Resolva, em $\mathbb{C}$, as equações
Números complexos: Infinito 12 A - Parte 3 Pág. 138 Ex. 36
Enunciado
Resolva, em $\mathbb{C}$, as equações:
- $\left( {3 – 4i} \right)z = 2 + i$
- $\left( {1 – i} \right)z + 3 + 4i = 5 – 2iz$
- ${\left( {1 – i} \right)^2}.\overline z = 3 – 2i$
- ${z^2} – 10z + 74 = 0$
Resolução
- Ora,
$$\begin{array}{*{20}{l}}
{\left( {3 – 4i} \right)z = 2 + i}& \Leftrightarrow &{z = \frac{{2 + i}}{{3 – 4i}}} \\
{}& \Leftrightarrow &{z = \frac{{2 + i}}{{3 – 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}} \\
{}& \Leftrightarrow &{z = \frac{{6 + 8i + 3i – 4}}{{9 + 16}}} \\
{}& \Leftrightarrow &{z = \frac{2}{{25}} + \frac{{11}}{{25}}i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{\left( {1 – i} \right)z + 3 + 4i = 5 – 2iz}& \Leftrightarrow &{\left( {1 – i + 2i} \right)z + 3 + 4i = 5 – 3 – 4i} \\
{}& \Leftrightarrow &{z = \frac{{2 – 4i}}{{1 + i}} \times \frac{{1 – i}}{{1 – i}}} \\
{}& \Leftrightarrow &{z = \frac{{2 – 2i – 4i – 4}}{{1 + 1}}} \\
{}& \Leftrightarrow &{z = – 1 – 3i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{{{\left( {1 – i} \right)}^2}.\overline z = 3 – 2i}& \Leftrightarrow &{\overline z = \frac{{3 – 2i}}{{{{\left( {1 – i} \right)}^2}}}} \\
{}& \Leftrightarrow &{\overline z = \frac{{3 – 2i}}{{1 – 2i – 1}} \times \frac{i}{i}} \\
{}& \Leftrightarrow &{\overline z = \frac{{3i + 2}}{2}} \\
{}& \Leftrightarrow &{\overline z = 1 + \frac{3}{2}i} \\
{}& \Leftrightarrow &{z = 1 – \frac{3}{2}i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{{z^2} – 10z + 74 = 0}& \Leftrightarrow &{z = \frac{{10 \pm \sqrt {100 – 296} }}{2}} \\
{}& \Leftrightarrow &{z = \frac{{10 \pm 14\sqrt { – 1} }}{2}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{z = 5 – 7i}& \vee &{z = 5 + 7i}
\end{array}}
\end{array}$$
