I am having trouble finding the cubed roots of $8$ as a complex number.
$$z^3 = 8+0i$$ $$z^3 = r^3 e^{3\theta i}=8e^{2i\pi k},\quad k\in \Bbb Z$$
$$\implies r=2,3\theta = 2\pi k\implies \theta = \frac{2\pi k}{3}$$
$$z=2e^{\frac{2i\pi k}{3}}$$
I would think I have then $$z=2,2e^{\frac{2i\pi}{3}},2e^{\frac{4i\pi}{3}}$$
But the answer says: $z=2,-1+\sqrt{3}i, -1-\sqrt{3}i$
What's going on?
$\endgroup$ 12 Answers
$\begingroup$$e^{i\theta}=\cos\theta+i\sin\theta$
$\endgroup$ 3 $\begingroup$$z^3-2^3=(z-2)(z^2+2z+4)=0$
So first term has solution $z=2$.
Now-
$z^2+2z+4=0$
$\frac{-2 \pm \sqrt{4-16}}{2}=z$
So, $z=-\sqrt{3}i-1 \ and\ z=\ \sqrt{3}i-1$
