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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. $\left| z \right|=6$.
B. $\left| z \right|=\sqrt{146}$.
C. $\left| z \right|=10$.
D. $\left| z \right|=\sqrt{58}$.
A. $\left| z \right|=6$.
B. $\left| z \right|=\sqrt{146}$.
C. $\left| z \right|=10$.
D. $\left| z \right|=\sqrt{58}$.
Đặt $z=a+bi \left( a,b\in \mathbb{R} \right)$.
Ta có: $z+2i.\overline{z}=1+17i\Leftrightarrow a+bi+2i(a-bi)=1+17\Leftrightarrow \left\{ \begin{aligned}
& a+2b=1 \\
& 2a+b=17 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=11 \\
& b=-5 \\
\end{aligned} \right.$.
Suy ra $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{11}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{146}$.
Ta có: $z+2i.\overline{z}=1+17i\Leftrightarrow a+bi+2i(a-bi)=1+17\Leftrightarrow \left\{ \begin{aligned}
& a+2b=1 \\
& 2a+b=17 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=11 \\
& b=-5 \\
\end{aligned} \right.$.
Suy ra $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{11}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{146}$.
Đáp án B.