12・19 If $a \pm b i(b \neq 0)$ are the complex roots of the equation $x^{3}+q x+r=0$, where $a$, $b, q$ and $r$ are real numbers. Then $q$ expressed in terms of $a, b$ is
(A) $a^{2}+b^{2}$.
(B) $2 a^{2}-b^{2}$.
(C) $b^{2}-a^{2}$.
(D) $b^{2}-2 a^{2}$.
(E) $b^{2}-3 a^{2}$.
(23rd American High School Mathematics Examination, 1972)
[Solution 1] Let the third root of the given equation be $s$. According to the problem (by Vieta's formulas), the sum of the roots is zero, i.e., $2a + s = 0$, which gives $s = -2a$.
By Vieta's formulas again, we have

i.e., $\square$
$$
\begin{aligned}
& (a+bi)(a-bi) + (a+bi)s + (a-bi)s = q, \\
& \left(a^{2} + b^{2}\right) + a(-2a) + bi(-2a) + a(-2a) - bi(-2a) \\
= & b^{2} - 3a^{2} = q.
\end{aligned}
$$

Therefore, the answer is $(E)$.
[Solution 2] Let $\alpha = a + bi$, then $\bar{\alpha} = a - bi$. According to the problem, we have
$$
\alpha^{3} + q\alpha + r = 0 \text{, and } \quad \bar{\alpha}^{3} + q\bar{\alpha} + r = 0.
$$

Subtracting the two equations, we get $\alpha^{3} - \bar{\alpha}^{3} + q(\alpha - \bar{\alpha}) = 0$.
Thus, $-q = \frac{\alpha^{3} - \bar{\alpha}^{3}}{\alpha - \bar{\alpha}} = \alpha^{2} + \alpha\bar{\alpha} + \bar{\alpha}^{2} = \left(\alpha^{2} + \bar{\alpha}^{2}\right) + \alpha\bar{\alpha}$
$$
= \left(2a^{2} - 2b^{2}\right) + \left(a^{2} + b^{2}\right) = 3a^{2} - b^{2}.
$$

Therefore, the answer is $(E)$.