Question

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

Answer 1

27(cos 279° + i sin 279°)
because it is a cube root we divide 360°/3 = 120°
so we add 120° to every root
and we need to find the third root
z1 = 27(cos 279° + i sin 279°)
z2 = 27(cos 39° + i sin 39°)
z3 = 27(cos 159° + i sin 159°)
(27 * cos 159°) + (27 * i sin 159°)
-25.20667152 + 9.675934638i

Answer 2

\[Let z=r \left( \cos \theta+\iota \sin \theta \right)=re ^{\iota \theta}=re ^{\iota \left( 2n \pi+\theta \right)} \]
\[z ^{\frac{ 1 }{ 3 }}=r ^{\frac{ 1 }{3 }}e ^{\frac{ \iota \left( 2n \pi+\theta \right) }{ 3 }} \]
\[z ^{\frac{ 1 }{3 }}=r ^{\frac{ 1 }{ 3 }}\left\{ \cos \frac{ 2n \pi+\theta }{ 3 } +\iota \sin \frac{ 2n \pi+\theta }{ 3 }\right\}\]
plug r=27,n=0,1,2,
\[\theta=279 \]
solve and get three solutions.