Escreva na forma $a + bi$ os números complexos seguintes
Números complexos: Infinito 12 A - Parte 3 Pág. 138 Ex. 35
Enunciado
Considere os números complexos $$\begin{array}{*{20}{c}}
{z = 1 – 2i}&{}&{\text{e}}&{}&{w = – 5 + 3i}
\end{array}$$ e escreva na forma $a + bi$ os números complexos seguintes:
- $z + w$
- $4z – 5w$
- $z.w$
- $\frac{z}{w}$
- ${z^2} – \frac{1}{z}$
- $\frac{2}{{{z^3}}}$
Resolução
- Ora,
$$\begin{array}{*{20}{l}}
{z + w}& = &{\left( {1 – 2i} \right) + \left( { – 5 + 3i} \right)} \\
{}& = &{ – 4 + i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{4z – 5w}& = &{4\left( {1 – 2i} \right) – 5\left( { – 5 + 3i} \right)} \\
{}& = &{4 – 8i + 25 – 15i} \\
{}& = &{29 – 23i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{z.w}& = &{\left( {1 – 2i} \right).\left( { – 5 + 3i} \right)} \\
{}& = &{ – 5 + 3i + 10i + 6} \\
{}& = &{1 + 13i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{\frac{z}{w}}& = &{\frac{{1 – 2i}}{{ – 5 + 3i}}} \\
{}& = &{\frac{{1 – 2i}}{{ – 5 + 3i}} \times \frac{{ – 5 – 3i}}{{ – 5 – 3i}}} \\
{}& = &{\frac{{ – 5 – 3i + 10i – 6}}{{25 + 9}}} \\
{}& = &{ – \frac{{11}}{{34}} + \frac{7}{{34}}i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{{z^2} – \frac{1}{z}}& = &{{{\left( {1 – 2i} \right)}^2} – \frac{1}{{1 – 2i}}} \\
{}& = &{1 – 4i – 4 – \frac{1}{{1 – 2i}} \times \frac{{1 + 2i}}{{1 + 2i}}} \\
{}& = &{ – 3 – 4i – \frac{{1 + 2i}}{{1 + 4}}} \\
{}& = &{ – 3 – 4i – \frac{1}{5} – \frac{2}{5}i} \\
{}& = &{ – \frac{{16}}{5} – \frac{{22}}{5}i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{\frac{2}{{{z^3}}}}& = &{\frac{2}{{{{\left( {1 – 2i} \right)}^3}}}} \\
{}& = &{\frac{2}{{\left( {1 – 4i – 4} \right)\left( {1 – 2i} \right)}}} \\
{}& = &{\frac{2}{{ – 3 + 6i – 4i – 8}}} \\
{}& = &{\frac{2}{{ – 11 + 2i}} \times \frac{{ – 11 – 2i}}{{ – 11 – 2i}}} \\
{}& = &{\frac{{ – 22 – 4i}}{{121 + 4}}} \\
{}& = &{ – \frac{{22}}{{125}} – \frac{4}{{125}}i}
\end{array}$$
