qual a raiz cubica de z= -8i
Soluções para a tarefa
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Olá Jucielly!
A radiciação de complexos pode ser dada por:
(1)
Para o problema proposto temos que:
![n=3 n=3](https://tex.z-dn.net/?f=n%3D3)
e
![\rho=\sqrt{0^2+(-8)^2}\to\rho=8 \rho=\sqrt{0^2+(-8)^2}\to\rho=8](https://tex.z-dn.net/?f=%5Crho%3D%5Csqrt%7B0%5E2%2B%28-8%29%5E2%7D%5Cto%5Crho%3D8)
e
![\left \{ {{\cos(\theta)=\frac{0}{8}=0} \atop {sin(\theta)=-\frac{8}{8}=-1}\right\to\theta=\frac{3\pi}{2} \left \{ {{\cos(\theta)=\frac{0}{8}=0} \atop {sin(\theta)=-\frac{8}{8}=-1}\right\to\theta=\frac{3\pi}{2}](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B%5Ccos%28%5Ctheta%29%3D%5Cfrac%7B0%7D%7B8%7D%3D0%7D+%5Catop+%7Bsin%28%5Ctheta%29%3D-%5Cfrac%7B8%7D%7B8%7D%3D-1%7D%5Cright%5Cto%5Ctheta%3D%5Cfrac%7B3%5Cpi%7D%7B2%7D)
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] 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]](https://tex.z-dn.net/?f=Z_k%3D%5Csqrt%5B3%5D%7B8%7D%5Cleft%5B%5Ccos%28%5Cfrac%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7B3%7D%2Bk%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%2Bi%5Ccdot%5Csin%28%5Cfrac%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7B3%7D%2Bk%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%5Cright%5D)
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_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](https://tex.z-dn.net/?f=Z_0%3D2%5Cleft%5B%5Ccos%28%5Cfrac%7B3%5Cpi%7D%7B6%7D%2B0%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%2Bi%5Ccdot%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B6%7D%2B0%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%5Cright%5D%5C%5C%0A%5C%5C%0AZ_0%3D2i)
![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_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}](https://tex.z-dn.net/?f=Z_1%3D2%5Cleft%5B%5Ccos%28%5Cfrac%7B3%5Cpi%7D%7B6%7D%2B1%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%2Bi%5Ccdot%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B6%7D%2B1%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%5Cright%5D%5C%5C%0A%5C%5C%0AZ_1%3D-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D-i%5Ccdot%5Cfrac%7B1%7D%7B2%7D)
![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} 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}](https://tex.z-dn.net/?f=Z_2%3D2%5Cleft%5B%5Ccos%28%5Cfrac%7B3%5Cpi%7D%7B6%7D%2B2%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%2Bi%5Ccdot%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B6%7D%2B2%5Ccdot%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%5Cright%5D%5C%5C%0A%5C%5C%0AZ_2%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D-i%5Ccdot%5Cfrac%7B1%7D%7B2%7D)
Abraço,
Douglas Joziel.
A radiciação de complexos pode ser dada por:
Para o problema proposto temos que:
e
e
Assim, de (1), vem:
Portanto:
Abraço,
Douglas Joziel.
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