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Câu hỏi: Cho hai số phức ${{z}_{1}},{{z}_{2}}$ thỏa mãn $\left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|$. Tính $A={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{4}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{4}}$.
A. 1.
B. 1 – i.
C. -1.
D. 1 + i.
A. 1.
B. 1 – i.
C. -1.
D. 1 + i.
Đặt ${{z}_{1}}=a+bi,{{z}_{2}}={a}'+{b}'i,$ với $a,{a}',b,{b}'\in \mathbb{R}$,
Ta có $\left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|\Leftrightarrow \left\{ \begin{aligned}
& \left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right| \\
& \left| {{z}_{1}} \right|=\left| {{z}_{2}} \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}} \\
& {{\left| {{z}_{1}} \right|}^{2}}={{\left| {{z}_{2}} \right|}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left( {{z}_{1}}-{{z}_{2}} \right)\left( \overline{{{z}_{1}}}-\overline{{{z}_{2}}} \right)={{z}_{1}}\overline{{{z}_{1}}} \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}\overline{{{z}_{1}}}-{{z}_{1}}\overline{{{z}_{2}}}-{{z}_{2}}\overline{{{z}_{1}}}+{{z}_{2}}\overline{{{z}_{2}}}={{z}_{1}}\overline{{{z}_{1}}} \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}}={{z}_{1}}\overline{{{z}_{1}}} \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}} \\
\end{aligned} \right..$
Lại có ${{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}}={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}}+\dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}}-2={{\left( \dfrac{{{z}_{1}}\overline{{{z}_{2}}}}{{{z}_{2}}{{{\bar{z}}}_{2}}}+\dfrac{{{z}_{2}}\overline{{{z}_{1}}}}{{{z}_{1}}{{{\bar{z}}}_{1}}} \right)}^{2}}-2={{\left( \dfrac{{{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}}}{{{z}_{1}}\overline{{{z}_{1}}}} \right)}^{2}}-2={{\left( \dfrac{{{z}_{1}}\overline{{{z}_{1}}}}{{{z}_{1}}\overline{{{z}_{1}}}} \right)}^{2}}-2=-1.$
Từ đó $A={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{4}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{4}}={{\left[ {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}} \right]}^{2}}-2={{\left( -1 \right)}^{2}}-2=-1.$
Ta có $\left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|\Leftrightarrow \left\{ \begin{aligned}
& \left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right| \\
& \left| {{z}_{1}} \right|=\left| {{z}_{2}} \right| \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}} \\
& {{\left| {{z}_{1}} \right|}^{2}}={{\left| {{z}_{2}} \right|}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left( {{z}_{1}}-{{z}_{2}} \right)\left( \overline{{{z}_{1}}}-\overline{{{z}_{2}}} \right)={{z}_{1}}\overline{{{z}_{1}}} \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}\overline{{{z}_{1}}}-{{z}_{1}}\overline{{{z}_{2}}}-{{z}_{2}}\overline{{{z}_{1}}}+{{z}_{2}}\overline{{{z}_{2}}}={{z}_{1}}\overline{{{z}_{1}}} \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}}={{z}_{1}}\overline{{{z}_{1}}} \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}} \\
\end{aligned} \right..$
Lại có ${{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}}={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}}+\dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}}-2={{\left( \dfrac{{{z}_{1}}\overline{{{z}_{2}}}}{{{z}_{2}}{{{\bar{z}}}_{2}}}+\dfrac{{{z}_{2}}\overline{{{z}_{1}}}}{{{z}_{1}}{{{\bar{z}}}_{1}}} \right)}^{2}}-2={{\left( \dfrac{{{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}}}{{{z}_{1}}\overline{{{z}_{1}}}} \right)}^{2}}-2={{\left( \dfrac{{{z}_{1}}\overline{{{z}_{1}}}}{{{z}_{1}}\overline{{{z}_{1}}}} \right)}^{2}}-2=-1.$
Từ đó $A={{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{4}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{4}}={{\left[ {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+{{\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}^{2}} \right]}^{2}}-2={{\left( -1 \right)}^{2}}-2=-1.$
Đáp án C.