Đặt $z=x+iy\,\,\left( x;y\in \mathbb{R} \right)\Rightarrow \bar{z}=x-iy$

Vậy phương trình trở thành:

$\left( 3-2i \right).\left( x-iy \right)-4\left( 1-i \right)=\left( 2+i \right).\left( x+iy \right)$

$\Leftrightarrow \left( 3x-2ix-3iy+2{{i}^{2}}y \right)-4+4i=2x+2iy+ix+{{i}^{2}}y$

$\Leftrightarrow 3x-2x+2{{i}^{2}}y-4-{{i}^{2}}y+\left( -2ix-3iy+4i-2iy-ix \right)=0$

$\Leftrightarrow \left( x-y-4 \right)+i\left( -3x-5y+4 \right)=0$

$\Leftrightarrow \left\{ \begin{align} & x-y-4=0 \\ & -3x-5y+4=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & x=3 \\ & y=-1 \\ \end{align} \right.$

$\Rightarrow z=3-i\Rightarrow \left| z \right|=\sqrt{{{3}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{10}$