Gọi $z = x + yi{\rm{ }}(x,y \in R)$\[|z| - 2\bar z = 3( - 1 + 2i) \Leftrightarrow \sqrt {{x^2} + {y^2}} - 2(x - yi) = 3( - 1 + 2i)\]\[ \Leftrightarrow \begin{cases} \sqrt {{x^2} + {y^2}} - 2x = - 3 \\ 2y=6 \end{cases}\]\[ \Leftrightarrow \begin{cases} \sqrt {{x^2} + 9} = 2x - 3 (1) \\ y= 3\end{cases}\]$(1) \Leftrightarrow \begin{cases} 2x - 3 \ge 0\\ {x^2} + 9 = 4{x^2} - 12x + 9 \end{cases} \Leftrightarrow \begin{cases}x \ge \frac{3}{2} \\ 3{x^2} - 12x = 0\end{cases} \Leftrightarrow x=4$Vậy $z = 4 + 3i$. $|z| + |z{|^2} = 5 + 25 = 30.$
Gọi $z = x + yi{\rm{ }}(x,y \in R)$\[|z| - 2\bar z = 3( - 1 + 2i) \Leftrightarrow \sqrt {{x^2} + {y^2}} - 2(x - yi) = 3( - 1 + 2i)\]\[ \Leftrightarrow \begin{cases} \sqrt {{x^2} + {y^2}} - 2x = - 3 \\ 2y=6 \end{cases}\]\[ \Leftrightarrow \begin{cases} \sqrt {{x^2} + 9} = 2x - 3 (1) \\ y= 3\end{cases}\]$(1) \Leftrightarrow \begin{cases} 2x - 3 \ge 0\\ {x^2} + 9 = 4{x^2} - 12x + 9 \end{cases} \Leftrightarrow \begin{cases}x \ge \frac{3}{2} \\ 3{x^2} - 12x = 0\end{cases} \Leftrightarrow x=2$Vậy $z = 2 + 3i$. $|z| + |z{|^2} = \sqrt {13} + 13.$
Gọi $z = x + yi{\rm{ }}(x,y \in R)$\[|z| - 2\bar z = 3( - 1 + 2i) \Leftrightarrow \sqrt {{x^2} + {y^2}} - 2(x - yi) = 3( - 1 + 2i)\]\[ \Leftrightarrow \begin{cases} \sqrt {{x^2} + {y^2}} - 2x = - 3 \\ 2y=6 \end{cases}\]\[ \Leftrightarrow \begin{cases} \sqrt {{x^2} + 9} = 2x - 3 (1) \\ y= 3\end{cases}\]$(1) \Leftrightarrow \begin{cases} 2x - 3 \ge 0\\ {x^2} + 9 = 4{x^2} - 12x + 9 \end{cases} \Leftrightarrow \begin{cases}x \ge \frac{3}{2} \\ 3{x^2} - 12x = 0\end{cases} \Leftrightarrow x=
4$Vậy $z =
4 + 3i$. $|z| + |z{|^2} =
5 + 25 = 3
0.$