Mostre que
Números complexos: Infinito 12 A - Parte 3 Pág. 143 Ex. 55
Mostre que $${\left( {1 + \sqrt 3 i} \right)^n} + {\left( {1 – \sqrt 3 i} \right)^n} = {2^{n + 1}}\operatorname{c} os\left( {\frac{{n\pi }}{3}} \right)$$ para todo o $n \in \mathbb{N}$.
Ora,
$$\begin{array}{*{20}{l}}
{{{\left( {1 + \sqrt 3 i} \right)}^n} + {{\left( {1 – \sqrt 3 i} \right)}^n}}& = &{{{\left[ {2\left( {\frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)} \right]}^n} + {{\left[ {2\left( {\frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)} \right]}^n}} \\
{}& = &{{{\left( {2\operatorname{cis} \frac{\pi }{3}} \right)}^n} + {{\left( {2\operatorname{cis} \left( { – \frac{\pi }{3}} \right)} \right)}^n}} \\
{}& = &{{2^n}\operatorname{cis} \left( {\frac{{n\pi }}{3}} \right) + {2^n}\operatorname{cis} \left( { – \frac{{n\pi }}{3}} \right)} \\
{}& = &{{2^n}\left[ {\cos \left( {\frac{{n\pi }}{3}} \right) + i\operatorname{sen} \left( {\frac{{n\pi }}{3}} \right) + \cos \left( { – \frac{{n\pi }}{3}} \right) + i\operatorname{sen} \left( { – \frac{{n\pi }}{3}} \right)} \right]} \\
{}& = &{{2^n}\left[ {\cos \left( {\frac{{n\pi }}{3}} \right) + i\operatorname{sen} \left( {\frac{{n\pi }}{3}} \right) + \cos \left( {\frac{{n\pi }}{3}} \right) – i\operatorname{sen} \left( {\frac{{n\pi }}{3}} \right)} \right]} \\
{}& = &{{2^n} \times 2\cos \left( {\frac{{n\pi }}{3}} \right)} \\
{}& = &{{2^{n + 1}}\cos \left( {\frac{{n\pi }}{3}} \right)}
\end{array}$$
Portanto, $${\left( {1 + \sqrt 3 i} \right)^n} + {\left( {1 – \sqrt 3 i} \right)^n} = {2^{n + 1}}\operatorname{c} os\left( {\frac{{n\pi }}{3}} \right),\forall n \in \mathbb{N}$$
