Trace no plano de Argand

Considerando $z = x + yi$, vem ${z^2} = \left( {{x^2} – {y^2}} \right) + 2xyi$.

1. ­$$\begin{array}{*{20}{l}}
   {\operatorname{Re} \left( {{z^2}} \right) = 0}& \Leftrightarrow &{{x^2} – {y^2}=0} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{c}}
   {y = x}& \vee &{y =  – x}
   \end{array}}
   \end{array}$$
2. ­$$\begin{array}{*{20}{l}}
   {\operatorname{Im} \left( {{z^2}} \right) = 2}& \Leftrightarrow &{2xy = 2} \\
   {}& \Leftrightarrow &{y = \frac{1}{x}}
   \end{array}$$
3. ­$$\begin{array}{*{20}{l}}
   {{z^2} = 2i}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {{x^2} – {y^2} = 0}& \wedge &{xy = 1}
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {\left( {y = x \vee y =  – x} \right)}& \wedge &{y = \frac{1}{x}}
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {\left\{ {\begin{array}{*{20}{l}}
   {x =  – 1} \\
   {y =  – 1}
   \end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
   {x = 1} \\
   {y = 1}
   \end{array}} \right.}
   \end{array}}
   \end{array}$$
   ­