Question

3.Considere z= (2 + 2i )y w=( 3−4i).Resolver z^10 · w^35

Answer 1

$z=2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i\right)\\ \\ \\ z=2\sqrt{2}\text{ cis }\dfrac{\pi}{4}\\ \\ \\ z^{10}=\left(2\sqrt{2}\right)^{10}\text{ cis }\dfrac{10\pi}{4}\\ \\ \\ z^{10}=\left(2\sqrt{2}\right)^{10}\text{ cis }\dfrac{5\pi}{2}\\ \\ \\ z^{10}=\left(2\sqrt{2}\right)^{10}\text{ cis }\dfrac{\pi}{2}\\ \\ \\ z^{10}=\left(2\sqrt{2}\right)^{10}i$

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$w=5\left(\dfrac{3}{5}-\dfrac{4}{5}i\right)\\ \\ \\ w=5\text{ cis }\theta~,~\theta \approx 2\pi -\dfrac{53\pi}{180}\\ \\ \\ w^{35}=5^{35}\text{ cis }35\theta$

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$z^{10}\cdot w^{35}=\left(2\sqrt{2}\right)^{10}\cdot 5^{35}~i\text{ cis }35\theta\\ \\ \\ z^{10}\cdot w^{35}=2^{15}\cdot 5^{35}~i\text{ cis }35\theta\\ \\ \\ z^{10}\cdot w^{35}\approx2^{15}\cdot 5^{35}~i(0.57-0.81i)\\ \\ \\ z^{10}\cdot w^{35}\approx2^{15}\cdot 5^{35}~(0.57i+0.81)\\ \\ \\ \texttt{Valor exacto}\\ \\ z^{10}\cdot w^{35}=2^{15}\cdot 5^{35}~\left(\frac{1474785254199855492093963}{2910383045673370361328125}i+\frac{2509051200461575078477484}{2910383045673370361328125}\right)$