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Câu hỏi: Cho số phức $z$ thỏa mãn $\left( 2+3i \right)\left| z \right|=\dfrac{\sqrt{26}}{\overline{z}}+3-2i$. Mệnh đề nào dưới đây đúng?
A. $1<\left| z \right|<2.$
B. $2<\left| z \right|<3.$
C. $0<\left| z \right|<2.$
D. $2<\left| z \right|<4.$
A. $1<\left| z \right|<2.$
B. $2<\left| z \right|<3.$
C. $0<\left| z \right|<2.$
D. $2<\left| z \right|<4.$
Ta có $\left( 2+3i \right)\left| z \right|=\dfrac{\sqrt{26}}{{\bar{z}}}+3-2i\Leftrightarrow 2\left| z \right|+3\left| z \right|i-3+2i=\dfrac{\sqrt{26}}{{\bar{z}}}$
$\Leftrightarrow 2\left| z \right|-3+\left( 3\left| z \right|+2 \right)i=\dfrac{\sqrt{26}}{{\bar{z}}}\Leftrightarrow \left| 2\left| z \right|-3+\left( 3\left| z \right|+2 \right)i \right|=\left| \dfrac{\sqrt{26}}{{\bar{z}}} \right|$
$\Leftrightarrow \sqrt{{{\left( 2\left| z \right|-3 \right)}^{2}}+{{\left( 3\left| z \right|+2 \right)}^{2}}}=\dfrac{\sqrt{26}}{\left| z \right|}\Leftrightarrow 13{{\left| z \right|}^{2}}+13=\dfrac{26}{{{\left| z \right|}^{2}}}\Rightarrow \left| z \right|=1.$
$\Leftrightarrow 2\left| z \right|-3+\left( 3\left| z \right|+2 \right)i=\dfrac{\sqrt{26}}{{\bar{z}}}\Leftrightarrow \left| 2\left| z \right|-3+\left( 3\left| z \right|+2 \right)i \right|=\left| \dfrac{\sqrt{26}}{{\bar{z}}} \right|$
$\Leftrightarrow \sqrt{{{\left( 2\left| z \right|-3 \right)}^{2}}+{{\left( 3\left| z \right|+2 \right)}^{2}}}=\dfrac{\sqrt{26}}{\left| z \right|}\Leftrightarrow 13{{\left| z \right|}^{2}}+13=\dfrac{26}{{{\left| z \right|}^{2}}}\Rightarrow \left| z \right|=1.$
Đáp án C.