Đặt $z=x+yi\ \ \left( x,y\in R \right)$ suy ra $\overline{z}=x-yi$. Khi đó ta được

$4x-2yi=2\sqrt{{{x}^{2}}+{{y}^{2}}}+\sqrt{3\left( {{x}^{2}}+{{y}^{2}} \right)i}$

$\Leftrightarrow \left\{ \begin{align}  & \sqrt{{{x}^{2}}+{{y}^{2}}}=2x \\  & \sqrt{3\left( {{x}^{2}}+{{y}^{2}} \right)}=-2y \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}  & x\ge 0,y\le 0 \\  & 3\left( {{x}^{2}}+{{y}^{2}} \right)=4{{y}^{2}} \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}  & x\ge 0,y\le 0 \\  & 3{{x}^{2}}={{y}^{2}} \\ \end{align} \right.\Leftrightarrow y=-\sqrt{3}x$.