Bài tập 5 trang 190 SGK Toán 12 NC

Ta có: \(|z| = \sqrt {{{\left( { - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}  = 1\)

Nên: \(\frac{1}{z} = \frac{{\overline z }}{{|z{|^2}}} = \overline z  =  - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{z^2} = {\left( { - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)^2}\\
 = \frac{1}{4} - \frac{{\sqrt 3 }}{2}i - \frac{3}{4} =  - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i
\end{array}\\
\begin{array}{l}
{\left( {\bar z} \right)^3} = \bar z{\left( {\bar z} \right)^2}\\
 = \left( { - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i} \right).{\left( { - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)^2}
\end{array}\\
\begin{array}{l}
 = \left( { - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i} \right).\left( { - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)\\
 = {\left( { - \frac{1}{2}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{2}i} \right)^2}
\end{array}\\
{ = \frac{1}{4} + \frac{3}{4} = 1}\\
\begin{array}{l}
1 + z + {z^2}\\
 = 1 + \left( { - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) + \left( { - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i} \right) = 0
\end{array}
\end{array}\)

-- Mod Toán 12 HỌC247