Question

raízes quadradas de 4i

Answer 1

raiz de 4i =2 *2 * i * i
= + ou - 2 raiz de i
resposta final =(+ ou - 2i)
ai pode ser também raiz de (2+i raiz de 2 e
-raiz de 2-i raiz de 2).
Pode ser desses dois jeitos que estar em parênteses .

Answer 2

✅ Após ter realizado todos os cálculos concluímos que as raízes quadradas do referido número complexo são:

      $\large\begin{cases}P_{0} = \sqrt{2} + i\sqrt{2}\\P_{1} = -\sqrt{2} - i\sqrt{2} \end{cases}$

Seja o número complexo "Z":

                 $\large\displaystyle\begin{gathered}Z = 4i \end{gathered}$

Calculando o módulo de "Z", temos:

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

                 $\large\displaystyle\text{ \begin{gathered}= \sqrt{0^{2} + 4^{2}} \end{gathered} }$

                 $\large\displaystyle\begin{gathered}= \sqrt{0 + 16} \end{gathered}$

                 $\large\displaystyle\begin{gathered}= \sqrt{16} \end{gathered}$

                 $\large\displaystyle\begin{gathered}= 4 \end{gathered}$

       $\large\displaystyle\begin{gathered}\therefore\:\:\:|Z| = 4 \end{gathered}$

Obtendo o argumento principal do referido número complexo:

    $\Large\begin{cases}sen\:\theta = \frac{b}{|Z|} = \frac{4}{4} = 1\\cos\:\theta = \frac{a}{|Z|} = \frac{0}{4} = 0 \end{cases}$

              $\large\displaystyle\begin{gathered}\therefore\:\:\:\theta = 90^\circ = \frac{\pi}{2} \end{gathered}$

Para encontrar as raízes de um número complexos podemos utilizar a segunda fórmula de Moivre, com suas restrições que é:

   $\large\displaystyle\begin{gathered}Z_{K} = \sqrt[n]{|Z|}\cdot\Bigg[cos\cdot\Bigg(\frac{\theta + 2k\pi}{n} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\theta + 2k\pi}{n} \Bigg)\Bigg] \end{gathered}$

          $\large\displaystyle\begin{gathered}Para\:\:k = 0, 1, 2, 3,\:\cdots,n - 1\:\:\:e\:\:\:\sqrt[n]{|Z|}\in\mathbb{R} \end{gathered}$

Então temos:

   $\large\displaystyle\text{ \begin{gathered}Z_{K} = \sqrt{4}\cdot\Bigg[cos\cdot\Bigg(\frac{\frac{\pi}{2} + 2k\pi}{2} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\frac{\pi}{2} + 2k\pi}{2} \Bigg)\Bigg] \end{gathered} }$

          $\large\displaystyle\text{ \begin{gathered}= 2\cdot\Bigg[cos\cdot\Bigg(\frac{\frac{\pi + 4k\pi}{2} }{2} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\frac{\pi + 4k\pi}{2} }{2}\Bigg)\Bigg] \end{gathered} }$

          $\large\displaystyle\text{ \begin{gathered}= 2\cdot\Bigg[cos\cdot\Bigg(\frac{\frac{\pi\cdot(1 + 4k)}{2} }{2} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\frac{\pi\cdot(1 + 4k)}{2} }{2}\Bigg)\Bigg] \end{gathered} }$

          $\large\displaystyle\begin{gathered}= 2\cdot\Bigg[cos\cdot\Bigg(\frac{\pi\cdot(1 + 4k)}{2}\cdot\frac{1}{2} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\pi\cdot(1 + 4k)}{2}\cdot\frac{1}{2}\Bigg)\Bigg] \end{gathered}$

         $\large\displaystyle\begin{gathered}= 2\cdot\Bigg[cos\cdot\Bigg(\frac{\pi\cdot(1 + 4k)}{4} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\pi\cdot(1 + 4k)}{4}\Bigg)\Bigg] \end{gathered}$

Portanto, chegamos à seguinte expressão:

    $\large\displaystyle\begin{gathered}Z_{K} = 2\cdot\Bigg[cos\cdot\Bigg(\frac{\pi\cdot(1 + 4k)}{4} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\pi\cdot(1 + 4k)}{4}\Bigg)\Bigg] \end{gathered}$

• K = 0

       $\large\displaystyle\begin{gathered}K_{0} = 2\cdot\Bigg[cos\cdot\Bigg(\frac{\pi\cdot(1 + 4\cdot0)}{4} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\pi\cdot(1 + 4\cdot0)}{4}\Bigg)\Bigg] \end{gathered}$

              $\large\displaystyle\begin{gathered}= 2\cdot\Bigg[cos\:\frac{\pi}{4} + i\cdot sen\:\frac{\pi}{4} \Bigg] \end{gathered}$

              $\large\displaystyle\text{ \begin{gathered}= 2\cdot\bigg[\frac{\sqrt{2}}{2} + i\cdot\frac{\sqrt{2}}{2} \bigg] \end{gathered} }$

              $\large\displaystyle\text{ \begin{gathered}= \frac{\!\diagup\!\!\!\!2\sqrt{2}}{\!\diagup\!\!\!\!2} + i\cdot\frac{\!\diagup\!\!\!\!2\sqrt{2}}{\!\diagup\!\!\!\!2} \end{gathered} }$

              $\large\displaystyle\begin{gathered}= \sqrt{2} + i\sqrt{2} \end{gathered}$

• k = 1

        $\large\displaystyle\begin{gathered}K_{1} = 2\cdot\Bigg[cos\cdot\Bigg(\frac{\pi\cdot(1 + 4\cdot1)}{4} \Bigg) + i\cdot sen\cdot\Bigg(\frac{\pi\cdot(1 + 4\cdot1)}{4}\Bigg)\Bigg] \end{gathered}$

              $\large\displaystyle\begin{gathered}= 2\cdot\Bigg[cos\:\frac{5\pi}{4} + i\cdot sen\:\frac{5\pi}{4} \Bigg] \end{gathered}$

              $\large\displaystyle\text{ \begin{gathered}= 2\cdot\bigg[-\frac{\sqrt{2}}{2} - i\cdot\frac{\sqrt{2}}{2} \bigg] \end{gathered} }$

              $\large\displaystyle\text{ \begin{gathered}= -\frac{\!\diagup\!\!\!\!2\sqrt{2}}{\!\diagup\!\!\!\!2} - i\cdot\frac{\!\diagup\!\!\!\!2\sqrt{2}}{\!\diagup\!\!\!\!2} \end{gathered} }$

              $\large\displaystyle\begin{gathered}= -\sqrt{2} - i\sqrt{2} \end{gathered}$

Portanto, os afixos são:

               $\large\begin{cases}P_{0} = \sqrt{2} + i\sqrt{2}\\P_{1} = -\sqrt{2} - i\sqrt{2} \end{cases}$

Geometricamente podemos interpretar os afixos como sendo as extremidades do diâmetro de uma circunferência com centro na origem do plano de Argand Gauss e raio igual a "2".

Saiba mais:

1. https://brainly.com.br/tarefa/46060617  
2. https://brainly.com.br/tarefa/48509167

Solução gráfica: