# Equivalence class with complex number

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?

## 1 Answer

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?

• 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? – BlackSky Nov 3 '18 at 17:06
• To the proposer: $a\sim b \iff x_a=x_b.$ So $[a]=\{b: x_b=x_a\}=\{x_a\}\times \Bbb C.$ – DanielWainfleet Nov 3 '18 at 21:32
• 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? – BlackSky Nov 4 '18 at 13:31