B

- Hướng dẫn giải

Ta có: 

\(z + 1 + 3i - \left| z \right|i = 0 \Leftrightarrow a + bi + 1 + 3i - i\sqrt {{a^2} + {b^2}}  = 0\)

\( \Leftrightarrow a + 1 + \left( {b + 3 - \sqrt {{a^2} + {b^2}} } \right)i = 0 \Leftrightarrow \left\{ \begin{array}{l}
a + 1 = 0\\
b + 3 = \sqrt {{a^2} + {b^2}} 
\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}
a =  - 1\\
\left\{ \begin{array}{l}
b \ge  - 3\\
{\left( {b + 3} \right)^2} = 1 + {b^2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a =  - 1\\
b =  - \frac{4}{3}
\end{array} \right. \Rightarrow S =  - 5\)