Cho ba số phức \({z_1},{z_2},{z_3}\) thỏa mãn \(\left\{ \begin{array}{l} {z_1} + {z_2} + {z_3} = 0\\ |{z_1}| = |{z_2}| = |{z_3}| = 1 \end{array

Cho ba số phức \({z_1},{z_2},{z_3}\) thỏa mãn \(\left\{ \begin{array}{l} {z_1} + {z_2} + {z_3} = 0\\ |{z_1}| = |{z_2}| = |{z_3}| = 1 \end{array} \right.\). Khẳng định nào sau đây là đúng?
A. \(|z_1^2 + z_2^2 + z_3^2| = |{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}|\)
B. \(|z_1^2 + z_2^2 + z_3^2| > |{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}|\)
C. \(|z_1^2 + z_2^2 + z_3^2| < |{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}|\)
D. \(3 = |z_1^2 + z_2^2 + z_3^2|.|{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}|\)

Ta có:
\({({z_1} + {z_2} + {z_3})^2} = z_1^2 + z_2^2 + z_3^2 + 2({z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1})\)
\(\Rightarrow z_1^2 + z_2^2 + z_3^2 = - 2({z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1})\)
Mặt khác \(|{z_1}| = 1 \Rightarrow |{z_1}{|^2} = 1 \Leftrightarrow {z_1}.\overline {{z_1}} = 1\), tương tự \({z_2}.\overline {{z_2}} = 1,{z_3}.\overline {{z_3}} = 1\) nên
\(\frac{1}{{{z_1}}} + \frac{1}{{{z_2} }} + \frac{1}{{{z_3}}} = \overline {{z_1}} + \overline {{z_2}} + \overline {{z_3}}\)
Khi đó:
\({z_1}^2 + {z_2}^2 + {z_3}^2 = - 2{z_1}{z_2}{z_3}\left( {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right)\)
\(=- 2z{z_2}{z_3}(\overline {{z_1}} + \overline {{z_2}} + \overline {{z_3}} ) = - 2z_1}{z_2}{z_3}(\overline {z_1 + {z_2} + {z_3}} ) = 0.\)
Vậy \(|z_1^2 + z_2^2 + z_3^2| = |{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}|.\)