Example 6 (Olympic Training Question from "Intermediate Mathematics" Issue 6, 2002) Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=1$, $\left|z_{2}\right|=2,3 z_{1}-z_{2}=2+\sqrt{3} \mathrm{i}$, then $2 z_{1}+z_{2}=$ $\qquad$ .
Solve: Fill in $3-\sqrt{3} \mathrm{i}$ or $-\frac{9}{7}+\frac{13 \sqrt{3}}{7} \mathrm{i}$. Reason: From $3 z_{1}-z_{2}=2+\sqrt{3} \mathrm{i}$, we get $\left|3-\frac{z_{2}}{z_{1}}\right|=$ $\frac{|2+\sqrt{3} i|}{\left|z_{1}\right|}=\sqrt{7}$.
$$
\begin{array}{l}
\text { Also, }\left|\frac{z_{2}}{z_{1}}\right|=\frac{\left|z_{2}\right|}{\left|z_{1}\right|}=2, \text { so } \frac{z_{2}}{z_{1}}=1 \pm \sqrt{3} \mathrm{i}, \\
\text { thus, } z_{1}=\frac{2+\sqrt{3} \mathrm{i}}{3-\frac{z_{2}}{z_{1}}}=1 \text { or } \frac{1+4 \sqrt{3} \mathrm{i}}{7} .
\end{array}
$$

Therefore, $\quad 2 z_{1}+z_{2}=5 z_{1}-\left(3 z_{1}-z_{2}\right)=3-\sqrt{3} \mathrm{i}$ or $-\frac{9}{7}+\frac{13 \sqrt{3}}{7} \mathrm{i}$.