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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?

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  • $\begingroup$ Question has already been asked (and answered) two days ago. $\endgroup$ – Wuestenfux Nov 4 '18 at 18:28
  • $\begingroup$ I'm not confident whether my answer is correct or not $\endgroup$ – BlackSky Nov 4 '18 at 20:39

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