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 $+$)

  • $\begingroup$ That $\cdot$ in $i \cdot b$ is formal. So is the $+$ in $a+i\cdot b$. $\endgroup$
    – Randall
    Oct 15, 2021 at 13:39
  • $\begingroup$ 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. $\endgroup$
    – user974557
    Oct 15, 2021 at 14:30
  • $\begingroup$ See also math.stackexchange.com/a/3556451/442 $\endgroup$
    – GEdgar
    Oct 18, 2021 at 10:46

2 Answers 2


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

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.


Given this fact, we take

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

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

  • $\begingroup$ Could you please explain in detail how does this apply to the line $(a,b)=(a,0)+(0,1) \times (b,0)=a+ib$? $\endgroup$ Oct 15, 2021 at 15:27
  • $\begingroup$ 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 $(0,b)$? $\endgroup$ Oct 15, 2021 at 15:33
  • $\begingroup$ @GabrielePrivitera you could do an analogous with the multiplicative operation. I'll develop it in my answer soon $\endgroup$
    – T.D
    Oct 15, 2021 at 15:49

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