Giả sử $z=x+yi;\left( x,y\in R \right)$

$\begin{align}   & \left| z \right|=2\leftrightarrow \left( {{x}^{2}}+{{y}^{2}} \right)=4 \\  & \left| {{z}^{2}}+\overline{z} \right|=2\leftrightarrow \left( {{x}^{2}}-{{y}^{2}}+x \right)+{{\left( 2xy-y \right)}^{2}}=4\leftrightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}+\left( {{x}^{2}}+{{y}^{2}} \right)-6x{{y}^{2}}+2{{x}^{3}}=4 \\ \end{align}$

$\leftrightarrow {{4}^{2}}+4-6x\left( 4-{{x}^{2}} \right)+2{{x}^{3}}=4\leftrightarrow 8{{x}^{3}}-24x+16=0\leftrightarrow \left[ \begin{align}  & x=1\to y=1\pm \sqrt{3} \\  & x=-2\to y=0 \\ \end{align} \right.$

Vậy $z=-2;z=1\pm \sqrt{3i}$