Answer
(a) The three cube roots are:
$3~(cos~0^{\circ}+i~sin~0^{\circ})$
$3~(cos~120^{\circ}+i~sin~120^{\circ})$
$3~(cos~240^{\circ}+i~sin~240^{\circ})$
(b) We can see the three vectors in the complex plane:
Work Step by Step
(a)
$z = 27$
$z = 27~(1+0~i)$
$z = 27~(cos~0^{\circ}+i~sin~0^{\circ})$
$r = 27$ and $\theta = 0^{\circ}$
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:
$z^{1/3} = 27^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = 3~(cos~0^{\circ}+i~sin~0^{\circ})$
When k = 1:
$z^{1/3} = 27^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = 3~(cos~120^{\circ}+i~sin~120^{\circ})$
When k = 2:
$z^{1/3} = 27^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = 3~(cos~240^{\circ}+i~sin~240^{\circ})$
(b) We can see the three vectors in the complex plane:
