Answer
(a) The three cube roots are:
$cos~30^{\circ}+i~sin~30^{\circ}$
$cos~150^{\circ}+i~sin~150^{\circ}$
$cos~270^{\circ}+i~sin~270^{\circ}$
(b) We can see the three vectors in the complex plane:
Work Step by Step
$z = cos~90^{\circ}+i~sin~90^{\circ}$
$r = 1$ and $\theta = 90^{\circ}$
We can use this equation to find the cube roots:
$z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
When k = 0:
$z^{1/3} = 1^{1/3}~[cos(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = 1~[cos~30^{\circ}+i~sin~30^{\circ}]$
$z^{1/3} = cos~30^{\circ}+i~sin~30^{\circ}$
When k = 1:
$z^{1/3} = 1^{1/3}~[cos(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = 1~[cos~150^{\circ}+i~sin~150^{\circ}]$
$z^{1/3} = cos~150^{\circ}+i~sin~150^{\circ}$
When k = 2:
$z^{1/3} = 1^{1/3}~[cos(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = 1~[cos~270^{\circ}+i~sin~270^{\circ}]$
$z^{1/3} = cos~270^{\circ}+i~sin~270^{\circ}$
(b) We can see the three vectors in the complex plane: