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Câu hỏi: Cho số phức z thỏa mãn $\left( 3+i \right)\left| z \right|=\dfrac{-2+14i}{z}+1-3i.$ Khẳng định nào sau đây là đúng?
A. $\left| z \right|\in \left( 1;3 \right).$
B. $\left| z \right|\in \left( 3;5 \right).$
C. $\left| z \right|\in \left( 4;8 \right).$
D. $\left| z \right|\in \left( 10;13 \right).$
A. $\left| z \right|\in \left( 1;3 \right).$
B. $\left| z \right|\in \left( 3;5 \right).$
C. $\left| z \right|\in \left( 4;8 \right).$
D. $\left| z \right|\in \left( 10;13 \right).$
Ta có $\left( 3+i \right)\left| z \right|=\dfrac{-2+14i}{z}+1-3i$
$\Leftrightarrow \left( 3+i \right)\left| z \right|-1+3i=\dfrac{-2+14i}{z}\Leftrightarrow \left( 3\left| z \right|-1 \right)+\left( 3+\left| z \right| \right)i=\dfrac{-2+14i}{z}$
$\Leftrightarrow \left| \left( 3\left| z \right|-1 \right)+\left( 3+\left| z \right| \right)i \right|=\left| \dfrac{-2+14i}{z} \right|\Leftrightarrow {{\left( 3\left| z \right|-1 \right)}^{2}}+{{\left( 3+\left| z \right| \right)}^{2}}=\dfrac{200}{{{\left| z \right|}^{2}}}$
$\Leftrightarrow {{\left| z \right|}^{4}}+{{\left| z \right|}^{2}}-20=0\Leftrightarrow {{\left| z \right|}^{2}}=4\Leftrightarrow \left| z \right|=2.$
$\Leftrightarrow \left( 3+i \right)\left| z \right|-1+3i=\dfrac{-2+14i}{z}\Leftrightarrow \left( 3\left| z \right|-1 \right)+\left( 3+\left| z \right| \right)i=\dfrac{-2+14i}{z}$
$\Leftrightarrow \left| \left( 3\left| z \right|-1 \right)+\left( 3+\left| z \right| \right)i \right|=\left| \dfrac{-2+14i}{z} \right|\Leftrightarrow {{\left( 3\left| z \right|-1 \right)}^{2}}+{{\left( 3+\left| z \right| \right)}^{2}}=\dfrac{200}{{{\left| z \right|}^{2}}}$
$\Leftrightarrow {{\left| z \right|}^{4}}+{{\left| z \right|}^{2}}-20=0\Leftrightarrow {{\left| z \right|}^{2}}=4\Leftrightarrow \left| z \right|=2.$
Đáp án A.