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Câu hỏi: Cho số phức $z=a+bi\left( a,b\in \mathbb{R} \right)$ thỏa mãn $z+1+3i-\left| z \right|i=0$. Tính $S=a+3b$.
A. $S=5$
B. $S=-5$
C. $S=\dfrac{7}{3}$
D. $S=-\dfrac{7}{3}$
A. $S=5$
B. $S=-5$
C. $S=\dfrac{7}{3}$
D. $S=-\dfrac{7}{3}$
Giả sử $z=a+bi\left( a,b\in \mathbb{R} \right)$
Ta có $z+1+3i-\left| z \right|i=0\Leftrightarrow a+bi+1+3i-i\sqrt{{{a}^{2}}+{{b}^{2}}}=0$
$\Leftrightarrow a+1+\left( b+3-\sqrt{{{a}^{2}}+{{b}^{2}}} \right)i=0$
$\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
a+1=0 \\
b+3=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
a=-1 \\
\left\{ \begin{array}{*{35}{l}}
b\ge -3 \\
{{\left( b+3 \right)}^{2}}=1+{{b}^{2}} \\
\end{array} \right. \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
a=-1 \\
b=-\dfrac{4}{3} \\
\end{array} \right.$
$\Rightarrow S=a+3b=-5$.
Ta có $z+1+3i-\left| z \right|i=0\Leftrightarrow a+bi+1+3i-i\sqrt{{{a}^{2}}+{{b}^{2}}}=0$
$\Leftrightarrow a+1+\left( b+3-\sqrt{{{a}^{2}}+{{b}^{2}}} \right)i=0$
$\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
a+1=0 \\
b+3=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
a=-1 \\
\left\{ \begin{array}{*{35}{l}}
b\ge -3 \\
{{\left( b+3 \right)}^{2}}=1+{{b}^{2}} \\
\end{array} \right. \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
a=-1 \\
b=-\dfrac{4}{3} \\
\end{array} \right.$
$\Rightarrow S=a+3b=-5$.
Đáp án B.