Resolva, em $\mathbb{C}$, as equações
Números complexos: Infinito 12 A - Parte 3 Pág. 103 Ex. 63
Enunciado
Resolva, em $\mathbb{C}$, as equações:
- ${z^2} = 1 + i$
- ${z^3} – iz = 0$
Resolução
- Como $w = 1 + i = \sqrt 2 \operatorname{cis} \frac{\pi }{4}$, temos:
$$\begin{array}{*{20}{l}}
{{z^2} = 1 + i}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = \sqrt {\sqrt 2 } \operatorname{cis} \left( {\frac{{\tfrac{\pi }{4}}}{2}} \right)}& \vee &{z = \sqrt {\sqrt 2 } \operatorname{cis} \left( {\frac{{\tfrac{\pi }{4}}}{2} + \frac{{2\pi }}{2}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = \sqrt[4]{2}\operatorname{cis} \left( {\frac{\pi }{8}} \right)}& \vee &{z = \sqrt[4]{2}\operatorname{cis} \left( {\frac{{9\pi }}{8}} \right)}
\end{array}}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{{z^3} – iz = 0}& \Leftrightarrow &{z\left( {{z^2} – i} \right) = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{{z^2} = i}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{{z^2} = \operatorname{cis} \frac{\pi }{2}}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{z = \sqrt 1 \operatorname{cis} \left( {\frac{{\tfrac{\pi }{2}}}{2}} \right)}& \vee &{z = \sqrt 1 \operatorname{cis} \left( {\frac{{\tfrac{\pi }{2}}}{2} + \frac{{2\pi }}{2}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{z = \operatorname{cis} \left( {\frac{\pi }{4}} \right)}& \vee &{z = \operatorname{cis} \left( {\frac{{5\pi }}{4}} \right)}
\end{array}}
\end{array}$$
