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Câu hỏi: Xét số phức z thỏa mãn $\left( 1+2i \right)\left| z \right|=\dfrac{\sqrt{10}}{z}-2+i$. Mệnh đề nào sau đây đúng?
A. $\dfrac{3}{2}<\left| z \right|<2$
B. $\left| z \right|>2$
C. $\left| z \right|<\dfrac{1}{2}$
D. $\dfrac{1}{2}<\left| z \right|<\dfrac{3}{2}$
A. $\dfrac{3}{2}<\left| z \right|<2$
B. $\left| z \right|>2$
C. $\left| z \right|<\dfrac{1}{2}$
D. $\dfrac{1}{2}<\left| z \right|<\dfrac{3}{2}$
Ta có ${{z}^{-1}}=\dfrac{1}{{{\left| z \right|}^{2}}}\overline{z}$. Vậy $\left( 1+2i \right)\left| z \right|=\dfrac{\sqrt{10}}{z}-2+i$
$\Leftrightarrow \left( \left| z \right|+2 \right)+\left( 2\left| z \right|-1 \right)i=\left( \dfrac{\sqrt{10}}{{{\left| z \right|}^{2}}} \right).\overline{z}\Rightarrow {{\left( \left| z \right|+2 \right)}^{2}}+{{\left( 2\left| z \right|-1 \right)}^{2}}=\left( \dfrac{10}{{{\left| z \right|}^{4}}} \right).{{\left| z \right|}^{2}}=\dfrac{10}{{{\left| z \right|}^{2}}}$
Đặt ${{\left| z \right|}^{2}}=a>0$
$\Rightarrow {{\left( a+2 \right)}^{2}}+{{\left( 2a-1 \right)}^{2}}=\dfrac{10}{{{a}^{2}}}\Leftrightarrow {{a}^{4}}+{{a}^{2}}-2=0\Leftrightarrow \left[ \begin{aligned}
& {{a}^{2}}=1 \\
& {{a}^{2}}=-2 \\
\end{aligned} \right.\Rightarrow a=1\Rightarrow \left| z \right|=1$
$\Leftrightarrow \left( \left| z \right|+2 \right)+\left( 2\left| z \right|-1 \right)i=\left( \dfrac{\sqrt{10}}{{{\left| z \right|}^{2}}} \right).\overline{z}\Rightarrow {{\left( \left| z \right|+2 \right)}^{2}}+{{\left( 2\left| z \right|-1 \right)}^{2}}=\left( \dfrac{10}{{{\left| z \right|}^{4}}} \right).{{\left| z \right|}^{2}}=\dfrac{10}{{{\left| z \right|}^{2}}}$
Đặt ${{\left| z \right|}^{2}}=a>0$
$\Rightarrow {{\left( a+2 \right)}^{2}}+{{\left( 2a-1 \right)}^{2}}=\dfrac{10}{{{a}^{2}}}\Leftrightarrow {{a}^{4}}+{{a}^{2}}-2=0\Leftrightarrow \left[ \begin{aligned}
& {{a}^{2}}=1 \\
& {{a}^{2}}=-2 \\
\end{aligned} \right.\Rightarrow a=1\Rightarrow \left| z \right|=1$
Đáp án D.