Equivalence Classes mod 2$\pi$, and a bijection to complex unit circle

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I'm trying to solve the following question, I've made some progress but I'm really struggling to understand the underlying conceptual idea.

Question: Define a relation $\tilde{}$ on $\mathbb{R}$ by $x \tilde{} y$ if and only if there exists $k \in \mathbb{Z}$ such that $x - y = 2k\pi$

a) Show that $\tilde{}$ is an equivalence relation.

b) Write $\mathbb{R}$ as the union of pairwise disjoint equivalence classes [$x$], for $x \in \mathbb{R}$.

c) Find a bijection $\phi:(\mathbb{R}/\tilde{})\to\mathbb{T}$ from the collection $(\mathbb{R}/\tilde{})$ of all equivalence classes $[x]$ of $\tilde{}$ onto the unit circle $$\mathbb{T} = \{z \in \mathbb{C}: |{z}| = 1\}$$ in $\mathbb{C}$. Show carefully that $\phi$ is both injective and surjective.

OK SO:

Part (a) is very straightforward, no problems showing reflexivity, transitivity, and symmetry. It's part (b) that's killing me. When we have to break up something like $\mathbb{Z}$ into equivalence classes, it's easy to see that under various equivalence relations we can find the associated classes. For example, mod 2, we have even and odd numbers, and mod 3 we have [0], [1], [2] etc. But we're dealing with the real numbers here, and so we can't split up $\mathbb{R}$ into a finite number of equivalence classes mod 2$\pi$. So do we generalise it somehow? I gather we draw on a function with period 2$\pi$ such as cos:$(\mathbb{R}/\tilde{})\to \mathbb{R}$ and then have some generalised equivalence class such as:

$[x'] = \{x \in \mathbb{R} | x = x' + 2k\pi\}$.

But this is only one class? And I'm not even sure it's written correctly!

I'm sure I can figure part (c) out if I can just understand what's going on in part (b). Thanks in advance!

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$(1.)\quad \forall x,y \in \mathbb R, x \thicksim y \iff \exists n \in \mathbb Z, y-x = 2n\pi$.
$(2.)\quad \thicksim$ is an equivalence relation on $\mathbb R$.
$(3.)\quad \forall x \in \mathbb R, [x] = \{y\in \mathbb R: y \thicksim x\} = \{x+2n\pi: n \in \mathbb Z\}$.
$(4.)\quad \mathbb R/\!\!\thicksim\;= \{[x] : x \in \mathbb R\}$.
$(5.)\quad \displaystyle\mathbb R = \bigcup_{0\le x < 2\pi}[x].$