5. Let the complex numbers $z_{1}, z_{2}$ correspond to points $A, B$ in the complex plane, and $\left|z_{1}\right|=4, 4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0$, 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. $12 \sqrt{3}$
$$
\begin{array}{l}
4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0 \Rightarrow 4\left(\frac{z_{1}}{z_{2}}\right)^{2}-2 \cdot \frac{z_{1}}{z_{2}}+1=0 \Rightarrow \frac{z_{1}}{z_{2}}=\frac{1 \pm \sqrt{3} \mathrm{i}}{4} \\
=\frac{1}{2}\left(\cos 60^{\circ} \pm \mathrm{i} \sin 60^{\circ}\right) \Rightarrow|O A|=4,|O B|=8, \angle A O B=60^{\circ} \Rightarrow S_{\triangle A O B}=8 \sqrt{3} .
\end{array}
$$

So the answer is $A$.