Bài tập 4.28 trang 206 SBT Toán 12

Ta có: \({z_1} + {z_2} =  - \frac{{\sqrt 3 }}{2};{z_1}.{z_2} = \frac{3}{2}\)

a) \(z_1^2 + z_2^2 = {\left( {{z_1} + {z_2}} \right)^2} - 2{z_1}{z_2} = {\left( {\frac{{ - \sqrt 3 }}{2}} \right)^2} - 2.\frac{3}{2} = \frac{3}{4} - 3 =  - \frac{9}{4}\)

b) \(z_1^3 + z_2^3 = {\left( {{z_1} + {z_2}} \right)^3} - 3{z_1}{z_2}\left( {{z_1} + {z_3}} \right) = {\left( {\frac{{ - \sqrt 3 }}{2}} \right)^3} - 3.\frac{3}{2}.\left( { - \frac{{\sqrt 3 }}{2}} \right) = \frac{{15\sqrt 3 }}{8}\)

c) \(z_1^4 + z_2^4 = {\left( {z_1^2 + z_2^2} \right)^2} - 2z_1^2.z_2^2 = {\left( { - \frac{9}{4}} \right)^2} - 2.{\left( {\frac{3}{2}} \right)^2} = \frac{9}{{16}}\)

d) \(\frac{{{z_1}}}{{{z_2}}} + \frac{{{z_2}}}{{{z_1}}} = \frac{{z_1^2 + z_2^2}}{{{z_1}{z_2}}} = \left( { - \frac{9}{4}} \right):\left( {\frac{3}{2}} \right) =  - \frac{3}{2}\)

-- Mod Toán 12 HỌC247