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Câu hỏi: Cho số phức z thỏa mãn $4(\overline{z}-i)-(3-i)z=-1-29i.$ Mođun của z bằng
A. $\left| z \right|=4.$
B. $\left| z \right|=\sqrt{5}.$
C. $\left| z \right|=1.$
D. $\left| z \right|=5.$
A. $\left| z \right|=4.$
B. $\left| z \right|=\sqrt{5}.$
C. $\left| z \right|=1.$
D. $\left| z \right|=5.$
Giả sử $z=x+yi(x,y\in )\Rightarrow 4(x-yi-i)-(3-i)(x+yi)=-1-29i$
$\begin{aligned}
& \Leftrightarrow 4x-4yi-4i-\!\![\!\!3x+y+(3y-x)i]=1-29i \\
& \Leftrightarrow x-y-(7y-x+4)i=-1-29i \\
& \Leftrightarrow \left\{ \begin{aligned}
& x-y=-1 \\
& 7y-x+4=29 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=3 \\
& y=4 \\
\end{aligned} \right.\Rightarrow \left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=5. \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow 4x-4yi-4i-\!\![\!\!3x+y+(3y-x)i]=1-29i \\
& \Leftrightarrow x-y-(7y-x+4)i=-1-29i \\
& \Leftrightarrow \left\{ \begin{aligned}
& x-y=-1 \\
& 7y-x+4=29 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=3 \\
& y=4 \\
\end{aligned} \right.\Rightarrow \left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=5. \\
\end{aligned}$
Đáp án D.