$\newcommand{\ms}{\mathscr}$Equivalence relations and partitions are very intimately related; indeed, it’s fair to say that they are two different ways of looking at basically the same thing.
Start with a set $A$. A partition $\ms P$ of $A$ is just a way of chopping $A$ up into pieces. More formally, it’s a collection of subsets of $A$ with a very simple property: every element of $A$ belongs to exactly one of the sets in $\ms P$. This is often expressed in a slightly more roundabout fashion: a collection $\ms P$ of non-empty subsets of $A$ is a partition of $A$ if
- $A=\bigcup_{P\in\ms P}P$, and
- if $P_1,P_2\in\ms P$ and $P_1\ne P_2$, then $P_1\cap P_2=\varnothing$, i.e., the members of $\ms P$ are pairwise disjoint.
The first of these conditions says that each element of $A$ belongs to at least one member of $\ms P$, and the second says that no element of $A$ belongs to more than one member of $\ms P$; put the two together, and you get my original definition.
We can use the partition $\ms P$ to define an associated relation $\overset{\ms P}\sim$ on $A$: for any $x,y\in A$, $x\overset{\ms P}\sim y$ if and only if $x$ and $y$ are in the same piece of the partition $\ms P$. For instance, if $A$ is a set of people, we can partition them according to their ages: the $20$-year-olds are one piece of the partition, the $50$-year-olds are another, and so on. The associated relation is simply has the same age as: $x\overset{\ms P}\sim y$ if and only if $x$ and $y$ are the same age. It’s easy to see in this case that $\overset{\ms P}\sim$ is an equivalence relation $-$ reflexive, symmetric, and transitive $-$ and it’s not hard to prove that this is always the case: if $\ms P$ is a partition of a set $A$, then $\overset{\ms P}\sim$ is an equivalence relation on $A$.
Now what are the equivalence classes of this relation $\overset{\ms P}\sim$? Fix $a\in A$. The equivalence class of $a$ is by definition $\{x\in A:a\overset{\ms P}\sim x\}$. But $a\overset{\ms P}\sim x$ just means that $a$ and $x$ are in the same piece of the partition $\ms P$, so $x$ is in the equivalence class of $a$ if and only if is in the same piece as $a$. In other words, the equivalence class of $a$ is the piece of $\ms P$ that contains $a$. And this is true for every $a\in A$, so the equivalence classes of the relation $\overset{\ms P}\sim$ are exactly the pieces of the partition $\ms P$, the ‘chunks’ into which it divides $A$.
Now set that aside for a moment, and let $R$ be an equivalence relation on $A$. For each $a\in A$ we set $[a]/R=\{x\in A:aRx\}$; this is the equivalence class of $a$, the set of things to which $a$ is related by $R$. It’s a subset of $A$. One of the first things that you prove about equivalence classes is that for any $a,b\in A$, either $aRb$, in which case $[a]/R=[b]/R$, or $a\not Rb$, in which case $[a]/R\cap[b]/R=\varnothing$: any two equivalence classes are either the same set or completely disjoint from each other. In other words, the equivalence classes chop up $A$ into pairwise disjoint pieces, and every element $a$ of $A$ belongs to exactly one of these pieces, namely $[a]/R$.
But this is exactly what it means to say that $A/R$, the collection of all of these equivalence classes, is a partition of $A$: each element of $A$ belongs to exactly one of the sets in the collection $A/R$. Just as a partition $\ms P$ of $A$ gives rise to an associated equivalence relation $\overset{\ms P}\sim$ on $A$, an equivalence relation $R$ on $A$ gives rise to an associated partition $A/R$ of $A$. What happens if we start with the partition $A/R$ and construct its associated equivalence relation $\overset{A/R}\sim$ on $A$? For any $x,y\in A$ we have by definition $x\overset{A/R}\sim y$ if and only if $x$ and $y$ are in the same piece of $A/R$. But the pieces of $A/R$ are the $R$-equivalence classes, so $x$ and $y$ are in the same piece of the partition $A/R$ if and only if $[x]/R=[y]/R$, i.e., if and only if $xRy$. That is, $x\overset{A/R}\sim y$ if and only if $xRy$, and $\overset{A/R}\sim$ and $R$ are exactly the same relation on $A$.
To recapitulate:
Each partition $\ms P$ of $A$ induces an associated equivalence relation $\overset{\ms P}\sim$ on $A$, and each equivalence relation $R$ on $A$ induces an associated partition $A/R$ of $A$ into equivalences classes.
These conversion operations from partition to equivalence relation and vice versa are inverses. If you start with a partition $\ms P$ of $A$, construct the equivalence relation $\overset{\ms P}\sim$ on $A$, and then construct the associated partition $A/\overset{\ms P}\sim$ of $A$, you’re back where you started: $A/\overset{\ms P}\sim=\ms P$. Similarly, if you start with an equivalence relation $R$ on $A$, construct the associated partition $A/R$ of $A$, and then build from that partition its associated equivalence relation $\overset{A/R}\sim$, you get the original relation $R$ back: $\overset{A/R}\sim=R$.
If I've understood correctly, you need to characterize the equivalence classes for the above equivalence relation on the set $X=\mathbb{C}$.
The definition of complex absolute value is very geometric. Given some complex number $a+bi$, $\lvert a+bi\rvert=\sqrt{a^2+b^2}$. It follows from here that an equivalence class on $\mathbb{C}$ with the relation of absolute value is actually a circle of radius corresponding to the absolute value of elements in the class, centered at the origin.
So, multiple different classes form a collection of concentric circles about the origin. Can you see why?
Best Answer
Partition of a set $X$ is a collection of pairwise disjoint subsets of $X$ such that union of all these subsets gives $X$. Any partition naturally sets an equivalence relation and vise versa. The partition in your question can be set by equivalence relation that can be described in many ways. For example like this: we say that $x$ is equivalent to $y$ if $x$ and $y$ have the same number of digits in binary representation. Check, that this an equivalence relation and that classes of equivalence form exactly your partition.