Consider the action of $\mathbb{Z}_3$ on the complex plane by rotations. Which of these pairs of complex numbers are in the same orbit?
A: $1$ and $e^{\frac{i\pi}{2}}$
B: $1$ and $e^{\frac{i\pi}{3}}$
C: $e^{\frac{i\pi}{3}}$ and $e^{\frac{-2i\pi}{3}}$
D: $e^{\frac{i\pi}{3}}$ and $e^{\frac{-i\pi}{3}}$
What I know so far:
For $\bar{k}\in\mathbb{Z}_3$ and $z\in\mathbb{C}$ $$\bar{k} \cdot z= e^{\frac{2ik\pi}{n}} \times z $$
Hence for $\bar{0}$ we get $1$ which rules out C and D. For $\bar{1}$ we get $e^{\frac{2i\pi}{3}}$ but that is not an option.
Any help is greatly appreciated!
Hint/solution: Each orbit has three elements, except for the orbit of the origin, which is a singleton. Two complex numbers $z$ and $w$ are in the same orbit if $|z|=|w|$ and the angle between them is $2\pi/3$. This immediately rules out everything except for option D, whose elements form an angle of $(\pi/3)-(-\pi/3)=2\pi/3$.