Don't know how to find all the roots So i got this problem :
 Find all the roots of  $r^{3}=(-1)$
 i can only think to use :
 $\sqrt[n]{z} =\sqrt[n]{r}\left[\cos \left(\dfrac{\theta + 2\pi{k}}{n}\right) + i \sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right] $
 i know that $\theta = \tan^{-1}(\dfrac{b}{a}) $as in from $ z=a + bi$
 $r=\sqrt[]{a^2 + b^2}$ , $k=0,1,2,3,4,.. n-1$
 so anyone could explain this to me ? 
lets say that $r^{3}=(-1) == z^{3}=(-1)$ a since i dont like it and it messes me up using the formula
 so far :
$ z^{3} = (-1) $ so $z = \sqrt[3]{(-1 + 0 \cdot i)}$
$ r= \sqrt[]{(-1)^{2} + 0^{2}}=1$
$\tan^{-1}(\dfrac{b}{a})=\tan^{-1}(\dfrac{0}{-1})=\tan^{-1}(0)=0$ <- not sure about his part
k=0;
$ z^{3}=\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 0}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 0}{3})$
$ =\cos(\dfrac{\pi}{3}) +i\cdot \sin(\dfrac{\pi}{3})= \dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2}$

k=1;
$ \sqrt[3]{1}(\cos(\dfrac{\pi+2\pi \cdot 1}{3})+i \cdot \sin(\dfrac{\pi+2\pi \cdot 1}{3})$
$=\cos(\pi) + i \cdot \sin(\pi)=(-1) + 0 = -1$

k=2;
$\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 2}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 2}{3})$
$=\cos(\dfrac{5\pi}{3})+i \cdot \sin(\dfrac{5\pi}{3})=-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2}$
 so is this correctet or im missing something here?
Roots $(\dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2};-1;-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2})$

Hints:
Using $\,cis(\theta):=\cos\theta+i\sin\theta=cis(\theta+2k\pi)\;,\;\;k\in\Bbb Z\;$ , we have:
$$z=r\,cis(\theta)\implies -1=cis(\pi+2k\pi)=z^3=r^3\,cis(3\theta)\;\;\text{(de Moivre's Theorem)}\implies$$
$$z_k:=r\,cis\left(\frac\pi3(1+2k)\right)\;,\;\;k=0,1,2\;\;\text{(why is it enough to take these values of}\;\;k)?$$
Now calculate the roots, taking into account that $\,r=1\;$ (why?)
