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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 stuck on this one. It's already given on the question that $a\sim b$ so I don't have to show the reflexive, symmetric and transitive.

I'm not sure how to find out the equivalence class of complex number, but I think I can get rid of option a. as it says $∀x∈\mathbb R$ which is incorrect. I'm not sure about the remaining 4.

Any help will be appreciated, thanks.

My working:

Working from $x_a+ix_b=x_b+ix_a$, we have $x_a=x_b$, $ix_b=ix_a$ 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?

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The condition $a\circ b = b\circ a$ means that $$(x_a+x_b, x_a + ix_b + z_a+z_b) = (x_b+x_a,x_b+ix_a+z_b+z_a),$$ i.e., $x_a+x_b = x_b+x_a$ and $$x_a+ix_b + z_a+z_b = x_b+ix_a+z_b+z_a.$$ The latter means that $x_a+ix_b = x_b+ix_a$. Since $x_a,x_b$ are real numbers, it follows that $x_a=x_b$. Could you complete it from here?

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  • $\begingroup$ I'm not sure if understand it correctly. So working from $x_a+ix_b=x_b+ix_a$, we have $x_a=x_b$, $ix_b=ix_a$ 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 $S=R \times C$, $a∈S$, $x_a∈R$ and $z_a ∈ C$,$ [a]=${$(x_a+I(z_a),z);∀z∈C$}. So d. is the answer, am I right? $\endgroup$ – BlackSky Nov 3 '18 at 17:06
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    $\begingroup$ To the proposer: $a\sim b \iff x_a=x_b.$ So $[a]=\{b: x_b=x_a\}=\{x_a\}\times \Bbb C.$ $\endgroup$ – DanielWainfleet Nov 3 '18 at 21:32
  • $\begingroup$ I am confused about the imagery part. If $[a]=${$b:x_b=x_a$}$=${$x_a$}$×C$ which I believe it is same as $[a]={(x_a,z);∀z∈C}$ but since $z_a+z_b=z_b+z_a$ $z_a$ and $z_b$ should be real numbers? If $[a]={(x_a,z);∀z∈C}$ is true, where does the real number $z$ goes? $\endgroup$ – BlackSky Nov 4 '18 at 13:31

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