Calcule

Forma trigonométrica da potência (Fórmula de Moivre):

Se $z = \rho \operatorname{cis} \theta $ é um número complexo não nulo, então $${z^n} = {\rho ^n}\operatorname{cis} \left( {n\theta } \right)$$

1. Ora,
   $$\begin{array}{*{20}{l}}
   {{{\left( { – 1 – \sqrt 3 i} \right)}^6}}& = &{{{\left[ {2\left( { – \frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)} \right]}^6}} \\
   {}& = &{{{\left( {2\operatorname{cis} \frac{{4\pi }}{3}} \right)}^6}} \\
   {}& = &{{2^6}\operatorname{cis} \left( {6 \times \frac{{4\pi }}{3}} \right)} \\
   {}& = &{64\operatorname{cis} \left( {8\pi } \right)} \\
   {}& = &{64\operatorname{cis} \left( 0 \right)}
   \end{array}$$
2. Ora,
   $$\begin{array}{*{20}{l}}
   {{{\left( {\frac{{2 + 2i}}{{2 – 2i}}} \right)}^4}}& = &{{{\left( {\frac{{2\sqrt 2 \operatorname{cis} \frac{\pi }{4}}}{{2\sqrt 2 \operatorname{cis} \left( { – \frac{\pi }{4}} \right)}}} \right)}^4}} \\
   {}& = &{{{\left( {\frac{{2\sqrt 2 }}{{2\sqrt 2 }}\operatorname{cis} \left( {\frac{\pi }{4} + \frac{\pi }{4}} \right)} \right)}^4}} \\
   {}& = &{{1^4}\operatorname{cis} \left( {4 \times \frac{\pi }{2}} \right)} \\
   {}& = &{\operatorname{cis} \left( {2\pi } \right)} \\
   {}& = &{\operatorname{cis} \left( 0 \right)}
   \end{array}$$
3. Ora,
   $$\begin{array}{*{20}{l}}
   {{{\left[ {3\operatorname{cis} \left( { – \frac{{4\pi }}{3}} \right)} \right]}^5}}& = &{{3^5}\operatorname{cis} \left( {5 \times \left( { – \frac{{4\pi }}{3}} \right)} \right)} \\
   {}& = &{243\operatorname{cis} \left( { – \frac{{20\pi }}{3} + 8\pi } \right)} \\
   {}& = &{243\operatorname{cis} \frac{{4\pi }}{3}}
   \end{array}$$