2. (1992 National High School Competition Question) Let the complex numbers $z_{1}$ and $z_{2}$ correspond to points $A$ and $B$ on the complex plane, respectively, and $\left|z_{1}\right|=4, 4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0$, with $O$ being the origin. Then the area of $\triangle O A B$ is ( ).
A. $8 \sqrt{3}$
B. $4 \sqrt{3}$
C. $6 \sqrt{3}$
D. $12 \sqrt{3}$
2. A. Reason: From the given, $\left(z_{2}-z_{1}\right)^{2}+3 z_{1}^{2}=0$, we get $z_{2}-z_{1}= \pm \sqrt{3} z_{1} \mathrm{i}$, hence in the complex plane, the vectors corresponding to the complex numbers of the equal sides are perpendicular to each other, i.e., $O A \perp A B$, so $S_{\triangle I A B}=\frac{1}{2}$. $\left.O A|\cdot| A B\left|=\frac{1}{2}\right| O A|\cdot| \sqrt{3} O A \right\rvert\,=8 \sqrt{3}$.