For me, the natural way to define complex numbers seems to be to take $\mathbb R^2$ and then define addition and multiplication on top of that an boom, you have complex numbers. You then pretty straightforwardly prove they have the required algebraic properties. Easy.

However for some reason, the more popular teaching method seems to be along the lines to introduce as an additional axiom a new value $i$ such that $i^2=-1$ (or worse: $i=\sqrt{-1}$) and then somehow silently expect the desired algebraic properties to hold and overall expect the theory to keep logical consistency somehow without proving it? I understand this may perhaps feel more familiar to students who are already used to solving quadratic equations etc. but it seems like cheating from a more rigorous perspective.

So my question is basically: Can this be made rigorous in the sense that you specify the required algebraic properties of complex numbers including the property $∃i. i^2=-1$ and then showing the set of complex numbers satisfying these properties is in fact unique up to isomorphism? Or alternatively: how do you rigorously define something like "closure for taking quadratic roots" while showing such definition is valid?

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    $\begingroup$ You might appreciate this relatively recent question, in particular Rob's top-rated answer, I think, might answer your question. $\endgroup$ Commented Jul 9 at 19:10
  • $\begingroup$ If the question linked to in @TheoBendit's comment doesn't help, you should edit your question to refer to a specific text that takes the approach you describe as "more popular". As things stand, I don't feel minded to second-guess what background is expected in the texts you consider to be "more popular". $\endgroup$
    – Rob Arthan
    Commented Jul 9 at 19:13
  • $\begingroup$ I found some interesting insights here: math.stackexchange.com/questions/4702068/… $\endgroup$ Commented Jul 9 at 19:20
  • $\begingroup$ Adding to the linked questions and answer here, we can use similar techniques with the rational numbers and the equation $x^2=2$ to get a field called $\mathbb{Q}(\sqrt{2})$, and then prove its elements can be described as $a+b\sqrt{2}$ where $a$ and $b$ are rational numbers. $\endgroup$
    – aschepler
    Commented Jul 9 at 19:38

2 Answers 2


introduce as an additional axiom a new value i such that $i^2=-1$ (or worse: $i=\sqrt{-1}$) and then somehow silently expect the desired algebraic properties to hold and overall expect the theory to keep logical consistency somehow without proving it?

So, that's not quite what's going on. We actually force the desired algebraic properties to hold; that is, it's not that we somehow "expect" multiplication to be associative and commutative but that we declare by fiat that multiplication is associative and commutative, and so forth. This can be formalized in ring theory using the idea of a presentation of a ring by generators and relations; formally, this approach amounts to constructing $\mathbb{C}$ as a quotient ring $\mathbb{R}[i]/(i^2 + 1)$.

The problem from this perspective is not "logical consistency" as such; rather, the problem from this perspective is that we don't know whether or not all the properties we're forcing will force every element to be equal to $0$. This is a real thing that can actually happen, for example if we tried to add a new value $\infty$ satisfying $0 \cdot \infty = 1$; this amounts to constructing the quotient ring $\mathbb{R}[\infty]/(0 \cdot \infty - 1)$, and in this quotient ring we have that

$$0 \cdot \infty + 0 \cdot \infty = (0 + 0) \cdot \infty = 0 \cdot \infty$$

forces $0 \cdot \infty = 0$ and hence $1 = 0$.

You could think of this as a "logical inconsistency" in the sense that it proves the "contradiction" $1 = 0$, but in modern mathematics we don't think of it that way, because the symbols $1$ and $0$ don't have their ordinary meanings anymore; they are actually an (extremely common and standard) abuse of notation referring to new entities $1'$ and $0'$ which take values in our new quotient ring $\mathbb{R}[\infty]/(0 \cdot \infty - 1)$, and the fact that $1' = 0'$ in this ring doesn't have any bearing on whether $1 = 0$ in the ordinary real numbers. It's just that this new ring is the zero ring which means computations in it aren't useful for anything.

