Đáp án

Giải thích

Ta có: \(z_1^2 + z_2^2 = {z_1}{z_2} \Leftrightarrow z_1^2 - {z_1}{z_2} + z_2^2 = 0\)

\(\left. { \Rightarrow z_1^3 + z_2^3 = \left. {\left( {{z_1} + {z_2}} \right.} \right)\left( {z_1^2 - {z_1}{z_2} + z_2^2} \right.} \right) = 0 \Leftrightarrow z_1^3 =  - z_2^3 \Rightarrow \left| {z_1^3} \right.\left| { = \left| {z_2^3} \right.} \right| \Rightarrow \left| \right.\left| { = {{\left| z \right.}_2}} \right| \Rightarrow OA = OB\)

Mặt khác \(\left. {{{\left( {\left. {{z_1} - {z_2}} \right)} \right.}^2} = \left( {z_1^2 - {z_1}{z_2} + z_2^2} \right.} \right) - {z_1}{z_2} =  - {z_1}{z_2} \Rightarrow {\left| {{z_1} - \left. \right|} \right.^2} = \left| \right.\left| {.\left| \right.} \right| \Rightarrow A{B^2} = OA.OB = O{A^2}\)

 đều.