12・20 Assume $z=a+b i$ is a solution to the polynomial equation
$$
c_{4} z^{4}+i c_{3} z^{3}+c_{2} z^{2}+i c_{1} z+c_{0}=0
$$

where $c_{0}, c_{1}, c_{2}, c_{3}, a$ and $b$ are real constants, and $i^{2}=-1$. Then which of the following numbers is also a solution to the equation?
(A) $-a-b i$.
(B) $a-b i$.
(C) $-a+b i$.
(D) $b+a i$.
(E) None of the above.
(33rd American High School Mathematics Examination, 1982)
［Solution］First, we take the conjugate of both sides of the equation:

That is $\square$
$$
\begin{aligned}
\overline{0} & =\overline{c_{4} z^{4}+i c_{3} z^{3}+c_{2} z^{2}+i c_{1} z+c_{0}} \\
0 & =c_{4} \bar{z}^{4}-i c_{3} \bar{z}^{3}+c_{2} \bar{z}^{2}-i c_{1} \bar{z}+c_{0} \\
& =c_{4}(-\bar{z})^{4}+i c_{3}(-\bar{z})^{3}+c_{2}(-\bar{z})^{2}+i c_{1}(-\bar{z})+c_{0} .
\end{aligned}
$$

This shows that $-\bar{z}=-a+b i$ is also a root of the original equation.
Therefore, the answer is $(C)$.