Google
Câu hỏi: Cho số phức $z=a+bi\left( a,b\in \mathbb{R} \right)$ thỏa mãn $z-4+2i=\left| z \right|i.$ Giá trị $S=a+2b$ bằng
A. $9$.
B. $11$.
C. $12$.
D. $10$.
A. $9$.
B. $11$.
C. $12$.
D. $10$.
Ta có: $z-4+2i=\left| z \right|i\Rightarrow \left( a-4 \right)+\left( b+2 \right)i=\sqrt{{{a}^{2}}+{{b}^{2}}}i$
$\Leftrightarrow \left\{ \begin{aligned}
& a-4=0 \\
& b+2=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=4 \\
& b+2=\sqrt{{{b}^{2}}+16}\text{ }\left( 1 \right) \\
\end{aligned} \right.$
Từ $\left( 1 \right)\Rightarrow b+2=\sqrt{{{b}^{2}}+16}\text{ }\Leftrightarrow \left\{ \begin{aligned}
& b+2>0 \\
& {{b}^{2}}+4b+4={{b}^{2}}+16 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b>-2 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow b=3$
Vậy: $S=a+2b=4+2.3=10.$
$\Leftrightarrow \left\{ \begin{aligned}
& a-4=0 \\
& b+2=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=4 \\
& b+2=\sqrt{{{b}^{2}}+16}\text{ }\left( 1 \right) \\
\end{aligned} \right.$
Từ $\left( 1 \right)\Rightarrow b+2=\sqrt{{{b}^{2}}+16}\text{ }\Leftrightarrow \left\{ \begin{aligned}
& b+2>0 \\
& {{b}^{2}}+4b+4={{b}^{2}}+16 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b>-2 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow b=3$
Vậy: $S=a+2b=4+2.3=10.$
Đáp án D.