Ta có: \({z_1} + {z_2} = - \dfrac{{\sqrt 3 }}{2},{z_1}.{z_2} = \dfrac{3}{2}\). Từ đó suy ra:
a) \(z_1^2 + z_2^2 = {\left( {{z_1} + {z_2}} \right)^2} - 2{z_1}{z_2}\)\( = \dfrac{3}{4} - 3 = - \dfrac{9}{4}\)
b) \(z_1^3 + z_2^3\)\( = \left( {{z_1} + {z_2}} \right)\left( {z_1^2 - {z_1}{z_2} + z_2^2} \right)\) \( = - \dfrac{{\sqrt 3 }}{2}\left( { - \dfrac{9}{4} - \dfrac{3}{2}} \right) = \dfrac{{15\sqrt 3 }}{8}\)
c) \(z_1^4 + z_2^4 = {\left( {z_1^2 + z_2^2} \right)^2} - 2z_1^2.z_2^2\)\( = {\left( { - \dfrac{9}{4}} \right)^2} - 2.{\left( {\dfrac{3}{2}} \right)^2} = \dfrac{9}{{16}}\)
d) \(\dfrac{{{z_1}}}{{{z_2}}} + \dfrac{{{z_2}}}{{{z_1}}} = \dfrac{{z_1^2 + z_2^2}}{{{z_1}.{z_2}}}\)\( = \dfrac{{ - \dfrac{9}{4}}}{{\dfrac{3}{2}}} = - \dfrac{3}{2}\)
