12・21 Let $a, b, c, d$ be real numbers. Suppose all the roots of the equation $z^{4}+a z^{3}+b z^{2}+c z+d=$ 0 lie on the circle in the complex plane centered at $0+0 i$ with radius 1, then the sum of the reciprocals of these roots must be
(A) $a$.
(B) $b$.
(C) $c$.
(D) $-a$.
(E) $-b$.
(38th American High School Mathematics Examination, 1987)
[Solution] From the problem, we know that the equation has real coefficients, and its roots are all on the unit circle (in the complex plane), i.e., $|z_1| = |z_2| = |z_3| = |z_4| = 1$.
Since the equation has real coefficients, complex roots appear in conjugate pairs. Without loss of generality, let $z_2 = \bar{z}_1, \quad z_4 = \bar{z}_3$.
Thus, $z_1 = \frac{1}{z_2}, \quad z_3 = \frac{1}{z_4}$. This is because $\frac{1}{z_2} = \frac{1}{\bar{z}_1} = \frac{z_1}{z_1 \bar{z}_1} = z_1$.

This means: if $z$ is a root of the given equation, then $\frac{1}{z}$ is also a root of the given equation. Thus, $d = 1, \quad a = c$.
By Vieta's formulas, we have:
$$
\left\{\begin{array}{l}
z_1 z_2 z_3 z_4 = d, \\
z_1 z_2 z_3 + z_2 z_3 z_4 + z_3 z_4 z_1 + z_4 z_1 z_2 = -c,
\end{array}\right.
$$

Therefore, $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4} = \frac{z_1 z_2 z_3 + z_2 z_3 z_4 + z_3 z_4 z_1 + z_4 z_1 z_2}{z_1 z_2 z_3 z_4}$
$$
= \frac{-c}{d} = \frac{-c}{1} = -c = -a.
$$

Hence, the answer is $(D)$.