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Câu hỏi: Cho số phức $z$ thỏa mãn $3\left( \overline{z}+i \right)-\left( 2-i \right)z=3+10i$. Mô đun của $z$ bằng
A. $3$.
B. $5$.
C. $\sqrt{5}$.
D. $\sqrt{3}$.
A. $3$.
B. $5$.
C. $\sqrt{5}$.
D. $\sqrt{3}$.
Gọi $z=x+yi$ $\left( x, y\in \mathbb{R} \right)$ $\Rightarrow \overline{z}=x-yi$.
Ta có $3\left( \overline{z}+i \right)-\left( 2-i \right)z=3+10i$ $\Leftrightarrow 3\left( x-yi \right)-\left( 2-i \right)\left( x+yi \right)=3+7i$
$\Leftrightarrow x-y+\left( x-5y \right)i=3+7i$ $\Leftrightarrow \left\{ \begin{aligned}
& x-y=3 \\
& x-5y=7 \\
\end{aligned} \right.$$\left\{ \begin{aligned}
& x=2 \\
& y=-1 \\
\end{aligned} \right.$.
Suy ra $z=2-i$. Vậy $\left| z \right|=\sqrt{5}$.
Ta có $3\left( \overline{z}+i \right)-\left( 2-i \right)z=3+10i$ $\Leftrightarrow 3\left( x-yi \right)-\left( 2-i \right)\left( x+yi \right)=3+7i$
$\Leftrightarrow x-y+\left( x-5y \right)i=3+7i$ $\Leftrightarrow \left\{ \begin{aligned}
& x-y=3 \\
& x-5y=7 \\
\end{aligned} \right.$$\left\{ \begin{aligned}
& x=2 \\
& y=-1 \\
\end{aligned} \right.$.
Suy ra $z=2-i$. Vậy $\left| z \right|=\sqrt{5}$.
Đáp án C.