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Ahlfors states "From elementary algebra the reader is acquainted with the imaginary unit $i$ with the property $i^2 = -1$." (Complex Analysis, Lars Ahlfors, page 1)

Kreyszig (Advanced Engineering Mathematics) first defines complex numbers “as an ordered pair $(x,y)$”, where $x, y \in \mathbb{R}$. He then defines the imaginary unit $i$ as $(0,1)$. Following this, he defines the addition and multiplication of complex numbers, and uses the definition of multiplication to arrive at $i^2 = -1$. (pages 652, 653)

Professor Paul Dawkins's online notes use both of the above approaches.

Various other places like MathWorld use similar definitions.

But these definitions only narrow down $i$ to two possible values (which are opposite in sign: $z_1$ and $-z_1$). So, when $i$ is used in an expression or in the representation of a complex number, how does one determine which one of the two values it actually is? Also, the sign of the imaginary part of the complex number will change if one person is using one of the two values given by the definition of $i$ and someone else is using the other one. This ambiguity will also affect the uniqueness of the complex plane.

Keeping the above points in mind, are there additional qualifications that help in narrowing it down to one value?

Or is it impossible to further narrow it down, and so we say that - we are using one of the square roots of -1 and representing that root as $i$, and we will all use $i$ to mean this same root, although we can't really define which one it is?

user
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  • @coreyman317 Can you please let me know its definition? The texts that I have read do not really define it other than stating this property, as far as I can understand. Can you please suggest a text that defines it differently? – user Jan 07 '20 at 18:12
  • We have a concept of the "primary root" of a number. i.e. $x^2 = 2$ has two roots i.e. $\pm \sqrt 2$ but $\sqrt 2$ is the primary root (i.e. the positive root is the primary root when real roots exist). In the complex numbers both $i$ and $-i$ are roots of $z^2 = 1$ but $i$ is the primary root. – Doug M Jan 08 '20 at 02:26

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The map $$f:\mathbb{C}\rightarrow\mathbb{C}: a+bi\mapsto a-bi$$ (conjugation) is an automorphism of $\mathbb{C}$: if $P$ is a true statement about complex numbers using $+$ and $\times$, then "$f(P)$" is also true, where $f(P)$ is the statement gotten by replacing $z$ with $f(z)$ for every complex number $z$ in $P$. E.g. "$i^3=-1$" is true, and $f($"$i^3=-1$") = "$(-1)^3=-1$" is also true.

So using addition and multiplication alone there is no way to distinguish $i$ from $-i$. That said, there are other operations on $\mathbb{C}$ which "add structure" - e.g. the "imaginary part" function $\mathfrak{Im}: a+bi\mapsto b$ (since we have $\mathfrak{Im}(i)=1$ but $\mathfrak{Im}(-i)=-1$).

Making this a bit more palatable (hopefully!):

We shouldn't really talk about $\mathbb{C}$ as if it were unique. Rather, we should talk about "algebraic closures of $\mathbb{R}$" (which is itself a bit wrong - reall really, we should talk about "algebraic closures of Dedekind-complete ordered fields"). The point is that any two such objects are isomorphic, and that means that when you're holding your version of $\mathbb{C}$ and I'm holding my version of $\mathbb{C}$ we'll never disagree about how they behave - even if they're quite different, or if they're the same except with $i$ and $-i$ "flipped."

The ordered-pairs-of-reals approach describes one way of producing a "version of $\mathbb{C}$."

Ultimately, thinking along these lines will take us to model theory (an area of logic that looks at the problem of what objects/functions/relations/etc. are "definable" in given contexts) and category theory (which can be thought of as studying "purely structural" behavior).

Noah Schweber
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  • "So using addition and multiplication alone there is no way to distinguish $i$ from $−i$." Please refer to this question which essentially asks whether it is OK to write $i=-i$. People there have strongly disagreed with writing this, in the answers and comments to that question. If they are indistinguishable, then, wouldn't writing $i=-i$ be OK? – user Jan 07 '20 at 18:30
  • @user No, "indistinguishable" does not mean "equal" - they have the same properties on their own, but they are different from each other. This can be made precise by noting that conjugation is not an automorphism of $(\mathbb{C}; +,\times, i)$. – Noah Schweber Jan 07 '20 at 18:46
  • Note that I've been rather careful to say in what sense they're indistinguishable: namely, that they're swapped by an automorphism. There are many kinds of indistinguishability in mathematics, and this is a very strong one but falls short of actual equality. (Along the lines of the middle part of my answer: any algebraic closure of a Dedekind-complete ordered field has two non-equal square roots of $-1$; that said, any such also has an automorphism swapping those two square roots. That may be easier to think about.) – Noah Schweber Jan 07 '20 at 18:48
  • With respect to algebra only, $\pi$ and $e$ both are transcendental over $\Bbb Q$, and are indistinguishable. But $\pi$ is not $e$. (Of course the supposed equation $i=-i$ leads to the conclusion that $i=0$.) – Lubin Jan 07 '20 at 23:03
  • @Lubin Careful - they're not indistinguishable (using addition and multiplication only) in $\mathbb{R}$. For example, $\pi-e$ has a square root but $e-\pi$ doesn't. They are indistinguishable (read: there is an automorphism switching them) in $\mathbb{C}$ with addition and multiplication only. – Noah Schweber Jan 08 '20 at 00:30
  • Right you are, @NoahSchweber: as reals. But I suppose I was thinking of the fields $\Bbb Q(\pi)$ and $\Bbb Q(e)$. – Lubin Jan 08 '20 at 03:58