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Câu hỏi: Cho số phức $z$ thỏa mãn $\left( 3-2i \right)\overline{z}-4\left( 1-i \right)=\left( 2+i \right)z$. Mô đun của $z$ là:
A. $\sqrt{10}$
B. $\dfrac{\sqrt{3}}{4}$
C. $\sqrt{5}$
D. $\sqrt{3}$
A. $\sqrt{10}$
B. $\dfrac{\sqrt{3}}{4}$
C. $\sqrt{5}$
D. $\sqrt{3}$
Gọi $z=x+yi,\ x,y\in \mathbb{R}$.
Ta có: $\begin{aligned}
& \left( 3-2i \right)\overline{z}-4\left( 1-i \right)=\left( 2+i \right)z\Leftrightarrow \left( 3-2i \right)\left( 2-i \right)\overline{z}-4\left( 1-i \right)\left( 2-i \right)=5z \\
& \Leftrightarrow \left( 4-7i \right)\left( x-yi \right)-5\left( x+yi \right)=4-12i\Leftrightarrow \left( -x-7y \right)-\left( 7x+9y \right)i=4-12i \\
\end{aligned}$
Ta có hệ: $\left\{ \begin{aligned}
& x+7y=-4 \\
& 7x+9y=12 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=3 \\
& y=-1 \\
\end{aligned} \right.$.
Vậy $z=3-i$ nên $\left| z \right|=\sqrt{{{3}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{10}$.
Ta có: $\begin{aligned}
& \left( 3-2i \right)\overline{z}-4\left( 1-i \right)=\left( 2+i \right)z\Leftrightarrow \left( 3-2i \right)\left( 2-i \right)\overline{z}-4\left( 1-i \right)\left( 2-i \right)=5z \\
& \Leftrightarrow \left( 4-7i \right)\left( x-yi \right)-5\left( x+yi \right)=4-12i\Leftrightarrow \left( -x-7y \right)-\left( 7x+9y \right)i=4-12i \\
\end{aligned}$
Ta có hệ: $\left\{ \begin{aligned}
& x+7y=-4 \\
& 7x+9y=12 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=3 \\
& y=-1 \\
\end{aligned} \right.$.
Vậy $z=3-i$ nên $\left| z \right|=\sqrt{{{3}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{10}$.
Đáp án A.