4. Given complex numbers $z_{1}, z_{2}$ correspond to points $A, B$ on the complex plane, and $\left|z_{2}\right|=4,4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0$, then the area of $\triangle A O B$ is
A. $\sqrt{3}$
B. 1
C. $4 \sqrt{3}$
D. $2 \sqrt{3}$
$\begin{array}{l}\text { 4. } \mathrm{D} \quad 4\left(\frac{z_{1}}{z_{2}}\right)^{2}-2\left(\frac{z_{1}}{z_{2}}\right)+1=0 \\ \frac{z_{1}}{z_{2}}=\frac{1}{2}\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}\right)=\frac{1}{2}\left(\cos 60^{\circ} \pm i \sin 60^{\circ}\right),\left|z_{1}\right|=2 \\ S=\frac{1}{2} \times 2 \times 4 \times \sin 60^{\circ}=2 \sqrt{3}\end{array}$

Translates to:

$\begin{array}{l}\text { 4. } \mathrm{D} \quad 4\left(\frac{z_{1}}{z_{2}}\right)^{2}-2\left(\frac{z_{1}}{z_{2}}\right)+1=0 \\ \frac{z_{1}}{z_{2}}=\frac{1}{2}\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}\right)=\frac{1}{2}\left(\cos 60^{\circ} \pm i \sin 60^{\circ}\right),\left|z_{1}\right|=2 \\ S=\frac{1}{2} \times 2 \times 4 \times \sin 60^{\circ}=2 \sqrt{3}\end{array}$