Question

Determine as raízes cúbicas do número complexo z = -8.

Answer 1

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Raízes enésimas de um número complexo

$\large\underbrace{\sf W_k=\sqrt[\sf n]{\sf \rho}\bigg[ cos\bigg(\dfrac{\theta+2k\pi}{n}\bigg)+i~sen\bigg(\dfrac{\theta+2k\pi}{n}\bigg)\bigg]}_{\sf 2^{\underline a}~ f\acute ormula~de~Moivre}$

$\sf z=-8\implies z=-8+0i\\\sf \rho=\sqrt{(-8)^2+0^2}=\sqrt{64}=8\\\sf cos(\theta)=\dfrac{-8}{8}\\\sf cos(\theta)=-1\implies \theta=arc~cos(-1)=\pi\\\sf z=\rho[cos(\theta)+i sen(\theta)]\\\sf z=8[cos(\pi)+i~sen(\pi)] \\\sf w_k=\sqrt[\sf 3]{\sf8}\bigg[cos\bigg(\dfrac{\pi+2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2k\pi}{3}\bigg)\bigg]\\\sf w_k=2\cdot\bigg[cos\bigg(\dfrac{\pi+2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2k\pi}{3}\bigg)\bigg]$

$\underline{\tt se~k=1:}\\\sf w_1= 2\cdot\bigg[cos\bigg(\dfrac{\pi+2\cdot 1\cdot\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2\cdot1\cdot \pi}{3}\bigg)\bigg]\\\sf w_1=2\cdot\bigg[cos\bigg(\dfrac{\diagup\!\!\!3\pi}{\diagup\!\!\!3}\bigg)+i~sen\bigg(\dfrac{\diagup\!\!\!3\pi}{\diagup\!\!\!3}\bigg)\bigg]\\\sf w_1=2[cos(\pi)+i~sen(\pi)]\\\sf w_1=2\cdot[-1+0]=-2$

$\underline{\tt se~k=2:}\\\sf w_2=2\bigg[cos\bigg(\dfrac{\pi+2\cdot2\cdot\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi+2\cdot2\cdot\pi}{3}\bigg)\bigg]\\\sf w_2=2cos\bigg[cos\bigg(\dfrac{5\pi}{3}\bigg)+i~sen\bigg(\dfrac{5\pi}{3}\bigg)\bigg]\\\sf w_2=2\bigg[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\bigg]\\\sf w_2=1-\sqrt{3}~i$

$\boxed{\begin{array}{c}\sf portanto~as~ra\acute izes~c\acute ubicas~de~z=-8~s\tilde ao\\\huge\boxed{\boxed{\boxed{\boxed{\sf-2}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf 1-\sqrt{3}~i}}}}\end{array}}$