Let us consider $S=\mathbb R\times\mathbb C$.
We write $a$ and $b$ two elements of $S$: $a=(x_a,z_a)$, and $b=(x_b,z_b)$.
We define the binary operation $∘$ as:
$a∘b=(x_a+x_b,x_a+ix_b+z_a+z_b)$, with $i$ the imaginary unit $i^2=−1$.
We say that $a\sim b$ if and only if $a∘b=b∘a$.
We write $[a]$ the equivalence class of $a$. Only one of the following is correct.
a. For $a=(x_a,z_a)$ one has $[a]=\{(x,z_a);∀x∈\mathbb R\}$
b. For $a=(x_a,z_a)$ one has $[a]=\{(x_a,z);∀z∈\mathbb C\}$
c. For $a=(x_a,z_a)$ one has $[a]=\{(x_a+R(z_a),z);∀z∈\mathbb C\}$
d. For $a=(x_a,z_a)$ one has $[a]=\{(x_a+I(z_a),z);∀z∈\mathbb C\}$
e. For $a=(x_a,z_a)$ one has $[a]=\{(x_a+|z_a|^2,z);∀z∈\mathbb C\}$
I am working on equivalence class questions but I'm so confused about this one. I'm not sure my understanding on equivalence class of complex number is correct.
My working:
For now I know $(x_a+x_b, x_a + ix_b + z_a+z_b) = (x_b+x_a,x_b+ix_a+z_b+z_a).$
Working from $x_a+ix_b + z_a+z_b = x_b+ix_a+z_b+z_a,$ we have $x_a+ix_b=x_b+ix_a$ and $x_a+x_b = x_b+x_a$.
I get $x_a=x_b$, $ix_b=ix_a$ and $z_a+z_b=z_b+z_a$. $z_a+z_b=z_b+z_a$ is just $z$ , $ix_b=ix_a$ where $ix_b,ix_a$ are imagery numbers. Since $S=R×C, a∈S, x_a∈R$ and $z_a∈C,[a]=${$(x_a+I(z_a),z);∀z∈C$}.
Am I right?
Any help will be appreciated, thanks.
Edit: I was told I should look for $[a]=${$b:x_b=x_a$}={$x_a$}$×C$ which I believe it is same as $[a]=${$(x_a,z);∀z∈C$} am I right? But if it is the case, where does real number $z$ goes ( from $z_a+z_b=z_b+z_a$)? Or it never exists?