Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
z = x + yi \Rightarrow \overline z = x - yi\\
z\left( {1 + i} \right) + \overline z - 3 = 0\\
\Leftrightarrow \left( {x + yi} \right)\left( {1 + i} \right) + x - yi - 3 = 0\\
\Leftrightarrow x + xi + yi + y{i^2} + x - yi - 3 = 0\\
\Leftrightarrow 2x + xi - y - 3 = 0\,\,\,\,\,\,\,\left( {{i^2} = - 1} \right)\\
\Leftrightarrow \left( {2x - y - 3} \right) + xi = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - y - 3 = 0\\
x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = - 3
\end{array} \right. \Rightarrow z = - 3i\\
d,\\
z = x + yi \Rightarrow \overline z = x - yi\\
{z^2} = {\left| z \right|^2} + \overline z \\
\Leftrightarrow {\left( {x + yi} \right)^2} = {\sqrt {{x^2} + {y^2}} ^2} + x - yi\\
\Leftrightarrow {x^2} + 2xyi + {y^2}{i^2} = {x^2} + {y^2} + x - yi\\
\Leftrightarrow {x^2} + 2xyi - {y^2} = {x^2} + {y^2} + x - yi\,\,\,\,\,\,\,\,\,\,\,\left( {{i^2} = - 1} \right)\\
\Leftrightarrow \left( {x + 2{y^2}} \right) - \left( {y + 2xy} \right)i = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 2{y^2} = 0\\
y\left( {1 + 2x} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 0\\
x = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - \frac{1}{2}\\
y = \pm \frac{1}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
z = 0\\
z = - \frac{1}{2} + \frac{1}{2}i\\
z = - \frac{1}{2} - \frac{1}{2}i
\end{array} \right.
\end{array}\)
