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Câu hỏi: Cho số phức $z=a+bi$, $\left( a,b \in \mathbb{R} \right)$ thỏa mãn $\left| z \right|\left( 2+i \right)=z-1+i\left( 2z+3 \right)$. Tính $S=a+b$.
A. $S=1$.
B. $S=-5$.
C. $S=-1$.
D. $S=7$.
Theo giả thiết: $\left| z \right|\left( 2+i \right)=z-1+i\left( 2z+3 \right)$
$\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}\left( 2+i \right)=a+bi-1+i\left[ 2\left( a+bi \right)+3 \right]$ $\Leftrightarrow \left\{ \begin{aligned}
& 2\sqrt{{{a}^{2}}+{{b}^{2}}}=a-1-2b\left( 1 \right) \\
& \sqrt{{{a}^{2}}+{{b}^{2}}}=b+2a+3\left( 2 \right) \\
\end{aligned} \right.$
$\Rightarrow 2b+4a+6=a-1-2b\Leftrightarrow b=-\dfrac{3}{4}a-\dfrac{7}{4}$
Thay vào $\left( 2 \right)$ ta được: $\sqrt{\dfrac{25{{a}^{2}}}{16}+\dfrac{42a}{16}+\dfrac{49}{16}}=\dfrac{5a}{4}+\dfrac{5}{4}\Leftrightarrow a=3\Rightarrow b=-4$.
Vậy $S=a+b=3-4=-1$.
A. $S=1$.
B. $S=-5$.
C. $S=-1$.
D. $S=7$.
Theo giả thiết: $\left| z \right|\left( 2+i \right)=z-1+i\left( 2z+3 \right)$
$\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}\left( 2+i \right)=a+bi-1+i\left[ 2\left( a+bi \right)+3 \right]$ $\Leftrightarrow \left\{ \begin{aligned}
& 2\sqrt{{{a}^{2}}+{{b}^{2}}}=a-1-2b\left( 1 \right) \\
& \sqrt{{{a}^{2}}+{{b}^{2}}}=b+2a+3\left( 2 \right) \\
\end{aligned} \right.$
$\Rightarrow 2b+4a+6=a-1-2b\Leftrightarrow b=-\dfrac{3}{4}a-\dfrac{7}{4}$
Thay vào $\left( 2 \right)$ ta được: $\sqrt{\dfrac{25{{a}^{2}}}{16}+\dfrac{42a}{16}+\dfrac{49}{16}}=\dfrac{5a}{4}+\dfrac{5}{4}\Leftrightarrow a=3\Rightarrow b=-4$.
Vậy $S=a+b=3-4=-1$.
Đáp án C.