Cho số phức z = a + bi (a, b thuộc R) thỏa mãn z + 1 +3i - |z|i = 0. Tính

Chọn D.

Ta có $z+1+3i-\left|z\right|i=0\iff a+bi+1+3i-i\sqrt{a^2+b^2}=0$

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

$\iff \left\{\begin{array}{l}a=-1\\ \sqrt{b^2+1}=b+3\end{array}\right.\iff \left\{\begin{array}{l}a=-1\\ b\ge -3\\ b^2+1={\left(b+3\right)}^2\end{array}\right.\iff \left\{\begin{array}{l}a=-1\\ b\ge -3\\ b=-\frac{4}{3}\end{array}\right.\iff \left\{\begin{array}{l}a=-1\\ b=-\frac{4}{3}\end{array}\right..$

Vậy $S=2a+3b=2.\left(-1\right)+3.\left(-\frac{4}{3}\right)=-6.$