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 \[{\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^4} + {\left( {\frac{{{z_2}}}{{{z_1}}}} \right)^4}\].

Đáp án C

Ta có \(\left| {\frac{{{z_1} - {z_2}}}{{{z_1}}}} \right| = \left| {\frac{{{z_1} - {z_2}}}{{{z_2}}}} \right| = 1 \Rightarrow \left| {1 - \frac{{{z_2}}}{{{z_1}}}} \right| = \left| {\frac{{{z_1}}}{{{z_2}}} - 1} \right| = 1.\)

Giả sử \(\frac{{{z_1}}}{{{z_2}}} = a + bi{\rm{\;}}\left( {a,b \in \mathbb{R}} \right)\), từ \(\left| {\frac{{{z_1}}}{{{z_2}}} - 1} \right| = 1 \Rightarrow {\left( {a - 1} \right)^2} + {b^2} = 1\) (1)

Ta có \(\frac{{{z_2}}}{{{z_1}}} = \frac{1}{{a + bi}} = \frac{{a - bi}}{{{a^2} + {b^2}}}\), từ \(\left| {1 - \frac{{{z_2}}}{{{z_1}}}} \right| \Rightarrow \left| {\frac{{{a^2} + {b^2} - a}}{{{a^2} + {b^2}}} + \frac{b}{{{a^2} + {b^2}}}i} \right| = 1\)

\( \Rightarrow {\left( {\frac{{{a^2} + {b^2} - a}}{{{a^2} + {b^2}}}} \right)^2} + {\left( {\frac{b}{{{a^2} + {b^2}}}} \right)^2} = 1 \Rightarrow {\left( {{a^2} + {b^2} - a} \right)^2} + {b^2} = {\left( {{a^2} + {b^2}} \right)^2}\)

Từ (1) \( \Rightarrow {b^2} = 2a - {a^2} \Rightarrow {a^2} + \left( {2a - {a^2}} \right) = {\left( {2a} \right)^2} \Rightarrow 2a = 4{a^2} \Rightarrow \left[ \begin{array}{l}a = 0\\a = \frac{1}{2}\end{array} \right.\)

Với \(a = 0 \Rightarrow b = 0 \Rightarrow \frac{{{z_1}}}{{{z_2}}} = 0 \Rightarrow \) không thỏa mãn.

Với \(a = \frac{1}{2} \Rightarrow \frac{1}{4} + {b^2} = 1 \Rightarrow b = \pm \frac{{\sqrt 3 }}{2} \Rightarrow \frac{{{z_1}}}{{{z_2}}} = \frac{1}{2} \pm \frac{{\sqrt 3 }}{2}i\)

Lưu ý \(P = {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^4} + {\left( {\frac{{{z_2}}}{{{z_1}}}} \right)^4} = {\left[ {{{\left( {\frac{{{z_1}}}{{{z_2}}}} \right)}^2} + {{\left( {\frac{{{z_2}}}{{{z_1}}}} \right)}^2}} \right]^2} - 2.\) Bấm máy tính được \(P = - 1.\)