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

Question

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 $+$)

Answer 1

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.

Answer 2

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

Operator overloading as a computer scientist might say

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 & $i$.