Answer
$3, -\frac{3}{2}+\frac{3\sqrt 3}{2}i,-\frac{3}{2}-\frac{3\sqrt 3}{2}i$
Work Step by Step
1. Use the given conditions, we have:
$z=27=27+0i=27(1+0i)=27(cos0+i\ sin0)$
2. Thus $z^{1/3}=[27(cos0+i\ sin0)]^{1/3}=3(cos\frac{2k\pi}{3}+i\ sin\frac{2k\pi}{3})$
3. For $k=0, z_0=3(cos0+i\ sin0)=3$
4. For $k=1, z_1=3(cos\frac{2\pi}{3}+i\ sin\frac{2\pi}{3})=3(-\frac{1}{2}+\frac{\sqrt 3}{2}i)=-\frac{3}{2}+\frac{3\sqrt 3}{2}i$
5. For $k=2, z_2=3(cos\frac{4\pi}{3}+i\ sin\frac{4\pi}{3})=3(-\frac{1}{2}-\frac{\sqrt 3}{2}i)=-\frac{3}{2}-\frac{3\sqrt 3}{2}i$
