Question

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

Answer 1

cube root for the magnitude, divide by 3 for the angle...

Answer 2

\(\large \bf {
\textit{De Moivre's theorem}\\
Z^{\color{blue}{ n}}=[r[cos(\theta)+i\ sin(\theta)]]^{\color{blue}{ n}}\iff r^{\color{blue}{ n}}[cos({\color{blue}{ n}}\theta)+i\ sin({\color{blue}{ n}}\theta)]
\\ \quad \\ \quad \\
thus\qquad \sqrt[{\color{blue}{ 3}}]{Z}\implies Z^{{\color{blue}{ \frac{1}{3}}}}\implies r^{\color{blue}{ \frac{1}{3}}}[cos({\color{blue}{ \frac{1}{3}}}\theta)+i\ sin({\color{blue}{ \frac{1}{3}}}\theta)]
}\)

Answer 3

to get the other 2 roots, add multiples of 2pi to angle before dividing by 3

\[\cos (\frac{279 + 360k}{3}) + i \sin (\frac{279+360k}{3})\]
k = 0,1,2