Chọn B
Giả sử \[z=x+iy{\rm \; }\left[x,y\in {\rm R}\right].\] Khi đó ta có
\[\left\{\begin{array}{l} {\left|z\right|^{2} =2\left|z+\overline{z}\right|+4} \\ {\left|z-1-i\right|=\left|z-3+3i\right|} \end{array}\right. \]
\[\Leftrightarrow \left\{\begin{array}{l} {x^{2} +y^{2} =4\left|x\right|+4} \\ {\left[x-1\right]^{2} +\left[y-1\right]^{2} =\left[x-3\right]^{2} +\left[y+3\right]^{2} } \end{array}\right.\]
\[\Leftrightarrow \left\{\begin{array}{l} {x^{2} +y^{2} =4\left|x\right|+4} \\ {x-2y=4} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {y=\frac{x-4}{2} } \\ {x^{2} +\left[\frac{x-4}{2} \right]^{2} =4\left|x\right|+4} \end{array}\right.
\]
\[\begin{array}{l} {\Rightarrow x^{2} +\left[\frac{x-4}{2} \right]^{2} =4\left|x\right|+4} \\ {\Leftrightarrow 5x^{2} -16\left|x\right|-8x=0} \end{array}\]
\[\Leftrightarrow \left[\begin{array}{l} {\left\{\begin{array}{l} {x\ge 0} \\ {5x^{2} -24x=0} \end{array}\right. } \\ {\left\{\begin{array}{l} {x