Question

Quais as raízes cúbicas de z = -128?

Answer 1

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$\sf z=-128\implies z=-128+0i\\\sf \rho=\sqrt{(-128)^2}=128\\\sf cos(\theta)=\dfrac{-128}{128}=-1\\\sf \theta=arc~cos(-1)=\pi\\\sf W_k=\sqrt[\sf3]{\sf128}\bigg[cos\bigg(\dfrac{\pi+2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2k\pi}{3}\bigg)\bigg]\\\sf W_k=4\sqrt[\sf3]{\sf 2}\bigg[cos\bigg(\dfrac{\pi+2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2k\pi}{3}\bigg)\bigg]$

$\tt fazendo ~k=0:\\\sf w_0=4\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{3}\bigg)\bigg]\\\sf W_0=4\sqrt[\sf3]{\sf2}\bigg[\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\bigg]=2\sqrt[\sf3]{\sf2}+2\sqrt[\sf3]{\sf2}\cdot\sqrt{3}i\checkmark$

$\tt fazendo~k=1:\\\sf W_1=4\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi+2\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2\pi}{3}\bigg)\bigg]\\\sf W_1=4\sqrt[\sf3]{\sf 2}[cos(\pi)+i~sen(\pi)]=-4\sqrt[\sf3]{\sf2}\checkmark$

$\tt fazendo~ k=2:\\\sf W_2=4\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{\pi+4\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+4\pi}{3}\bigg)\bigg]\\\sf 4\sqrt[\sf3]{\sf2}\bigg[cos\bigg(\dfrac{5\pi}{3}\bigg) +i~sen\bigg(\dfrac{5\pi}{3}\bigg)\bigg]\\\sf W_2=4\sqrt[\sf3]{\sf2}\bigg[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}~\bigg]=2\sqrt[\sf3]{\sf2}-2\sqrt{3}\cdot\sqrt[\sf3]{\sf2}i\checkmark$

$\boxed{\begin{array}{c}\underline{\rm portanto~as~ra\acute izes~c\acute ubicas~de~ z=-128~s\tilde ao:}\\\huge\boxed{\boxed{\boxed{\boxed{\sf2\sqrt[\sf3]{\sf2}+2\sqrt[\sf3]{\sf2}\cdot\sqrt{3}i}}}}\checkmark\\\huge\boxed{\boxed{\boxed{\boxed{\sf-4\sqrt[\sf3]{\sf2}}}}}\checkmark\\\huge\boxed{\boxed{\boxed{\boxed{\sf 2\sqrt[\sf3]{\sf2}-2\sqrt{3}\cdot\sqrt[\sf3]{\sf2}i}}}}\checkmark\end{array}}$