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Câu hỏi: Cho số phức z thỏa mãn $\left( 2-i \right)z+3+16i=2\left( \overrightarrow{z}+i \right)$. Môđun của z bằng
A. $\sqrt{5}.$
B. 13.
C. $\sqrt{13}.$
D. 5.
A. $\sqrt{5}.$
B. 13.
C. $\sqrt{13}.$
D. 5.
Gọi $z=x+yi$. Ta có $\left( 2-i \right)z+3+16i=2\left( \overline{z}+i \right)$
$\Leftrightarrow \left( 2-i \right)\left( x+yi \right)+3+16i=2\left( x-yi+i \right)$
$\Leftrightarrow 2x+2yi-xi+y+3+16i=2x-2yi+2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 2x+y+3=2x \\
& 2y-x+16=-2y+2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y+3=0 \\
& -x+4y=-14 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=2 \\
& y=-3 \\
\end{aligned} \right.$
Suy ra $z=2-3i$. Vậy $\left| z \right|=\sqrt{13}$
$\Leftrightarrow \left( 2-i \right)\left( x+yi \right)+3+16i=2\left( x-yi+i \right)$
$\Leftrightarrow 2x+2yi-xi+y+3+16i=2x-2yi+2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 2x+y+3=2x \\
& 2y-x+16=-2y+2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y+3=0 \\
& -x+4y=-14 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=2 \\
& y=-3 \\
\end{aligned} \right.$
Suy ra $z=2-3i$. Vậy $\left| z \right|=\sqrt{13}$
Đáp án C.