Cho số phức z thỏa mãn $\left( 1-i \right)z+4z=7-7i.$ Khi đó...

Gọi $z=a+bi\left( a,b\in \mathbb{R} \right)\Rightarrow \overline{z}=a-bi.$
Ta có
$\begin{aligned}
& \left( 1-i \right)z+4\overline{z}=7-7i\Leftrightarrow \left( 1-i \right)\left( a+bi \right)+4\left( a-bi \right)=7-7i\Leftrightarrow \left( 5a+b-7 \right)+\left( -a-3b+7 \right)i=0 \\
& \Leftrightarrow \left\{ \begin{aligned}
& 5a+b=7 \\
& a+3b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=2 \\
\end{aligned} \right.\Rightarrow z=1+2i\Rightarrow \left| z \right|=\sqrt{{{1}^{2}}+{{2}^{2}}}=\sqrt{5}. \\
\end{aligned}$