Cho số phức $z$ thỏa mãn $\left| z \right|-2\bar{z}=-7+3i+z$. Mô...

Cách 1: Đặt $z=a+bi\left( a,b\in \mathbb{R} \right)$. Ta có:
$\left| z \right|-2\bar{z}=-7+3i+z\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}-2\left( a-bi \right)=-7+3i+a+bi$
$\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{a}^{2}}+{{b}^{2}}}-3a+7=0 (*) \\
& b=3 \\
\end{aligned} \right.$
Thay $b=3$ vào (*) ta có:
$\sqrt{{{a}^{2}}+9}=3a-7\Leftrightarrow \left\{ \begin{aligned}
& 3a-7\ge 0 \\
& {{a}^{2}}+9={{\left( 3a-7 \right)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge \dfrac{7}{3} \\
& 8{{a}^{2}}-42a+40=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge \dfrac{7}{3} \\
& \left[ \begin{aligned}
& a=4 \\
& a=\dfrac{5}{4} \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow a=4$
Vậy $z=4+3i\Rightarrow \left| z \right|=5$
Cách 2: Ta có $\left| z \right|-2\bar{z}=-7+3i+z\Leftrightarrow z+2\bar{z}=7+\left| z \right|-3i$ (1)
Suy ra $\overline{z+2\bar{z}}=\overline{7+\left| z \right|-3i}\Leftrightarrow 2z+\bar{z}=7+\left| z \right|+3i$ (2)
Kết hợp (1) và (2) ta có hệ $\left\{ \begin{aligned}
& z+2\bar{z}=7+\left| z \right|-3i \\
& 2z+\bar{z}=7+\left| z \right|+3i \\
\end{aligned} \right.$
$\Rightarrow 3z=7+\left| z \right|+9i\Rightarrow \left| 3z \right|=\left| 7+\left| z \right|+9i \right|\Rightarrow 9{{\left| z \right|}^{2}}={{\left( 7+\left| z \right| \right)}^{2}}+81$
$\Leftrightarrow 8{{\left| z \right|}^{2}}-14\left| z \right|-130=0\Rightarrow \left| z \right|=5$