Phương pháp giải:

Lời giải chi tiết:

Giả sử ${z_1} = a + bi\left( {a,b \in \mathbb{R},{a^2} + {b^2} > 0} \right) \Rightarrow {z_2} = a - bi$

$\begin{array}{l} + )\,\,{z_1} - {z_2} = 2bi \Rightarrow \left| {{z_1} - {z_2}} \right| = \left| {2bi} \right| = \left| {2b} \right| = 2\sqrt 3  \Leftrightarrow {b^2} = 3\\ + )\,\,\dfrac{{{z_1}}}{{z_2^2}} = \dfrac{{a + bi}}{{{{\left( {a - bi} \right)}^2}}} = \dfrac{{{{\left( {a + bi} \right)}^3}}}{{{{\left( {{a^2} + {b^2}} \right)}^2}}} = \dfrac{{{a^3} - 3a{b^2}}}{{{{\left( {{a^2} + {b^2}} \right)}^2}}} + \dfrac{{3{a^2}b - {b^3}}}{{{a^2} + {b^2}}}i\,\, \in \mathbb{R}\\ \Rightarrow \dfrac{{3{a^2}b - {b^3}}}{{{a^2} + {b^2}}} = 0 \Leftrightarrow b\left( {3{a^2} - {b^2}} \right) \Leftrightarrow \left[ \begin{array}{l}b = 0\\{b^2} = 3{a^2}\end{array} \right.\end{array}$

+) $b = 0 \Rightarrow {z_1} = \,\,{z_2} = a \Rightarrow$$\left| {{z_1} - {z_2}} \right| = 0 \ne 2\sqrt 3  \Rightarrow$ Loại

+) ${b^2} = 3{a^2} \Leftrightarrow {a^2} = 1 \Rightarrow \left| z \right| = \sqrt {{a^2} + {b^2}}  = \sqrt {1 + 3}  = 2$

Chọn: C