$z= x+iy \Rightarrow \overline{z} =x-iy; |z| =\sqrt{x^2 +y^2}$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$
$z= x+iy \Rightarrow \overline{z} =x-iy; |z| =\sqrt{x^2 +y^2}$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=3;\ y=2 \Rightarrow z=3+2i$
$z= x+iy \Rightarrow \overline{z} =x-iy; |z| =\sqrt{x^2 +y^2}$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$