So, to make this approach rigorous it's necessary to prove that $\mathbb{R}[i]/(i^2 + 1)$ is not the zero ring, but instead that it is $2$-dimensional over $\mathbb{R}$, with the expected basis $\{ 1, i \}$; equivalently, you need to prove that $a + bi = c + di$ iff $a = c, b = d$. This can be done using the standard construction in terms of ordered pairs, and it can also be done using a matrix representation of the complex numbers, which sends

$$\mathbb{R}[i]/(i^2 + 1) \ni a + bi \mapsto \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \in M_2(\mathbb{R}).$$

One way or another you have to do something to prove that $\mathbb{R}[i]/(i^2 + 1)$ is really $2$-dimensional as expected. So, summarizing:

  • In the ordered pair approach the complex numbers are $2$-dimensional by definition but you have to work to prove that multiplication is commutative, associative, has a unit, and distributes over addition.
  • In the quotient ring approach multiplication is commutative, associative, has a unit, and distributes over addition by definition but you have to work to prove that the complex numbers are $2$-dimensional.

So you have to do some work somewhere but different approaches shift around where the work is.

However, because the complex numbers are in fact $2$-dimensional, pedagogically you can get away with not showing this in the quotient ring approach, because your calculations will never run into any problems. So, if you just wanted to introduce how to do calculations with the complex numbers with a minimum of technical baggage, and you also wanted to avoid using ordered pairs because the definition of multiplication is kind of awkward and unintuitive in that approach, you might use this algebraic approach and just not bother fully justifying it.

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    $\begingroup$ Incidentally there's another approach which I've never seen anywhere but which avoids what I see as pedagogical issues with both of the above approaches. Namely, here is an amazing purely geometric fact about the Euclidean plane: if $v, w$ are any two nonzero vectors, there is a unique way to rotate and scale one of them to the other one. This means you can define new objects which are "formal quotients" $\frac{v}{w}$ of vectors, representing these rotations-and-scalings, and these objects turn out to be exactly the complex numbers. $\endgroup$ Commented Jul 9 at 19:44
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    $\begingroup$ In this approach there's no thorny philosophical questions about what $i$ "really is": it is any quotient $\frac{v}{w}$ where $v$ is a $90^{\circ}$ counterclockwise rotation from $w$. Or said another way, it is the relationship of "being a $90^{\circ}$ counterclockwise rotation" between two vectors. It squares to $-1$ because when you do a $90^{\circ}$ counterclockwise rotation twice you get a $180^{\circ}$ rotation. This is the most geometric way to do it by far, and what I learned recently is that it can be done without matrices! $\endgroup$ Commented Jul 9 at 19:45
  • $\begingroup$ Moreover this idea generalizes immediately to higher dimensions: in any Euclidean space, if $v, w$ are two nonzero vectors there is a unique way to rotate and scale one of them to the other one in a way that leaves the vectors orthogonal to both $v, w$ alone. And if you try to work out what the theory of the resulting formal quotients $\frac{v}{w}$ is you'll get the Clifford algebra / geometric algebra, and this was apparently how Grassmann and Clifford thought about them: en.wikipedia.org/wiki/Geometric_algebra $\endgroup$ Commented Jul 9 at 19:47
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    $\begingroup$ Perhaps it's worth mention that already Kronecker did have misgivings about assuming that various irrational (etc.) "numbers" existed "out there (somewhere)"... and could be "adjoined" ... and conceived of the idea of (irrefutably?) constructing field extensions (etc.) as quotients (e.g., of polynomial rings...) $\endgroup$ Commented Jul 9 at 19:54
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    $\begingroup$ @Damian: what I mean by "generalize" is that there is a general construction that can be applied to Euclidean spaces of any dimension which when applied to the Euclidean plane produces the complex numbers. When applied to higher-dimensional Euclidean spaces it produces Clifford algebras, which are not commutative and not division algebras. The rotation-and-scaling property is true in all dimensions with the extra orthogonality condition imposed and it reduces immediately to the $2$-dimensional case. I am also not giving full details here due to lack of space. $\endgroup$ Commented Jul 10 at 6:54

You could try the following axiomatic definition of the complex numbers that I just made up.

Definition. A quadruple $(K,+,\cdot,i)$ is a complex field if the following axioms are satisfied:

  1. $(K,+,\cdot)$ is a field containing $\mathbb R$ as a subfield.
  2. $i\in K$ and $i^2=-1$.
  3. For all $z\in K$, there exists $a,b\in\mathbb R$ such that $z=a+bi$.

