Is this an equivalence relation (reflexivity, symmetry, transitivity)

Question

Let $\theta(s):\mathbb{C}\to \mathbb{R}$ be a well defined function. I define the following relation in $\mathbb{C}$.

$\forall s,q \in \mathbb{C}: s\mathbin{R}q\iff\theta(s)\ne 0 \pmod {2\pi}$ (and)

$\theta(q)\ne 0 \pmod {2\pi}$

The function $\pmod {2\pi}$ is the addition $\pmod {2\pi}$

My question: Is this an equivalence relation (reflexivity, symmetry, transitivity)?

The formula of $\theta(s)$ is not important for this question.

Answer 1

Your relation is $$sRq\iff \theta(s)\not \equiv 0\text{ and }\theta(q)\not \equiv 0 \mod 2\pi$$ for $s,q\in \mathbb{C}$.

For symmetry: $$sRq\iff \theta(s)\not \equiv 0\text{ and }\theta(q)\not \equiv 0 \mod 2\pi \iff qRs$$

For transitivity: $$sRq\text{ and }qRp\iff \theta(s)\not \equiv 0\text{ and }\theta(q)\not \equiv 0\text{ and }\theta(p)\not \equiv 0 \mod 2\pi\implies sRp$$ Reflexitivity is: $$sRs\iff \theta(s)\not \equiv 0\mod 2\pi$$ That clearly depends on your choice of $\theta$. Therefore, $R$ is an equivalence relation iff $$\theta(\mathbb{C})\cap \left\{2k\pi:k\in \mathbb{Z}\right\}=\emptyset$$

Answer 2

It is not an equivalence relation if there exists $x$ such that $\theta(x) = 0 \mod 2\pi $.

You need reflexivity, so if $x$ satisfies $\theta(x) = 0 \mod 2\pi $, then you do not have $x R x$, hence it is not an equivalence relation.