Consider the action of $\mathbb{Z}_4$ on the complex plane by rotations. Which of these pairs of complex numbers are in the same orbit?
A: $1$ and $e^{i\pi/3}$
B: $1$ and $e^{-i\pi/4}$
C: $e^{-i\pi/2}$ and $e^{i\pi/2}$
D: $e^{i\pi/2}$ and $e^{-3i\pi/4}$
So I know for them to be in the same orbit they must satisfy the equation:
$$
e^{2ik\pi/n} \times z
$$
where $n$ is $4$ in my case and z is each complex number in such a way so that the results are connected. But I'm not sure what to do exactly...
I'll say $z \sim w$ if they lie in the same orbit under this action. Per your definition, $z \sim w$ iff there is some $k$ for which $z e^{2 \pi i k/4} = w$. Let $S = \{e^{2 \pi i k/4} : k \in \mathbb Z\}$. $S$ only has 4 elements, as it only depends on $k$ modulo 4. Indeed, $S = \{1, e^{\pi i /2}, e^{\pi i}, e^{\pi i 3/2}\}=\{1, i, -1, -i\}$.
Now, we have equivalently that $z \sim w$ iff $z/w \in S$. This gives us a way to approach the problem: divide the two values and see if they are any of the 4 elements in $S$.
a) $e^{i \pi / 3}$ is not in $S$ so these are not in the same orbit.
b) $e^{-i \pi / 4}$ is not in $S$ so these are not in the same orbit.
c) $e^{i \pi / 2}/e^{-i \pi / 2} = e^{i \pi} \in S$ so these are in the same orbit.
d) $e^{i \pi / 2}/e^{-3 i \pi / 4} = e^{i \pi/4} \notin S$ so these are not in the same orbit.