}$ is not $2 \alpha$, so (A) is incorrect. Therefore, $(D)$ is correct.
In fact, the set $S$ in the complex plane is a ray with the origin as its vertex and an angle of $-2 \alpha$ with the real axis, but it cannot be written as the ray $\arg z=-2 \alpha$.
Thus, the correct choice is $(D)$.