Question

Determine a raíz cúbica do complexo Z=√2+√2i

Answer 1

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Raízes enésimas de um número complexo

$\small\boxed{\boxed{\boxed{\boxed{\sf w_k=\sqrt[\sf n]{\sf\rho}\bigg[cos\bigg(\dfrac{\theta+2k\pi}{n}\bigg)+i~sen\bigg(\dfrac{\theta+2k\pi}{n}\bigg)\bigg]}}}}$

$\sf z=\sqrt{2}+\sqrt{2}i\\\sf\rho=\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=\sqrt{2+2}=\sqrt{4}=2\\\sf tg(\theta)=\dfrac{\sqrt{2}}{\sqrt{2}}\\\sf tg(\theta)=1\implies \theta=arctg(1)=\dfrac{\pi}{4}\\\sf w_k=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\frac{\pi}{4}+2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\frac{\pi}{4}+2k\pi}{3}\bigg)\bigg]\\\sf w_k=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi}{12}+\dfrac{2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{12}+\dfrac{2k\pi}{3}\bigg)\bigg]$

$\sf se~k=0:\\\sf w_0=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi}{12}\bigg)+i~sen\bigg(\dfrac{\pi}{12}\bigg)\bigg]$

$\sf se~k=1:\\\sf w_1=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi}{12}+\dfrac{2\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{12}+\dfrac{2\pi}{3}\bigg)\bigg]\\\sf w_1=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{9\pi}{12}\bigg)+i~sen\bigg(\dfrac{9\pi}{12}\bigg)\bigg]\\\sf w_1=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{3\pi}{4}\bigg)+i~sen\bigg(\dfrac{3\pi}{4}\bigg)\bigg]$

$\sf se~k=2\\\sf w_3=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi}{12}+\dfrac{4\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{12}+\dfrac{4\pi}{3}\bigg)\bigg]\\\sf w_3=\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{17\pi}{12}\bigg)+i~sen\bigg(\dfrac{17\pi}{12}\bigg)\bigg]$

$\Large\boxed{\begin{array}{l}\sf portanto~as~ra\acute izes~c\acute ubicas~de~z=\sqrt{2}+\sqrt{2}i\\\sf s\tilde ao~\huge\boxed{\boxed{\boxed{\boxed{\sf \sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi}{12}\bigg)+i~sen\bigg(\dfrac{\pi}{12}\bigg)\bigg]}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{3\pi}{4}\bigg)+i~sen\bigg(\dfrac{3\pi}{4}\bigg)\bigg]}}}}\\\rm e\end{array}}$

$\Large\boxed{\begin{array}{l}\huge\boxed{\boxed{\boxed{\boxed{\sf\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{17\pi}{12}\bigg)+i~sen\bigg(\dfrac{17\pi}{12}\bigg)\bigg]}}}}\end{array}}$