Respostas

2014-02-03T13:10:09-02:00
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.