Question

Find the cube roots of 27(cos 279° + i sin 279°).

Answer 1

`\bf \textit{ roots of complex numbers, DeMoivre's theorem} \\\\ \sqrt[n]{z}=\sqrt[n]{r}\left[ cos\left( \cfrac{\theta+2\pi k}{n} \right) +i\ sin\left( \cfrac{\theta+2\pi k}{n} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3,... \end{array}\\\\ -------------------------------\\\\ 27[cos(279^o)+i~sin(279^o)]`


`\bf \stackrel{\textit{first root, k = 0}}{\sqrt[3]{27}\left[cos\left( (279+360(0))/(3) \right)+i~ sin\left( (279+360(0))/(3) \right) \right]} \\\\\\ 3\left[cos\left( (279)/(3) \right)+i~ sin\left( (279)/(3) \right) \right]\implies 3[cos(93^o)+i~sin(93^o)] \\\\\\ \boxed{-0.15700787+29958886i}`


`\bf \stackrel{\textit{second root, k = 1}}{\sqrt[3]{27}\left[cos\left( (279+360(1))/(3) \right)+i~ sin\left( (279+360(1))/(3) \right) \right]} \\\\\\ 3\left[cos\left( (639)/(3) \right)+i~ sin\left( (639)/(3) \right) \right]\implies 3[cos(213^o)+i~sin(213^o)] \\\\\\ \boxed{-2.5160117-1.633917i}`


`\bf \stackrel{\textit{third root, k = 2}}{\sqrt[3]{27}\left[cos\left( (279+360(2))/(3) \right)+i~ sin\left( (279+360(2))/(3) \right) \right]} \\\\\\ 3\left[cos\left( (999)/(3) \right)+i~ sin\left( (999)/(3) \right) \right]\implies 3[cos(333^o)+i~sin(333^o)] \\\\\\ \boxed{2.67301957-1.3619715i}`