Given the complex number:
[tex]z=r\cdot(\cos \theta+i\cdot\sin \theta)\text{.}[/tex]
The n n-th roots of the complex number z are given by:
[tex]w_k=\sqrt[n]{r}\cdot\lbrack\cos (\frac{\theta+k\cdot360\degree}{n})+i\cdot\sin (\frac{\theta+k\cdot360\degree}{n})\rbrack\text{ where }k=0,1,2,\ldots,n-1.[/tex]
We must find all the cube roots of the following complex number:
[tex]z=64\cdot(\cos (219\degree)+i\cdot\sin (219\degree)).[/tex]
For this number, we identify:
[tex]\begin{gathered} r=64, \\ \theta=219\degree. \end{gathered}[/tex]
Using the formula above with these numbers, we get:
[tex]\begin{gathered} w_0=\sqrt[3]{64}\cdot\lbrack\cos (\frac{219\degree+0\cdot360\degree}{3})+i\cdot\sin (\frac{219\degree+0\cdot360\degree}{3})\rbrack=4\cdot\lbrack\cos (73\degree)+i\cdot\sin (73\degree)\rbrack, \\ w_1=\sqrt[3]{64}\cdot\lbrack\cos (\frac{219\degree+1\cdot360\degree}{3})+i\cdot\sin (\frac{219\degree+1\cdot360\degree}{3})\rbrack=4\cdot\lbrack\cos (193\degree)+i\cdot\sin (193\degree)\rbrack, \\ w_2=\sqrt[3]{64}\cdot\lbrack\cos (\frac{219\degree+2\cdot360\degree}{3})+i\cdot\sin (\frac{219\degree+2\cdot360\degree}{3})\rbrack=4\cdot\lbrack\cos (313\degree)+i\cdot\sin (313\degree)\rbrack\text{.} \end{gathered}[/tex]
Answer
[tex]\begin{gathered} w_0=4\cdot\lbrack\cos (73\degree)+i\cdot\sin (73\degree)\rbrack \\ w_1=4\cdot\lbrack\cos (193\degree)+i\cdot\sin (193\degree)\rbrack \\ w_2=4\cdot\lbrack\cos (313\degree)+i\cdot\sin (313\degree)\rbrack \end{gathered}[/tex]
