Cho số phức $z$ thỏa mãn $\left( 2+i \right)z-4\left(...

Gọi $z=x+yi,\left( x,y\in \mathbb{R} \right)$.
$\begin{aligned}
& \left( 2+i \right)z-4\left( \overline{z}-i \right)=-8+19i \\
& \Leftrightarrow \left( 2+i \right)\left( x+yi \right)-4\left( x-yi-i \right)=-8+19i \\
& \Leftrightarrow 2x-y+xi+2yi-4x+4yi+4i=-8+19i \\
& \Leftrightarrow -2x-y+\left( x+6y+4 \right)i=-8+19i \\
& \left\{ \begin{aligned}
& -2x-y=-8 \\
& x+6y+4=19 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=3 \\
& y=2 \\
\end{aligned} \right. \\
\end{aligned}$
Vậy $z=3+2i$ và $\left| z \right|=\sqrt{13}.$