12.31 Let the complex numbers $z_{1}, z_{2}$ correspond to points $A, B$ in the complex plane, respectively, and $\left|z_{1}\right|$ $=4, 4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0, O$ is the origin. Then the area of $\triangle O A B$ is
(A) $8 \sqrt{3}$.
(B) $4 \sqrt{3}$.
(C) $6 \sqrt{3}$.
(D) $12 \sqrt{3}$.
(China High School Mathematics League, 1992)
[Solution 1] From the given, we have $\left(z_{2}-z_{1}\right)^{2}=-3 z_{1}^{2}$, hence $z_{2}-z_{1}= \pm \sqrt{3} z_{1} i$, so the vectors corresponding to the complex numbers on both sides of the equation in the plane are perpendicular, i.e., $O A \perp A B$.
Therefore, $\left.S_{\triangle A B O}=\frac{1}{2}|O A| \cdot \right| \cdot A B$ |
$$
\begin{array}{l}
=\frac{1}{2} \cdot 4 \cdot 4 \sqrt{3} \\
=8 \sqrt{3} .
\end{array}
$$

Thus, the answer is $(A)$.
[Solution 2] From the given, we have $z_{2}=(1 \pm \sqrt{3} i) z_{1}=2\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) z_{1}$
$$
=2\left(\cos 60^{\circ}+i \sin 60^{\circ}\right) z_{1},
$$

thus $|O B|=2|O A|$. By the cosine rule, we have $|A B|=\sqrt{3}|O A|$.
Therefore, $|A B|=4 \sqrt{3}$. Hence, $\triangle O A B$ is a right triangle.
So, $S=\frac{1}{2}|O A| \cdot|O B|=8 \sqrt{3}$.
Thus, the answer is $(A)$.