Cho các số phức \({z_1}, {z_2}, {z_3}\) thỏa mãn 2 điều kiện \(\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = 2017\) và \({z_1} + {z_2} + {z_3} \ne 0.\) Tính \(P = \left| {\frac{{{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}}}{{{z_1} + {z_2} + {z_3}}}} \right|.\)
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Lời giải:
Báo sai\(\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = 2017 \Rightarrow \left\{ \begin{array}{l}{z_1}{{\bar z}_1} = {2017^2}\\{z_2}{{\bar z}_2} = {2017^2}\\{z_3}{{\bar z}_3} = {2017^2}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{{\bar z}_1} = \frac{{{{2017}^2}}}{{{z_1}}}\\{{\bar z}_2} = \frac{{{{2017}^2}}}{{{z_2}}}\\{{\bar z}_3} = \frac{{{{2017}^2}}}{{{z_3}}}\end{array} \right..\)
Ta có \({P^2} = {\left| {\frac{{{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}}}{{{z_1} + {z_2} + {z_3}}}} \right|^2} = \left( {\frac{{{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}}}{{{z_1} + {z_2} + {z_3}}}} \right)\left( {\frac{{{{\bar z}_1}{{\bar z}_2} + {{\bar z}_2}{{\bar z}_3} + {{\bar z}_3}{{\bar z}_1}}}{{{{\bar z}_1} + {{\bar z}_2} + {{\bar z}_3}}}} \right)\)
\( = \left( {\frac{{{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}}}{{{z_1} + {z_2} + {z_3}}}} \right)\left( {\frac{{\frac{{{{2017}^2}}}{{{z_1}}}.\frac{{{{2017}^2}}}{{{z_2}}} + \frac{{{{2017}^2}}}{{{z_2}}}.\frac{{{{2017}^2}}}{{{z_3}}} + \frac{{{{2017}^2}}}{{{z_3}}}.\frac{{{{2017}^2}}}{{{z_1}}}}}{{\frac{{{{2017}^2}}}{{{z_1}}} + \frac{{{{2017}^2}}}{{{z_2}}} + \frac{{{{2017}^2}}}{{{z_3}}}}}} \right) = {2017^2}.\)
\( \Rightarrow P = 2017.\)
