Question

Hallar las raices cubicas del siguente numero complejo -1-i√3 y el producto de las raices.

Answer 1

$c=-1-i\sqrt3\\ \\ |c|=\sqrt{(-1)^2+(-\sqrt{3})^2}=2\\ \\ \vec{u}_c=-\dfrac{1}{2}-\dfrac{\sqrt3}{2}i \sim \left(-\dfrac{1}{2},-\dfrac{\sqrt3}{2}\right)\in IIIC\to \theta =\dfrac{4\pi}{3}\\ \\ \\ \text{Entonces:}\\ \\ c=2\left(\cos \dfrac{4\pi}{3}+i\sin\dfrac{4\pi}{3}\right)$

$w_0=c^{1/3}=\sqrt[3]{2}\left(\cos \dfrac{4\pi}{9}+i\sin\dfrac{4\pi}{9}\right)\\ \\ \\ w_1=\sqrt[3]{2}\left[\cos (\frac{4\pi}{9}+\frac{2\pi}{3})+i\sin(\frac{4\pi}{9}+\frac{2\pi}{3})\right]\\ \\ \\ w_2=\sqrt[3]{2}\left[\cos (\frac{4\pi}{9}+\frac{4\pi}{3})+i\sin(\frac{4\pi}{9}+\frac{4\pi}{3})\right]$


==================================


                                 $w_0=\sqrt[3]{2}\left(\cos \dfrac{4\pi}{9}+i\sin\dfrac{4\pi}{9}\right)\\ \\ \\ w_1=-\sqrt[3]{2}\left(\cos \dfrac{\pi}{9}+i\sin\dfrac{\pi}{9}\right)\\ \\ \\ w_2=\sqrt[3]{2}\left(\cos \dfrac{2\pi}{9}-i\sin\dfrac{2\pi}{9}\right)$