Google
Câu hỏi: Cho số phức z thỏa mãn điều kiện: $\left( 1-2i \right)\left| z \right|=\dfrac{\sqrt{5}}{z}-1+2i$. Mệnh đề nào dưới đây là đúng?
A. $0<\left| z \right|<1.$
B. $\left| z \right|>\sqrt{5}.$
C. $1<\left| z \right|<2.$
D. $2<\left| z \right|<\sqrt{5}.$
A. $0<\left| z \right|<1.$
B. $\left| z \right|>\sqrt{5}.$
C. $1<\left| z \right|<2.$
D. $2<\left| z \right|<\sqrt{5}.$
Theo đề ta có $\left( 1-2i \right)\left| z \right|=\dfrac{\sqrt{5}}{z}-1+2i\Leftrightarrow \left( 1-2i \right)\left( \left| z \right|+1 \right)=\dfrac{\sqrt{5}}{z}\left( 1 \right).$
$\left( 1 \right)\Rightarrow \left| \left( 1-2i \right)\left( \left| z \right|+1 \right) \right|=\left| \dfrac{\sqrt{5}}{z} \right|\Leftrightarrow \left| \left( 1-2i \right) \right|\left| \left| z \right|+1 \right|=\dfrac{\sqrt{5}}{\left| z \right|}\Leftrightarrow \sqrt{5}.\left( \left| z \right|+1 \right)=\dfrac{\sqrt{5}}{\left| z \right|}$
$\Leftrightarrow \left| z \right|+1=\dfrac{1}{\left| z \right|}\Leftrightarrow {{\left| z \right|}^{2}}+\left| z \right|-1=0\Leftrightarrow \left| z \right|=\dfrac{-1+\sqrt{5}}{2}$ ( do $\left| z \right|>0$ ) $\Rightarrow 0<\left| z \right|<1$.
$\left( 1 \right)\Rightarrow \left| \left( 1-2i \right)\left( \left| z \right|+1 \right) \right|=\left| \dfrac{\sqrt{5}}{z} \right|\Leftrightarrow \left| \left( 1-2i \right) \right|\left| \left| z \right|+1 \right|=\dfrac{\sqrt{5}}{\left| z \right|}\Leftrightarrow \sqrt{5}.\left( \left| z \right|+1 \right)=\dfrac{\sqrt{5}}{\left| z \right|}$
$\Leftrightarrow \left| z \right|+1=\dfrac{1}{\left| z \right|}\Leftrightarrow {{\left| z \right|}^{2}}+\left| z \right|-1=0\Leftrightarrow \left| z \right|=\dfrac{-1+\sqrt{5}}{2}$ ( do $\left| z \right|>0$ ) $\Rightarrow 0<\left| z \right|<1$.
Đáp án A.