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Câu hỏi: Cho số phức $z=a+bi$ ( $a,b\in \mathbb{R}$ ) thỏa mãn $z+1+3i-\left| z \right|i=0$. Tính $S=a+3b$.
A. $S=\dfrac{7}{3}$
B. $S=-5$
C. $S=5$
D. $S=-\dfrac{7}{3}$
A. $S=\dfrac{7}{3}$
B. $S=-5$
C. $S=5$
D. $S=-\dfrac{7}{3}$
Ta có $(a+1)+(b+3)i-i\sqrt{{{a}^{2}}+{{b}^{2}}}=0\Leftrightarrow \left\{ \begin{aligned}
& a+1=0 \\
& b+3-\sqrt{{{a}^{2}}+{{b}^{2}}}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& \sqrt{{{b}^{2}}+1}=b+3 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b\ge -3 \\
& {{b}^{2}}+1={{(b+3)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b=-\dfrac{4}{3} \\
\end{aligned} \right.\Rightarrow S=-5$.
& 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.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b\ge -3 \\
& {{b}^{2}}+1={{(b+3)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b=-\dfrac{4}{3} \\
\end{aligned} \right.\Rightarrow S=-5$.
Đáp án B.