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 $z= {\dfrac{\sqrt{2}}{1+i}}$ , temos: \begin{matrix} z &=& {\dfrac{\sqrt{2}}{1+i} \cdot \color{royalblue}{\dfrac{(1-i)}{(1-i)}}} &=& {\dfrac{\sqrt{2}}{2} + \dfrac{i\sqrt{2}}{2}} &=& { \left( \dfrac{\sqrt{2}}{2} \ , \ \dfrac{\sqrt{2}}{2}\right)} \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} &=& {\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}} &=& 1 &\Rightarrow& \fbox{$\varphi = { \dfrac{\pi}{4} }$} \end{matrix}Voltando a nossa expressão, temos:\begin{matrix} z^{93} &=& |z|^{93} \cdot \left[\cos{\left(24\pi - \dfrac{3\pi}{4}\right)} + i\sin{\left(24\pi - \dfrac{3\pi}{4}\right)} \right] \\ \\ \end{matrix}Portanto, \begin{matrix} z^{93} &=& {\dfrac{-\sqrt{2} + i\sqrt{2}}{2} \cdot \color{royalblue}{\dfrac{(\sqrt{2})}{(\sqrt{2})}}} &=& {\dfrac{-1 + i}{\sqrt{2}}} \end{matrix} \begin{matrix} Letra \ (A) \end{matrix}$\color{orangered}{Obs:}$ \begin{matrix} \color{}{|z| = \sqrt{ \left(\dfrac{\sqrt{2}}{2}\right)^2 + \left(\dfrac{\sqrt{2}}{2}\right)^2 } = 1} \end{matrix}
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