$z= x+iy \Rightarrow \overline{z} =x-iy; |z| =\sqrt{x^2 +y^2}$
Ta có $x-iy +\sqrt{x^2+y^2} =6-2i$
$\Leftrightarrow (2-y)i=6-x-\sqrt{x^2+y^2}$
$\Leftrightarrow \begin{cases} 2-y=0 \\6-x-\sqrt{x^2+y^2}=0 \end{cases} \Leftrightarrow x=\dfrac{8}{3};\ y=2 \Rightarrow z=\dfrac{8}{3}+2i$