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Câu hỏi: Cho số phức $z$ thỏa mãn $z+2i.\overline{z}=1+17i$. Khi đó $\left| z \right|$ bằng
A. $\sqrt{146}$.
B. $12$.
C. $\sqrt{148}$.
D. $\sqrt{142}$.
A. $\sqrt{146}$.
B. $12$.
C. $\sqrt{148}$.
D. $\sqrt{142}$.
Đặt $z=a+bi$, $\left( a , b\in \mathbb{R} \right)$, khi đó ta có
$z+2i.\overline{z}=1+17i\Leftrightarrow \left( a+bi \right)+2i\left( a-bi \right)=1+17i$
$\Leftrightarrow \left( a+2b \right)+\left( 2a+b \right)i=1+17i$ $\Leftrightarrow \left\{ \begin{aligned}
& a+2b=1 \\
& 2a+b=17 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=11 \\
& b=-5 \\
\end{aligned} \right.$
Vậy $\left| z \right|=\sqrt{{{11}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{146}$.
$z+2i.\overline{z}=1+17i\Leftrightarrow \left( a+bi \right)+2i\left( a-bi \right)=1+17i$
$\Leftrightarrow \left( a+2b \right)+\left( 2a+b \right)i=1+17i$ $\Leftrightarrow \left\{ \begin{aligned}
& a+2b=1 \\
& 2a+b=17 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=11 \\
& b=-5 \\
\end{aligned} \right.$
Vậy $\left| z \right|=\sqrt{{{11}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{146}$.
Đáp án A.