# Complex number in equivalence class

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?

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