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Números complexos: Infinito 12 A - Parte 3 Pág. 72 Ex. 38
Enunciado
Sendo ${z_1} = {a_1} + {b_1}i$ e ${z_2} = {a_2} + {b_2}i$, mostre que:
- $\overline {{z_1} + {z_2}} = \overline {{z_1}} + \overline {{z_2}} $
- $\overline {{z_1}.{z_2}} = \overline {{z_1}} .\overline {{z_2}} $
- $\overline {{z_1} – {z_2}} = \overline {{z_1}} – \overline {{z_2}} $
- $\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} = \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}$, para ${z_2} \ne 0$.
Resolução
- Ora,
$$\begin{array}{*{20}{l}}
{\overline {{z_1} + {z_2}} }& = &{\overline {\left( {{a_1} + {b_1}i} \right) + \left( {{a_2} + {b_2}i} \right)} }&{}&{}&{\overline {{z_1}} + \overline {{z_2}} }& = &{\overline {\left( {{a_1} + {b_1}i} \right)} + \overline {\left( {{a_2} + {b_2}i} \right)} } \\
{}& = &{\overline {\left( {{a_1} + {a_2}} \right) + \left( {{b_1} + {b_2}} \right)i} }&{}&{}&{}& = &{\left( {{a_1} – {b_1}i} \right) + \left( {{a_2} – {b_2}i} \right)} \\
{}& = &{\left( {{a_1} + {a_2}} \right) – \left( {{b_1} + {b_2}} \right)i}&{}&{}&{}& = &{\left( {{a_1} + {a_2}} \right) – \left( {{b_1} + {b_2}} \right)i}
\end{array}$$
Portanto, $\overline {{z_1} + {z_2}} = \overline {{z_1}} + \overline {{z_2}} $.
- Ora,
$$\begin{array}{*{20}{l}}
{\overline {{z_1}.{z_2}} }& = &{\overline {\left( {{a_1} + {b_1}i} \right).\left( {{a_2} + {b_2}i} \right)} }&{}&{}&{\overline {{z_1}} .\overline {{z_2}} }& = &{\overline {\left( {{a_1} + {b_1}i} \right)} .\overline {\left( {{a_2} + {b_2}i} \right)} } \\
{}& = &{\overline {{a_1}{a_2} + {a_1}{b_2}i + {a_2}{b_1}i – {b_1}{b_2}} }&{}&{}&{}& = &{\left( {{a_1} – {b_1}i} \right).\left( {{a_2} – {b_2}i} \right)} \\
{}& = &{\overline {\left( {{a_1}{a_2} – {b_1}{b_2}} \right) + \left( {{a_1}{b_2} + {a_2}{b_1}} \right)i} }&{}&{}&{}& = &{{a_1}{a_2} – {a_1}{b_2}i – {a_2}{b_1}i – {b_1}{b_2}} \\
{}& = &{\left( {{a_1}{a_2} – {b_1}{b_2}} \right) – \left( {{a_1}{b_2} + {a_2}{b_1}} \right)i}&{}&{}&{}& = &{\left( {{a_1}{a_2} – {b_1}{b_2}} \right) – \left( {{a_1}{b_2} + {a_2}{b_1}} \right)i}
\end{array}$$
Portanto, $\overline {{z_1}.{z_2}} = \overline {{z_1}} .\overline {{z_2}} $.
- Ora,
$$\begin{array}{*{20}{l}}
{\overline {{z_1} – {z_2}} }& = &{\overline {\left( {{a_1} + {b_1}i} \right) – \left( {{a_2} + {b_2}i} \right)} }&{}&{}&{\overline {{z_1}} – \overline {{z_2}} }& = &{\overline {\left( {{a_1} + {b_1}i} \right)} – \overline {\left( {{a_2} + {b_2}i} \right)} } \\
{}& = &{\overline {\left( {{a_1} – {a_2}} \right) + \left( {{b_1} – {b_2}} \right)i} }&{}&{}&{}& = &{\left( {{a_1} – {b_1}i} \right) – \left( {{a_2} – {b_2}i} \right)} \\
{}& = &{\left( {{a_1} – {a_2}} \right) – \left( {{b_1} – {b_2}} \right)i}&{}&{}&{}& = &{\left( {{a_1} – {a_2}} \right) – \left( {{b_1} – {b_2}} \right)i}
\end{array}$$
Portanto, $\overline {{z_1} – {z_2}} = \overline {{z_1}} – \overline {{z_2}} $.
- Ora,
$$\begin{array}{*{20}{l}}
{\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} }& = &{\overline {\left( {\frac{{{a_1} + {b_1}i}}{{{a_2} + {b_2}i}}} \right)} }&{}&{}&{\frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}}& = &{\frac{{\overline {{a_1} + {b_1}i} }}{{\overline {{a_2} + {b_2}i} }}} \\
{}& = &{\overline {\left( {\frac{{{a_1} + {b_1}i}}{{{a_2} + {b_2}i}} \times \frac{{{a_2} – {b_2}i}}{{{a_2} – {b_2}i}}} \right)} }&{}&{}&{}& = &{\frac{{{a_1} – {b_1}i}}{{{a_2} – {b_2}i}} \times \frac{{{a_2} + {b_2}i}}{{{a_2} + {b_2}i}}} \\
{}& = &{\overline {\left( {\frac{{\left( {{a_1}{a_2} + {b_1}{b_2}} \right) + \left( {{a_2}{b_1} – {a_1}{b_2}} \right)i}}{{{{\left( {{a_2}} \right)}^2} + {{\left( {{b_2}} \right)}^2}}}} \right)} }&{}&{}&{}& = &{\frac{{\left( {{a_1}{a_2} + {b_1}{b_2}} \right) – \left( {{a_2}{b_1} – {a_1}{b_2}} \right)i}}{{{{\left( {{a_2}} \right)}^2} + {{\left( {{b_2}} \right)}^2}}}} \\
{}& = &{\frac{{\left( {{a_1}{a_2} + {b_1}{b_2}} \right) – \left( {{a_2}{b_1} – {a_1}{b_2}} \right)i}}{{{{\left( {{a_2}} \right)}^2} + {{\left( {{b_2}} \right)}^2}}}}&{}&{}&{}&{}&{}
\end{array}$$
Portanto, $\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} = \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}$, para ${z_2} \ne 0$.
