We must find the complex cubic roots of the complex number:
[tex]z=64\cdot(\cos(219\degree)+i\sin(219\degree)).[/tex]We identify the modulus r and angle θ as:
• modulus r = 64,
,• angle θ = 219°.
The k = 0, 1, ..., n - 1 roots of a complex number 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})][/tex]In this case three roots, we have k = 0, 1, 2. Using the data and formula from above, we have:
[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})]=4\cdot\lbrack\cos(73\degree)+i\cdot\sin(73\degree)], \\ 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})]=4\cdot\lbrack\cos(193\degree)+i\cdot\sin(193\degree)], \\ 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})]=4\cdot\lbrack\cos(313\degree)+i\cdot\sin(313\degree)]. \end{gathered}[/tex]AnswerThere are three cubic roots:
[tex]\begin{gathered} w_0=4\cdot\lbrack\cos(73\degree)+i\cdot\sin(73\degree)] \\ w_1=4\cdot\lbrack\cos(193\degree)+i\cdot\sin(193\degree)] \\ w_2=4\cdot\lbrack\cos(313\degree)+i\cdot\sin(313\degree)] \end{gathered}[/tex]