Resolva, em $\mathbb{C}$, as equações
Números complexos: Infinito 12 A - Parte 3 Pág. 105 Ex. 65
Enunciado
Resolva, em $\mathbb{C}$, as equações:
- $z – \frac{{2i}}{z} = 0$
- ${z^3} – i{z^2} – z + i = 0$
Resolução
- Ora,
$$\begin{array}{*{20}{l}}
{z – \frac{{2i}}{z} = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{{z^2} – 2i = 0}& \wedge &{z \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{{z^2} = 2i}& \wedge &{z \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{{z^2} = 2\operatorname{cis} \left( {\frac{\pi }{2}} \right)}& \wedge &{z \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = \sqrt 2 \operatorname{cis} \left( {\frac{{\tfrac{\pi }{2}}}{2}} \right)}& \vee &{z = \sqrt 2 \operatorname{cis} \left( {\frac{{\tfrac{\pi }{2}}}{2} + \frac{{2\pi }}{2}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = \sqrt 2 \operatorname{cis} \left( {\frac{\pi }{4}} \right)}& \vee &{z = \sqrt 2 \operatorname{cis} \left( {\frac{{5\pi }}{4}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 1 + i}& \vee &{z = – 1 – i}
\end{array}}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{{z^3} – i{z^2} – z + i = 0}& \Leftrightarrow &{i\left( {1 – {z^2}} \right) – z\left( {1 – {z^2}} \right) = 0} \\
{}& \Leftrightarrow &{\left( {1 – {z^2}} \right)\left( {i – z} \right) = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{1 – {z^2} = 0}& \vee &{i – z = 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{{z^2} = 1}& \vee &{z = i}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = – 1}& \vee &{z = 1}& \vee &{z = i}
\end{array}}
\end{array}$$
