1. Given complex numbers $z_{1}, z_{2}$ corresponding to points $A, B$ on the complex plane, and $\left|z_{1}\right|=2, z_{1}^{2}-2 z_{1} z_{2}+4 z_{2}^{2}=0, O$ is the origin, then the perimeter of $\triangle O A B$ is $\qquad$ .
Solve:
$$
\left(\frac{z_{1}}{z_{2}}\right)^{2}-2 \cdot \frac{z_{1}}{z_{2}}+4=0 \Rightarrow \frac{z_{1}}{z_{2}}=1 \pm \sqrt{3} \mathrm{i}=2\left(\cos 60^{\circ} \pm \mathrm{i} \sin 60^{\circ}\right),
$$

Thus, $\left|\frac{z_{1}}{z_{2}}\right|=2 \Rightarrow\left|z_{2}\right|=1$ and $\angle A O B=60^{\circ}$, hence $|A B|^{2}=1+4-2=3$ $\Rightarrow|A B|=\sqrt{3}$. Therefore, the perimeter of $\triangle O A B$ is $3+\sqrt{3}$.