(For simplicity, I will yield to the usual abuse of notation whereby an algebraic structure is referred by the name of its underlying set.)

We can easily deduce a few basic properties of complex fields from these axioms: for instance, if $a+bi=0$ with $a,b\in\mathbb R$, then $(a+bi)(a-bi)=0$, hence $a^2+b^2=0$ and so $a=b=0$. It follows that if $a+bi=c+di$, then $a=c$ and $b=d$. Continuing in this manner, we can deduce all of the basic results about complex fields, and then move onto proving results in complex analysis. This is in the same way that all results in real analysis can be deduced on the basis of the fact that $\mathbb R$ is a complete ordered field – at no point does the precise construction of $\mathbb R$ play any role. The constructions of $\mathbb R$ and $\mathbb C$ respectively are arguably only important to the extent that they tell us that complete ordered fields and complex fields exist – without them, there is a danger that everything we deduce is in fact vacuous. On the other hand, once we have carried out the respective existence proofs, and shown that $\mathbb R$ and $\mathbb C$ are unique up to unique isomorphism, we can mentally discard their set-theoretic constructions. It is best to think of $\mathbb R$ as denoting an arbitrary (but fixed) complete ordered field, rather than a collection of Dedekind cuts, say. The same goes for $\mathbb C$.

The proof that there is a unique up to unique isomorphism complete ordered field is well-known, and is covered in most textbooks in real analysis, say Spivak's Calculus. Here I would like to prove that complex fields are similarly unique. There is a natural notion of "complex field homomorphism" between complex fields $K$ and $K'$: it is a field homomorphism $\varphi:K\to K'$ which fixes $\mathbb R$ in place and sends $i_K$ to $i_{K'}$. If $K$ and $K'$ are complex fields, and $\varphi:K\to K'$ is a homomorphism, then for all $a,b\in\mathbb R$, $$ \varphi(a+bi_K)=\varphi(a)+\varphi(b)\varphi(i_K)=a+bi_{K'} \, . $$ It follows that there is just one homomorphism between $K$ and $K'$, and it is an isomorphism sending $a+bi_K$ to $a+bi_{K'}$. Thus, there is no real harm in speaking of the complex field. Note that to ensure the isomorphism is unique, we had to specify a choice of square root of $-1$, and insist that complex field homomorphisms leave every real number alone.

From the ordered pair construction of $\mathbb C$, we know that complex fields exist*, and the above paragraph shows that they are unique in a suitable sense. Hence, we can indeed do complex analysis simply on the basis of the axioms of complex fields mentioned above.

*There is a slight technical issue with the ordered pair construction of $\mathbb C$: if we define $\mathbb C$ as $\mathbb R^2$, then $\mathbb R$ is not literally a subfield of $\mathbb C$, and hence $\mathbb C$ is not a complex field. (Similarly, $\mathbb R$ is not literally a subfield of $\mathbb R[x]/\langle x^2+1\rangle$.)

This issue can be patched in a number of ways. For instance, we could take $\mathbb C$ to be $\bigl(\mathbb R^2\setminus \{(a,0)\mid a\in\mathbb R\}\bigr)\cup\mathbb R$ instead. Alternatively, we could modify axiom (1) by simply requiring that there is an embedding $\xi$ of $\mathbb R$ in $\mathbb C$. To ensure that complex fields are unique up to unique isomorphism, we would then have to insist that the embedding of $\mathbb R$ in $\mathbb C$ is part of the structure of a complex field, and that homomorphisms of complex fields respect the emeddings.

  • $\begingroup$ The last part of axiom 3 is unnecessary; it follows from the others via calculating that $(a + bi)(a - bi) = a^2 + b^2$, meaning that if $a, b \neq 0$ then $a + bi$ is nonzero (because $a^2 + b^2 > 0$). $\endgroup$ Commented Jul 9 at 19:54
  • $\begingroup$ @QiaochuYuan: Good point! I have modified my answer. $\endgroup$
    – Joe
    Commented Jul 10 at 12:15

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