Chọn B
Thế z=a+bi vào \(z+1+3i-\left|z\right|i=0\), ta thu được:
\(a+bi+1+3i-i\sqrt{a^{2} +b^{2} } =0\)
\(\Leftrightarrow a+1+\left(b+3-\sqrt{a^{2} +b^{2} } \right)i=0 \)
\(\Leftrightarrow \left\{\begin{array}{l} {a+1=0} \\ {b+3-\sqrt{a^{2} +b^{2} } =0} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {a=-1} \\ {b+3-\sqrt{b^{2} +1} =0\, \, \, \left(1\right)} \end{array}\right. \)
Xét \(\left(1\right)\Leftrightarrow b+3=\sqrt{b^{2} +1} .\)
\(\Leftrightarrow \left\{\begin{array}{l} {b^{2} +6b+9=b^{2} +1} \\ {b\ge -3} \end{array}\right. \Leftrightarrow b=-\frac{4}{3} .\)
Vậy: \(S=a+3b=-1+3.\left(-\frac{4}{3} \right)=-5.\)
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