Cho số phức $z=a+bi, \left( a,b\in \mathbb{R} \right)$ thỏa mãn...

$z+5+3i=\left| z \right|\Leftrightarrow a+bi+5+3i=\sqrt{{{a}^{2}}+{{b}^{2}}}\Leftrightarrow \left\{ \begin{aligned}
& a+5=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
& b+3=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a+5=\sqrt{{{a}^{2}}+9} \\
& b=-3 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{-8}{5} \\
& b=-3 \\
\end{aligned} \right.$
Do đó: $5a+b=-11$.