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Câu hỏi: Cho hai số phức ${{z}_{1}},{{z}_{2}}$ thỏa mãn $\left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|>0$. Tính ${{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{4}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{4}}$.
A. 1
B. $1-i$
C. $-1$
D. $1+i$
A. 1
B. $1-i$
C. $-1$
D. $1+i$
Ta có $\left| \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}} \right|=\left| \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{2}}} \right|=1\Rightarrow \left| 1-\dfrac{{{z}_{2}}}{{{z}_{1}}} \right|=\left| \dfrac{{{z}_{1}}}{{{z}_{2}}}-1 \right|=1.$
Giả sử $\dfrac{{{z}_{1}}}{{{z}_{2}}}=a+bi\!\!~\!\!\left( a,b\in \mathbb{R} \right)$, từ $\left| \dfrac{{{z}_{1}}}{{{z}_{2}}}-1 \right|=1\Rightarrow {{\left( a-1 \right)}^{2}}+{{b}^{2}}=1$ (1)
Ta có $\dfrac{{{z}_{2}}}{{{z}_{1}}}=\dfrac{1}{a+bi}=\dfrac{a-bi}{{{a}^{2}}+{{b}^{2}}}$, từ $\left| 1-\dfrac{{{z}_{2}}}{{{z}_{1}}} \right|\Rightarrow \left| \dfrac{{{a}^{2}}+{{b}^{2}}-a}{{{a}^{2}}+{{b}^{2}}}+\dfrac{b}{{{a}^{2}}+{{b}^{2}}}i \right|=1$
$\Rightarrow {{\left( \dfrac{{{a}^{2}}+{{b}^{2}}-a}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}+{{\left( \dfrac{b}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}=1\Rightarrow {{\left( {{a}^{2}}+{{b}^{2}}-a \right)}^{2}}+{{b}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Từ (1) $\Rightarrow {{b}^{2}}=2a-{{a}^{2}}\Rightarrow {{a}^{2}}+\left( 2a-{{a}^{2}} \right)={{\left( 2a \right)}^{2}}\Rightarrow 2a=4{{a}^{2}}\Rightarrow \left[ \begin{aligned}
& a=0 \\
& a=\dfrac{1}{2} \\
\end{aligned} \right.$
Với $a=0\Rightarrow b=0\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=0\Rightarrow $ không thỏa mãn.
Với $a=\dfrac{1}{2}\Rightarrow \dfrac{1}{4}+{{b}^{2}}=1\Rightarrow b=\pm \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i$
Lưu ý $P={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{4}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{4}}={{\left[ {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}} \right]}^{2}}-2.$ Bấm máy tính được $P=-1.$
Giả sử $\dfrac{{{z}_{1}}}{{{z}_{2}}}=a+bi\!\!~\!\!\left( a,b\in \mathbb{R} \right)$, từ $\left| \dfrac{{{z}_{1}}}{{{z}_{2}}}-1 \right|=1\Rightarrow {{\left( a-1 \right)}^{2}}+{{b}^{2}}=1$ (1)
Ta có $\dfrac{{{z}_{2}}}{{{z}_{1}}}=\dfrac{1}{a+bi}=\dfrac{a-bi}{{{a}^{2}}+{{b}^{2}}}$, từ $\left| 1-\dfrac{{{z}_{2}}}{{{z}_{1}}} \right|\Rightarrow \left| \dfrac{{{a}^{2}}+{{b}^{2}}-a}{{{a}^{2}}+{{b}^{2}}}+\dfrac{b}{{{a}^{2}}+{{b}^{2}}}i \right|=1$
$\Rightarrow {{\left( \dfrac{{{a}^{2}}+{{b}^{2}}-a}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}+{{\left( \dfrac{b}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}=1\Rightarrow {{\left( {{a}^{2}}+{{b}^{2}}-a \right)}^{2}}+{{b}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Từ (1) $\Rightarrow {{b}^{2}}=2a-{{a}^{2}}\Rightarrow {{a}^{2}}+\left( 2a-{{a}^{2}} \right)={{\left( 2a \right)}^{2}}\Rightarrow 2a=4{{a}^{2}}\Rightarrow \left[ \begin{aligned}
& a=0 \\
& a=\dfrac{1}{2} \\
\end{aligned} \right.$
Với $a=0\Rightarrow b=0\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=0\Rightarrow $ không thỏa mãn.
Với $a=\dfrac{1}{2}\Rightarrow \dfrac{1}{4}+{{b}^{2}}=1\Rightarrow b=\pm \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i$
Lưu ý $P={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{4}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{4}}={{\left[ {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}} \right]}^{2}}-2.$ Bấm máy tính được $P=-1.$
Đáp án C.