Quais são os números complexos cujos quadrados são iguais ao seu conjugado?

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

Assim, temos:

$$\begin{array}{*{20}{l}}
{{z^2} = \overline z }& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{{x^2} – {y^2} = x} \\
{2xy =  – y}
\end{array}} \right.} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{{x^2} – {y^2} = x} \\
{y\left( {2x + 1} \right) = 0}
\end{array}} \right.} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{{x^2} – {y^2} = x} \\
{y = 0 \vee x =  – \frac{1}{2}}
\end{array}} \right.} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{y = 0} \\
{{x^2} = x}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x =  – \frac{1}{2}} \\
{\frac{1}{4} – {y^2} =  – \frac{1}{2}}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{y = 0} \\
{x = 0 \vee x = 1}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x =  – \frac{1}{2}} \\
{y =  – \frac{{\sqrt 3 }}{2} \vee y = \frac{{\sqrt 3 }}{2}}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x = 0} \\
{y = 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x = 1} \\
{y = 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x =  – \frac{1}{2}} \\
{y =  – \frac{{\sqrt 3 }}{2}}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x =  – \frac{1}{2}} \\
{y = \frac{{\sqrt 3 }}{2}}
\end{array}} \right.}
\end{array}}
\end{array}$$

Portanto, são apenas quatro os números complexos cujos quadrados são iguais ao seu conjugado:
$$\begin{array}{*{20}{l}}
{{z^2} = \overline z }& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{z = 1}& \vee &{z =  – \frac{1}{2} – \frac{{\sqrt 3 }}{2}i}& \vee &{z =  – \frac{1}{2} + \frac{{\sqrt 3 }}{2}i}
\end{array}}
\end{array}$$

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