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Câu hỏi: Cho số phức z thỏa mãn $\left( 2+i \right)z-4\left( \overline{z}-i \right)=-8+19i$. Môdun của z bằng
A. 13
B. 5
C. $\sqrt{13}$
D. $\sqrt{5}$
A. 13
B. 5
C. $\sqrt{13}$
D. $\sqrt{5}$
Gọi $z=a+bi;\overline{z}=a-b{{i}^{{}}}a,b\in \mathbb{R}$.
Ta có: $\left( 2+i \right)z-4\left( \overline{z}-i \right)=-8+19i\Leftrightarrow \left( 2+i \right)\left( a+bi \right)-4\left( a-bi-i \right)=-8+19i$
$\Leftrightarrow -2a-b+\left( a+6b+4 \right)i=-8+19i\Leftrightarrow \left\{ \begin{aligned}
& -2a-b=-8 \\
& a+6b+4=19 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=2 \\
\end{aligned} \right.$
Vậy $z=3+2i\Rightarrow \left| z \right|=\sqrt{13}$
Ta có: $\left( 2+i \right)z-4\left( \overline{z}-i \right)=-8+19i\Leftrightarrow \left( 2+i \right)\left( a+bi \right)-4\left( a-bi-i \right)=-8+19i$
$\Leftrightarrow -2a-b+\left( a+6b+4 \right)i=-8+19i\Leftrightarrow \left\{ \begin{aligned}
& -2a-b=-8 \\
& a+6b+4=19 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b=2 \\
\end{aligned} \right.$
Vậy $z=3+2i\Rightarrow \left| z \right|=\sqrt{13}$
Đáp án C.