Resolva, em $\mathbb{C}$, as equações
Números complexos: Infinito 12 A - Parte 3 Pág. 144 Ex. 63
Enunciado
Resolva, em $\mathbb{C}$, as equações:
- ${z^4}.\overline z = 32i$
- ${z^3} + \left( {\sqrt 3 + i} \right)z = 0$
Resolução
- $${z^4}.\overline z = 32i$$
Considerando $z = \rho \operatorname{cis} \theta $, temos:
$$\begin{array}{*{20}{l}}
{{{\left( {\rho \operatorname{cis} \theta } \right)}^4} \times \overline {\rho \operatorname{cis} \theta } = 32i}& \Leftrightarrow &{{\rho ^4}\operatorname{cis} \left( {4\theta } \right) \times \rho \operatorname{cis} \left( { – \theta } \right) = 32\operatorname{cis} \left( {\frac{\pi }{2}} \right)} \\
{}& \Leftrightarrow &{{\rho ^5}\operatorname{cis} \left( {3\theta } \right) = 32\operatorname{cis} \left( {\frac{\pi }{2}} \right)} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{{\rho ^5} = 32} \\
{3\theta = \frac{\pi }{2} + 2k\pi ,k \in \mathbb{Z}}
\end{array}} \right.} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{\rho = \sqrt[5]{{32}}} \\
{\theta = \frac{\pi }{6} + \frac{{2k\pi }}{3},k \in \mathbb{Z}}
\end{array}} \right.} \\
{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}
{\rho = 2} \\
{\theta = \frac{\pi }{6} + \frac{{2k\pi }}{3},k \in \mathbb{Z}}
\end{array}} \right.}
\end{array}$$
Logo, $$\begin{array}{*{20}{l}}
{{z^4}.\overline z = 32i}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 2\operatorname{cis} \frac{\pi }{6}}& \vee &{z = 2\operatorname{cis} \frac{{5\pi }}{6}}& \vee &{z = 2\operatorname{cis} \frac{{3\pi }}{2}}
\end{array}}
\end{array}$$
- Ora,
$$\begin{array}{*{20}{l}}
{{z^3} + \left( {\sqrt 3 + i} \right)z = 0}& \Leftrightarrow &{\left( {{z^2} + \sqrt 3 + i} \right)z = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{{z^2} + \sqrt 3 + i = 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{{z^2} = – \sqrt 3 – i}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{{z^2} = 2\operatorname{cis} \frac{{7\pi }}{6}}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{z = \sqrt 2 \operatorname{cis} \left( {\frac{{\tfrac{{7\pi }}{6}}}{2}} \right)}& \vee &{z = \sqrt 2 \operatorname{cis} \left( {\frac{{\tfrac{{7\pi }}{6}}}{2} + \frac{{2\pi }}{2}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 0}& \vee &{z = \sqrt 2 \operatorname{cis} \left( {\frac{{7\pi }}{{12}}} \right)}& \vee &{z = \sqrt 2 \operatorname{cis} \left( {\frac{{19\pi }}{{12}}} \right)}
\end{array}}
\end{array}$$
