Ta có \(\left| z \right| = \sqrt {{{\left( { - {1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}} = 1\)
Nên \({1 \over z} = {{\overline z } \over {{{\left| z \right|}^2}}} = \overline z = - {1 \over 2} - {{\sqrt 3 } \over 2}i\)
\({z^2} = {\left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^2} = {1 \over 4} - {{\sqrt 3 } \over 2}i - {3 \over 4} = - {1 \over 2} - {{\sqrt 3 } \over 2}i\)
\({\left( {\overline z } \right)^3} = \overline z .{\left( {\overline z } \right)^2} = \left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right).{\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^2}\)
\( = \left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right).\left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right) = {\left( { - {1 \over 2}} \right)^2} - {\left( {{{\sqrt 3 } \over 2}i} \right)^2} = {1 \over 4} + {3 \over 4} = 1\)
\(1 + z + {z^2} = 1 + \left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right) + \left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right) = 0\)
