Answer
(a) The three cube roots are:
$2^{1/3}~(cos~20^{\circ}+i~sin~20^{\circ})$
$2^{1/3}~(cos~140^{\circ}+i~sin~140^{\circ})$
$2^{1/3}~(cos~260^{\circ}+i~sin~260^{\circ})$
(b) We can see the three vectors in the complex plane:
Work Step by Step
(a)
$z = 1+i~\sqrt{3}$
$z = 2~(cos~60^{\circ}+i~sin~60^{\circ})$
$r = 2$ and $\theta = 60^{\circ}$
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:
$z^{1/3} = 2^{1/3}~[cos(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = 2^{1/3}~(cos~20^{\circ}+i~sin~20^{\circ})$
When k = 1:
$z^{1/3} = 2^{1/3}~[cos(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = 2^{1/3}~(cos~140^{\circ}+i~sin~140^{\circ})$
When k = 2:
$z^{1/3} = 2^{1/3}~[cos(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = 2^{1/3}~(cos~260^{\circ}+i~sin~260^{\circ})$
(b) We can see the three vectors in the complex plane:
