Question

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

Answer 1

See below for all the cube roots

Explanation:

DeMoivre's Theorem

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

Nth Root of a Complex Number

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

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

for
`k=0,1,2,...\:,n-1`

Calculation

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

First cube root where k=2

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

Second cube root where k=1

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

Third cube root where k=0

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