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I just started studying Complex analysis and have in fact just switched field to mathematics recently and so please forgive me if this is question seems trivial for a mathematics student to ask.

Question: Why do we use Complex numbers instead of another algebraic field or number system? i.e. It is "natural" to have the hierarchy $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ and not others.

Attempt: I have often been told that $\mathbb{C}$ is algebraically closed (every polynomial with complex coefficients have a root in $\mathbb{C}$) and that it contains $\mathbb{R}$ as a isomorphic subfield.

But should there be other fields or number systems with the same properties as well? Did I miss anything "extra" about $\mathbb{C}$ that makes it special?

Jean Valjean
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    If an algebraically closed field contains (a copy of) the real numbers, then it contains (a copy of) the complex numbers, so $\mathbb{C}$ is the minimum such field. – egreg Jan 04 '14 at 18:24
  • @egreg - That was the key insight that I was missing! Thanks! +1 – Jean Valjean Jan 04 '14 at 18:25
  • Because these real numbers are of two kinds: rational and irrational; and the latter are also of two kinds: algebraic and transcendental. But these two are self-coherent and self-contained within $\mathbb{C}$, not within $\mathbb{R}$, since $\mathbb{C}=\mathbb{A}\cup\mathbb{T}$. – Lucian Jan 05 '14 at 00:30

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If an algebraically closed field $F$ contains (a copy of) the real numbers, then $F$ also contains (a copy of) the complex numbers.

Indeed, if we have $\mathbb{R}\subset F$ and $F$ is algebraically closed, then $F$ contains a root $j$ of the polynomial $X^2+1\in \mathbb{R}[X]$. Therefore it contains $\mathbb{R}[j]$, but this field is isomorphic to $\mathbb{C}$: just define $a+bi\in\mathbb{C}\to a+bj\in \mathbb{R}[j]$ and check it's an isomorphism.

The key point is that $\mathbb{C}$ is algebraically closed and algebraic over $\mathbb{R}$.

egreg
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  • I upvoted this answer because it's interesting. I really think of it in terms of Polynomials of degree more than 1 and points where the derivative is 0. When you look at the complex plane and see an image where what appears at each point is the point on graph paper where the function would have taken that point to, it seems not that hard to see why there would be 1 less spot of derivative 0 than the order of the polynomial. Now it's not that hard to prove its equivalence to the roots of its derivative. – Timothy Nov 16 '22 at 02:50
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Reading up further, thanks to prompting in the comments, I discovered this: http://en.wikipedia.org/wiki/Algebraic_closure

Using Zorn's lemma, it can be shown that every field has an algebraic closure,[1] and that the algebraic closure of a field K is unique up to an isomorphism that fixes every member of K. Because of this essential uniqueness, we often speak of the algebraic closure of K, rather than an algebraic closure of K.

$\mathbb{C}$ is the unique algebraic closure of $\mathbb{R}$.

Jean Valjean
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"Algebraically closed" is a heavy loaded notion. Think of the extension ${\mathbb R}\rightsquigarrow{\mathbb C}$ this way instead: We already have the real numbers ${\mathbb R}$ and are extremely successful with them in analysis, geometry, and physics. Unfortunately the equation $x^2+1=0$, and similar ones which turn up all the time don't have a solution in ${\mathbb R}$. Therefore we need a field $K\supset{\mathbb R}$ containing a number $i$ with $i^2=-1$ as everyday computing environment. It turns out that there are such fields. The simplest one of these is ${\mathbb C}$, consisting of no more than the obviously necessary numbers $x+ y\,i$, where $x$, $y\in{\mathbb R}$.

J. W. Tanner
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Christian Blatter
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  • And $x + yi$ is obviously necessary because the field has to be closed under addition and multiplication? That is a very nice way to put it. Thanks! +1 – Jean Valjean Jan 04 '14 at 18:54
  • Besides $\mathbb{C}$ there are no such finite-dimensional fields any more. – Anixx Mar 02 '21 at 22:43
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I feel it is worth mentioning that there are other number systems that are of interest to mathematicians, although for different reasons, such as the $p$-adics and the quaternions. The complex numbers are most widely used though because of the algebraic properties you have already discussed, along with the fact that they play nicely with the Euclidean norm.

cbishop
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There are split-complex numbers and dual numbers, and they are used in certain fields. But the both are "worse" than $\mathbb{C}$: they are not algebraically closed and not fields (have zero divisors).

Still, they have many properties of numbers, allowing full-fledged calculus on them.

Anixx
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