Cho số phức $z=a+bi\left( a,b\in \mathbb{R} \right)$ thỏa mãn $z+1+3i-\left| z \right|i=0.$ Tính $S=2a+3b.$

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