Question

o valor da potência (√3 + i)¹⁸ é:

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Answer 1

✅ Após ter resolvido todos os cálculos, concluímos que a décima oitava potência do referido número complexo é:

        $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf (\sqrt{3} + i)^{18} = -262144}} \end{gathered} }$

Se nos foi dado a seguinte expressão:

             $\Large\displaystyle\begin{gathered}(\sqrt{3} + i)^{18} \end{gathered}$

Significa que o número complexo é:

             $\Large\displaystyle\begin{gathered}z = \sqrt{3} + i \end{gathered}$

Onde:

                 $\Large\begin{cases}a = \sqrt{3}\\b = 1 \end{cases}$

Calculando o módulo do número complexo temos:

          $\Large\displaystyle\text{ \begin{gathered}|z| = \sqrt{a^{2} + b^{2}} \end{gathered} }$

                 $\Large\displaystyle\text{ \begin{gathered}= \sqrt{(\sqrt[\!\diagup\!\!]{3} )^{\!\diagup\!\!\!\!2} + 1^{2}} \end{gathered} }$

                 $\Large\displaystyle\begin{gathered}= \sqrt{3 + 1} \end{gathered}$

                 $\Large\displaystyle\begin{gathered}= \sqrt{4} \end{gathered}$

                 $\Large\displaystyle\begin{gathered}= 2 \end{gathered}$

        $\Large\displaystyle\begin{gathered}\therefore\:\:\:|z| = 2 \end{gathered}$

Encontrando o argumento principal de "z":

             $\Large\begin{cases}sen\:\theta = \frac{b}{|z|} = \frac{1}{2}\\cos\:\theta = \frac{a}{|z|} = \frac{\sqrt{3}}{2} \end{cases}$

Sabendo que:

 $\Large\displaystyle\text{ \begin{gathered}tg\:\theta = \frac{\frac{1}{2} }{\frac{\sqrt{3}}{2} } = \frac{1}{\!\diagup\!\!\!\!2} \cdot\frac{\!\diagup\!\!\!\!2}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{gathered} }$

Então:

     $\Large\displaystyle\text{ \begin{gathered}\therefore\:\:\:\theta = arctg\Bigg(\frac{\sqrt{3}}{3} \Bigg) = 30^{\circ} = \frac{\pi}{6}rad \end{gathered} }$

Sabendo que:

          $\Large\displaystyle\begin{gathered}z^{n} = |z|^{n}\cdot[cos\:(n\theta) + i\cdot sen\:(n\theta)]\end{gathered}$

Calculando a décima oitava potência de "z", temos:

      $\Large\displaystyle\begin{gathered}z^{18} = 2^{18}\cdot\Bigg[cos\:\Bigg(18\cdot\frac{\pi}{6} \Bigg) + i\cdot\:sen\:\Bigg(18\cdot\frac{\pi}{6} \Bigg)\Bigg] \end{gathered}$

              $\Large\displaystyle\begin{gathered} = 262144\cdot\Bigg[cos\:\Bigg(\frac{18\pi}{6} \Bigg) + i\cdot sen\:\Bigg(\frac{18\pi}{6} \Bigg)\Bigg] \end{gathered}$

              $\Large\displaystyle\begin{gathered}= 262144\cdot[cos\:3\pi + i\cdot sen\:3\pi] \end{gathered}$

    $\Large\displaystyle\begin{gathered}\therefore\:\:\: z^{18} = 262144\cdot[cos\:3\pi + i\cdot sen\:3\pi]\end{gathered}$

Como 3π é congruente à π, ou seja:

                              $\Large\displaystyle\begin{gathered}3\pi \equiv \pi \end{gathered}$

Então, temos:

            $\Large\displaystyle\begin{gathered}z^{18} = 262144\cdot[cos\:\pi + i\cdot sen\:\pi] \end{gathered}$

                    $\Large\displaystyle\begin{gathered}= 262144\cdot[-1 + i\cdot0] \end{gathered}$

                    $\Large\displaystyle\begin{gathered} = -262144 \end{gathered}$

✅ Portanto, a décima oitava potência de "z" é:

        $\Large\displaystyle\begin{gathered}(\sqrt{3} + i)^{18} = -262144 \end{gathered}$

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