A partir de ${i^2} = – 1$
Números complexos: Infinito 12 A - Parte 3 Pág. 68 Ex. 27
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Enunciado
A partir de ${i^2} = – 1$
- Calcule: ${i^3}$, ${i^4}$, ${i^6}$, ${i^{10}}$, ${i^{96}}$ e ${i^{105}}$.
- Para todo o $n \in \mathbb{N}$, calcule: ${i^{4n}}$, ${i^{4n + 1}}$, ${i^{4n + 2}}$ e ${i^{4n + 3}}$.
Resolução
- Ora,
$$\begin{array}{*{20}{l}}
{{i^3} = {i^2} \times i = – 1 \times i = – i} \\
{{i^4} = {{\left( {{i^2}} \right)}^2} = {{( – 1)}^2} = 1} \\
{{i^6} = {{\left( {{i^2}} \right)}^3} = {{( – 1)}^3} = – 1} \\
{{i^7} = {{\left( {{i^2}} \right)}^3} \times i = {{( – 1)}^3} \times i = – i} \\
{{i^{10}} = {{\left( {{i^2}} \right)}^4} \times {i^2} = {{( – 1)}^4} \times ( – 1) = – 1} \\
{{i^{96}} = {{\left( {{i^2}} \right)}^{48}} = {{( – 1)}^{48}} = 1} \\
{{i^{105}} = {{\left( {{i^2}} \right)}^{52}} \times i = {{( – 1)}^{52}} \times i = i}
\end{array}$$ - Ora,
$$\begin{array}{*{20}{l}}
{{i^{4n}} = {{\left( {{i^4}} \right)}^n} = {1^n} = 1,\forall n \in \mathbb{N}} \\
{{i^{4n + 1}} = {{\left( {{i^4}} \right)}^n} \times i = {1^n} \times i = i,\forall n \in \mathbb{N}} \\
{{i^{4n + 2}} = {{\left( {{i^4}} \right)}^n} \times {i^2} = {1^n} \times ( – 1) = – 1,\forall n \in \mathbb{N}} \\
{{i^{4n + 3}} = {{\left( {{i^4}} \right)}^n} \times {i^3} = {1^n} \times ( – i) = – i,\forall n \in \mathbb{N}}
\end{array}$$
