Does it make sense to consider the set of complex numbers without the operations defined on it?

Question

I was thinking about how the imaginary unit $i$ is in fact not defined as "$\sqrt{-1}$" since the square root function is only defined on positive real numbers, but (roughly) as an object such that $i^2=-1$. Then I realised that, in all rigor, it does not make much sense to define $i$ by the value of its square if we haven't defined what it means to square a complex number, i.e. we're defining $i$ according to some operation on it that we haven't defined. In light of that, I reason that if we want to define the set of complex numbers, we must at least say that:

1. $\mathbb R \subset \mathbb C$

2. There exists a non-real complex number, namely $i$.

3. We're defining two operations on the complex numbers, namely $+$ and $\ast$, with such and such properties, in particular the property that $i\ast i=-1$.

In that case, it seems to me that whenever one talks about complex numbers, the notion of the operations defined on it always come with it, especially the notion of complex multiplication. Therefore, does it make sense to talk about the set of complex numbers "on its own", i.e. not $(\mathbb C,+,\ast)$ but just $\mathbb C$, completely independently of any operations defined on its elements ?

Answer 1

You are right. The expressions $i=\sqrt{-1}$ and $i^2=-1$ are "symbolic" and have no validity as definitions.

A very simple way is to define the complex numbers as pairs of reals, and the definitions of arithmetic operations

$$(a,b)+(c,d)=(a+c,b+d)$$ and $$(a,b)\cdot(c,d)=(ac-bd,ad+bc).$$

In particular, this implies

$$(0,1)^2=(-1,0).$$

It is no big deal to show that $(a,b)$ can equivalently be denoted $a+ib$ where $i$ is a reserved symbol.

In this sense, $\mathbb C=\{(a,b):a,b\in\mathbb R\}$, with no operation defined. $\mathbb R\subset\mathbb C$ does not hold, but $\mathbb R\times\{0\}\subset\mathbb C$ does. (These real complex numbers are also denoted $a+i0$, or $a$, for short.)

Answer 2

====== full answer =======

Well technically we don't define $i =\sqrt{-1}$.

We define $i$ a number where $i^2 = -1$ and we technically don't do that either.

Technically we define two operations on $\mathbb R\times \mathbb R$ which I will label as $+_c$ and $\cdot_c$ and define

$(a,b)+_c (c,d) = (a+c,b+d)$ (where $+$ means addition on real numbers)

And $(a,b)\cdot_c(c,d) = (ac-bd,bc+ad)$ (where $\cdot$ means multiplication on real numbers) and we note:

1. $\mathbb R^* = \{(a,0)|a\in \mathbb R\},+_c, \cdot_c \subset \mathbb R^2$ is equivalent to $\mathbb R, +, \cdot$ where $(a,0)\mapsto a$ then $(a,0)+_c(b,0) \mapsto a+b$ (as $(a,0)+_c(b,0)= (a+b,0)$ and $(a,0)\cdot_c(b,0)\mapsto ab$ (as $(a,0)\cdot_c(b,0)= (ab-0\cdot 0, 0\cdot b + a\cdot 0) = (ab,0)$.

And

2. if we define $z^k= \underbrace{z\cdot_c z\cdot_c z\cdot_c.....}_{k\ times}$ then $(0,1)^2 = (0\cdot 0 - 1\cdot 1, 0\cdot 1 - 0\cdot 1) = (-1, 0)$. And if we note that $(a,b) = (a,0) +_c (0,b)= (a,0) +_c (b,0)\cdot_c(0,1)$

Then if we used the notation $i = (0,1)$ and we extend the real number field so that $(a,0)\mapsto a$ and we allow $(0,1)\mapsto i$ which is a symbol for some sort of number that isn't a real number (and if it isn't "real" we can call it.... let's see "imaginary"! .... ha! we mathematicians are so witty!) with the property that $i^2 = -1$, and if we assume this extension to obey the field function (namely the distributive law) then $(a,b)= (a,0) + (b,0)(0,1)\mapsto a+bi$ will be an equivalent system

So $\mathbb C = \{a + bi|a,b\in \mathbb R; i^2 = 1; $ addition and multiplication extend and distribute$\} \sim \mathbb R \times \mathbb R, +_c, \cdot_c$.

====== shorter tl;dr answer =======

yes.