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

Ta có $z+1+3i-\left| z \right|i=0$ $\Leftrightarrow \left( a+1 \right)+\left( b+3-\sqrt{{{a}^{2}}+{{b}^{2}}} \right)i=0$.
$\Leftrightarrow \left\{ \begin{aligned}
& a+1=0 \\
& b+3-\sqrt{{{a}^{2}}+{{b}^{2}}}=0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& \sqrt{1+{{b}^{2}}}=b+3 \left( * \right) \\
\end{aligned} \right.$.
$\left( * \right)\Leftrightarrow \left\{ \begin{aligned}
& b\ge -3 \\
& 1+{{b}^{2}}={{\left( b+3 \right)}^{2}} \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& b\ge -3 \\
& b=-\dfrac{4}{3} \\
\end{aligned} \right. $ $ \Leftrightarrow b=-\dfrac{4}{3}$.
Vậy $\left\{ \begin{aligned}
& a=-1 \\
& b=-\dfrac{4}{3} \\
\end{aligned} \right. $ $ \Rightarrow S=2a+3b=-6$.