# Product in $\mathbb{R}$ vs product in $\mathbb{R^2}$

The set of complex numbers is the set of ordered pairs $$\{ (a,b): a,b \in \mathbb{R} \}$$ together with the operations of sum and product thus defined: $$(a,b)+(c,d)=(a+c, b+d)$$, $$(a,b) \times (c,d)=(ac-bd,ad+bc)$$ (I'm using '$$\times$$' for product in $$\mathbb{R^2}$$, '$$\cdot$$' for product in $$\mathbb{R}$$). Also, considering the set $$\mathbb{C_0}=\{ (a,0): a \in \mathbb{R} \}$$ we know that the set of complex numbers can be regarded as an extension of $$\mathbb{R}$$ since there is a bijective function $$\phi: \mathbb{C_0} \rightarrow \mathbb{R}$$ such that $$\phi[(a,0)]=a$$, that allows us to identify $$(a,0)$$ with $$a$$. Now, when we introduce the algebraic form of complex numbers, we write $$(a,b)=(a,0)+(0,b)=(a,0)+(0,1) \times (b,0)=a+i \cdot b$$, but who allows us to turn the operation of product $$\times$$ in $$\mathbb{R^2}$$ into the operation $$\cdot$$ defined in $$\mathbb{R}$$? (The same question could be asked for $$+$$)

• That $\cdot$ in $i \cdot b$ is formal. So is the $+$ in $a+i\cdot b$. Oct 15, 2021 at 13:39
• You should already be asking this about the addition sign, as $a+_{\mathbb c}b\equiv(a,0)+_{\mathbb c}(b,0)=(a+b,0+0)$ is the addition of two complex, which is defined via the ordinary addition of reals. As the two operators enjoy the same group properties, overloading is not a problem.
– user974557
Oct 15, 2021 at 14:30
• Oct 18, 2021 at 10:46

I think this is a classic case of abuse of notation. If you write $$(a,b)+_{\mathbb{R}^2}(c,d)=(a+_{\mathbb{R}}b, c+_{\mathbb{R}}d)$$, the confusion disappears. More specifically, this is how we define addition in $$\mathbb{R}^2$$. Indeed, you add component by component and then write it up as a vector. The same is true for $$\mathbb{C}$$, indeed $$z+_\mathbb{C}w$$ = $$(z_1+_\mathbb{R}w_1) + i(z_2+_\mathbb{R}w_2)$$, where $$z=z_1+iz_2$$ and $$w=w_1 + iw_2$$. Thus when we write $$+$$, we usually have different kinds of addition in mind.

One could argue that $$\phi$$ as defined in your question is a group morphism between $$\left(\mathbf R,+\raise-1.5ex\mathbf R\right)~\&~\left(\mathbf C^*,+\kern-1pt{\raise-1.5ex{\mathbf C_\left|\mathbf C^*\right.\!}}\right)$$,

where $$\mathbf C^*$$ is the set of complex numbers with a zero imaginary part, which justifies the use of the same symbol ($$+$$) in both cases

As rigour was at stake here, I denoted $$+\kern-1pt\raise-1.5ex{\mathbf C_\left|\mathbf C^*\right.\!}$$ the restriction of the complex addition mapping to the $$\mathbf C^*$$ set, as we should technically use it to define a morphism.

This goes to show that sometimes, abuse of notation is desirable.

EDIT

Given this fact, we take

$$(a,b)=(a,0)+(0,1) \times (b,0):= a+ib$$

as granted, because of the property of sum & $$i$$.

• Could you please explain in detail how does this apply to the line $(a,b)=(a,0)+(0,1) \times (b,0)=a+ib$? Oct 15, 2021 at 15:27
• I mean, I do understand that for numbers like $(a,0)$, $+_\mathbb{C}$ is the same as $+_\mathbb{R}$, but what happens when we deal with numbers like $(0,b)$? Oct 15, 2021 at 15:33
• @GabrielePrivitera you could do an analogous with the multiplicative operation. I'll develop it in my answer soon
– T.D
Oct 15, 2021 at 15:49