Question

qual a raiz cubica de z= -8i

Answer 1

Olá Jucielly!

A radiciação de complexos pode ser dada por:

$Z_k=\sqrt[n]{\rho}\left[\cos(\frac{\theta}{n}+k\cdot\frac{2\pi}{n})+i\cdot\sin(\frac{\theta}{n}+k\cdot\frac{2\pi}{n})\right]$   (1)

Para o problema proposto temos que:

$n=3$

e

$\rho=\sqrt{0^2+(-8)^2}\to\rho=8$

e

$\left \{ {{\cos(\theta)=\frac{0}{8}=0} \atop {sin(\theta)=-\frac{8}{8}=-1}\right\to\theta=\frac{3\pi}{2}$

Assim, de (1), vem:

$Z_k=\sqrt[3]{8}\left[\cos(\frac{\frac{3\pi}{2}}{3}+k\cdot\frac{2\pi}{3})+i\cdot\sin(\frac{\frac{3\pi}{2}}{3}+k\cdot\frac{2\pi}{3})\right]$

Portanto:

$Z_0=2\left[\cos(\frac{3\pi}{6}+0\cdot\frac{2\pi}{3})+i\cdot\sin(\frac{3\pi}{6}+0\cdot\frac{2\pi}{3})\right]\\ \\ Z_0=2i$


$Z_1=2\left[\cos(\frac{3\pi}{6}+1\cdot\frac{2\pi}{3})+i\cdot\sin(\frac{3\pi}{6}+1\cdot\frac{2\pi}{3})\right]\\ \\ Z_1=-\frac{\sqrt{3}}{2}-i\cdot\frac{1}{2}$


$Z_2=2\left[\cos(\frac{3\pi}{6}+2\cdot\frac{2\pi}{3})+i\cdot\sin(\frac{3\pi}{6}+2\cdot\frac{2\pi}{3})\right]\\ \\ Z_2=\frac{\sqrt{3}}{2}-i\cdot\frac{1}{2}$

Abraço,

Douglas Joziel.