Question

Cho ba số phức \(z_{1} , z_{2} , z_{3}\)  thỏa mãn điều kiện  \(\left\{\begin{array}{l} {\left|z_{1} \right|=\left|z_{2} \right|=\left|z_{3} \right|=1} \\ {z_{1}^{2} =z_{2} z_{3} } \\ {\left|z_{1} -z_{2} \right|=\frac{\sqrt{6} +\sqrt{2} }{2} } \end{array}\right. \). Tính \(M=\left|z_{2} -z_{3} \right|-\left|z_{3} -z_{1} \right|.\)

\(A.-\sqrt{6} -\sqrt{3} -\sqrt{2} . \)

\(B. \sqrt{6} +\sqrt{3} -\sqrt{2} . \)

\(C. \frac{\sqrt{6} +\sqrt{2} -2}{2} . \)

\(D. \frac{-\sqrt{6} -\sqrt{2} +2}{2} .\)

Answer 1

Chọn D

Ta có
\(\left|z_{1} -z_{2} \right|^{2} +\left|z_{1} +z_{2} \right|^{2} =2\left(\left|z_{1} \right|^{2} +\left|z_{2} \right|^{2} \right)\)
\(\Leftrightarrow \left|z_{1} +z_{2} \right|^{2} =4-\frac{\left(\sqrt{6} +\sqrt{2} \right)^{2} }{4} \Leftrightarrow \left|z_{1} +z_{2} \right|=\sqrt{4-\frac{\left(\sqrt{6} +\sqrt{2} \right)^{2} }{4} }  \)
Khi đó, \(M=\left|z_{2} -z_{3} \right|-\left|z_{3} -z_{1} \right|=\left|z_{2} -\frac{z_{1}^{2} }{z_{2} } \right|-\left|z_{1} \right|\left|z_{3} -z_{1} \right|\)
\(\begin{array}{l} {=\left|\frac{z_{2}^{2} -z_{1}^{2} }{z_{2} } \right|-\left|z_{1} z_{3} -z_{1}^{2} \right|=\frac{\left|z_{1} -z_{2} \right|.\left|z_{1} +z_{2} \right|}{\left|z_{2} \right|} -\left|z_{1} z_{3} -z_{2} z_{3} \right|} \\ {=\frac{\left|z_{1} -z_{2} \right|.\left|z_{1} +z_{2} \right|}{\left|z_{2} \right|} -\left|z_{3} \right|.\left|z_{1} -z_{2} \right|} \end{array} \)
\(=\frac{\sqrt{6} +\sqrt{2} }{2} .\sqrt{4-\frac{\left(\sqrt{6} +\sqrt{2} \right)^{2} }{4} } -\frac{\sqrt{6} +\sqrt{2} }{2} =\frac{-\sqrt{6} -\sqrt{2} +2}{2} . \)