O valor da potência $\left(\dfrac{\sqrt{2}}{1+i}\right)^{93}$ é:


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ITA IIIT 28/02/2022 22:28
$-$ Seja $z= \large{\frac{\sqrt{2}}{1+i}}$ , temos: \begin{matrix} z &=& \large{\frac{\sqrt{2}}{1+i} \ .\ \color{royalblue}{\frac{(1-i)}{(1-i)}}} &=& \large{\frac{\sqrt{2}}{2} + \frac{i\sqrt{2}}{2}} &=& \large{( \frac{\sqrt{2}}{2} \ , \ \frac{\sqrt{2}}{2})} \end{matrix} $-$ Pela $\text{Lei de Moivre}$, podemos escrever: \begin{matrix} z^{93} &=& |z|^{93} \ . \ [\cos{(93.\varphi)} + i\sin{(93.\varphi)}] \end{matrix} $-$ Agora, encontrando o argumento $(\varphi)$ de $z$ \begin{matrix} \tan{\varphi} &=& \Large{\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}} &=& 1 &\Rightarrow& \fbox{$\varphi = \large{ \frac{\pi}{4} }$} \end{matrix} $-$ Voltando a nossa expressão, temos: \begin{matrix} z^{93} &=& |z|^{93} \ . \ [\cos{(24\pi - \frac{3\pi}{4})} + i\sin{(24\pi - \frac{3\pi}{4})}] \\ \\ \end{matrix} Portanto, \begin{matrix} z^{93} &=& \large{\frac{-\sqrt{2} + i\sqrt{2}}{2} \ .\ \color{royalblue}{\frac{(\sqrt{2})}{(\sqrt{2})}}} &=& \large{\frac{-1 + i}{\sqrt{2}}} \end{matrix} \begin{matrix} Letra \ (A) \end{matrix} $\color{orangered}{Obs:}$ \begin{matrix} \color{gray}{|z| = \sqrt{(\frac{\sqrt{2}}{2})^2 +(\frac{\sqrt{2}}{2})^2 } = 1} \end{matrix}
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