Find the cube roots of 27(cos 330° + i sin 330°)

Let [tex]z=r(cos\theta+isin\theta)[/tex] be a complex number in polar form, where [tex]n[/tex] is an integer and [tex]n\geq1[/tex]. If [tex]z^n=r^n(cos\theta+isin\theta)^n[/tex], then [tex]z^n=r^n(cos(n\theta)+isin(n\theta))[/tex].

Nth Root of a Complex Number

If [tex]n[/tex] is any positive integer, the nth roots of [tex]z=rcis\theta[/tex] are given by [tex]\sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}}[/tex] where the nth roots are found with the formulas:

[tex]\sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr][/tex] for degrees (the one applicable to this problem)
[tex]\sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr][/tex] for radians

for [tex]k=0,1,2,...\:,n-1[/tex]

Calculation

[tex]z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)[/tex]

First cube root where k=2

[tex]\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr][/tex]

Second cube root where k=1

[tex]\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr][/tex]

Third cube root where k=0

[tex]\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr][/tex]