5. Let the complex numbers $z_{1}, z_{2}$ correspond to points $A, B$ in the complex plane, 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) $12 \sqrt{3}$.
5. (A)

From the given information, we have
$$
\begin{array}{l}
\left(z_{2}-z_{1}\right)^{2}+3 z_{1}^{2}=0, \\
z_{2}-z_{1}= \pm \sqrt{3} z_{1} \mathrm{i} .
\end{array}
$$

Therefore, in the complex plane, the vectors corresponding to both sides of the equation are perpendicular, i.e., $O A$ $\perp A B$, as shown in the figure.
$$
\begin{array}{l}
\text { Hence } S_{\triangle A B C}=\frac{1}{2}|O A| \cdot|A B| \\
=\frac{1}{2}|O A| \cdot|\sqrt{3} O A| \\
=\frac{1}{2} \times 4 \times 4 \sqrt{3}=8 \sqrt{3} .
\end{array}
$$