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Câu hỏi: Cho ${{z}_{1}}, {{z}_{2}}\in \mathbb{C} , \left| {{z}_{1}} \right|=3, \left| {{z}_{2}} \right|=4 , \left| {{z}_{1}}-{{z}_{2}} \right|=5$. Giá trị $A=\left| {{\left( {{z}_{1}}\overline{{{z}_{2}}} \right)}^{2}}+{{\left( \overline{{{z}_{1}}}{{z}_{2}} \right)}^{2}} \right|$ bằng
A. $288$.
B. $144$.
C. $0$.
D. $24$.
A. $288$.
B. $144$.
C. $0$.
D. $24$.
Ta có $\left| {{z}_{1}}-{{z}_{2}} \right|=5\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}=25\Leftrightarrow \left( {{z}_{1}}-{{z}_{2}} \right)\left( \overline{{{z}_{1}}}-\overline{{{z}_{2}}} \right)=25\Leftrightarrow {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-\left( \overline{{{z}_{1}}}{{z}_{2}}+{{z}_{1}}\overline{{{z}_{2}}} \right)=25$
$\Rightarrow \overline{{{z}_{1}}}{{z}_{2}}+{{z}_{1}}\overline{{{z}_{2}}}=0$.
$A=\left| {{\left( {{z}_{1}}\overline{{{z}_{2}}} \right)}^{2}}+{{\left( \overline{{{z}_{1}}}{{z}_{2}} \right)}^{2}} \right|=\left| {{\left( \overline{{{z}_{1}}}{{z}_{2}}+{{z}_{1}}\overline{{{z}_{2}}} \right)}^{2}}-2{{\left| {{z}_{1}} \right|}^{2}}.{{\left| {{z}_{2}} \right|}^{2}} \right|=2{{\left| {{z}_{1}} \right|}^{2}}.{{\left| {{z}_{2}} \right|}^{2}}=288$.
$\Rightarrow \overline{{{z}_{1}}}{{z}_{2}}+{{z}_{1}}\overline{{{z}_{2}}}=0$.
$A=\left| {{\left( {{z}_{1}}\overline{{{z}_{2}}} \right)}^{2}}+{{\left( \overline{{{z}_{1}}}{{z}_{2}} \right)}^{2}} \right|=\left| {{\left( \overline{{{z}_{1}}}{{z}_{2}}+{{z}_{1}}\overline{{{z}_{2}}} \right)}^{2}}-2{{\left| {{z}_{1}} \right|}^{2}}.{{\left| {{z}_{2}} \right|}^{2}} \right|=2{{\left| {{z}_{1}} \right|}^{2}}.{{\left| {{z}_{2}} \right|}^{2}}=288$.
Đáp án A.