${{z}_{1}}^{2}-{{z}_{1}}{{z}_{2}}+{{z}_{2}}^{2}=0$(*)

Ta có:

${{z}_{1}}^{3}+{{z}_{2}}^{2}=({{z}_{1}}+{{z}_{2}})({{z}_{1}}^{2}-{{z}_{1}}{{z}_{2}}+{{z}_{2}}^{2})=0<=>{{z}_{1}}^{3}=-{{z}_{2}}^{3}$

Ta có

OA=|${{z}_{1}}$|

OB=|${{z}_{2}}$|

AB=|${{z}_{1}}$-${{z}_{2}}$|

$\begin{align}  & {{z}_{1}}^{3}=-{{z}_{2}}^{3} \\  & =>|{{z}_{1}}^{3}|=|-{{z}_{2}}^{3}| \\ \end{align}$

$\begin{align}  & <=>|{{z}_{1}}{{|}^{3}}=|{{z}_{2}}{{|}^{3}} \\  & <=>|{{z}_{1}}|=|{{z}_{2}}|=>OA=OB(1) \\ \end{align}$

$\begin{align}  & (*)<=>{{({{z}_{1}}-{{z}_{2}})}^{2}}+{{z}_{1}}{{z}_{2}}=0 \\  & <=>{{({{z}_{1}}-{{z}_{2}})}^{2}}=-{{z}_{1}}{{z}_{2}} \\ \end{align}$

$\begin{align}  & =>|({{z}_{1}}-{{z}_{2}}){{|}^{2}}=|{{z}_{1}}{{z}_{2}}|=|{{z}_{1}}||{{z}_{2}}| \\  & <=>A{{B}^{2}}=OA.OB(2) \\ \end{align}$

$(1);(2)<=>OA=OB=AB$

Vậy đáp án là A