China Post 2: The complex numbers

by t0rajir0u, Jul 22, 2008, 6:09 AM

The study of the complex numbers is so fundamental to so many areas of modern mathematics that it is perhaps strange to ask why the complex numbers are special. One might respond that the fundamental theorem of algebra or similar nearly demands it, or simply "what else is there to study?"

An interesting example of an extension of the real numbers that does not reside in the complex numbers is a construction known as the split-complex numbers. These are the numbers of the form $ a + bj$ where $ j$ is defined with the property that $ j^2 = 1$. This might seem like an extremely curious definition: since it implies $ (j - 1)(j + 1) = 0$, the split-complex numbers have zero divisors, hence are not an integral domain (and hence not a field). Nevertheless, the split-complex numbers turn out to have interesting properties and to be useful in their own right (and I refer the reader to the article for more details). So now that we have an example of an extension of the real numbers to study that is not the complex numbers, perhaps the first question does not look so inane.

Rather than attempt to answer it in full generality, I will instead provide a few constructions of the complex numbers that hopefully shed some light on their nature from various viewpoints.


Formulation 1: The complex numbers are (isomorphic to) the quotient ring $ \mathbb{R}[x]/(x^2 + 1)$.

This requires some explanation. You may recall my post about polynomial modular arithmetic earlier. Essentially, the basic tool of modular arithmetic on the integers is the equivalence relation

$ a \equiv b \bmod m \Leftrightarrow a - b \in m \mathbb{Z}$

where $ m\mathbb{Z}$ denotes the set of integer multiples of $ m$. In abstract terms, this is an example of an ideal and can also be written $ (m)$, and the above equivalence relation can be used to describe the "quotient of $ \mathbb{Z}$ by $ (m)$," which essentially means that we consider two numbers equivalent if they are equivalent $ \bmod m$ and consider operations on the equivalence classes generated. This notion generalizes: over the real polynomials, we can write

$ A(x) \equiv B(x) \bmod x^2 + 1 \Leftrightarrow a - b \in (x^2 + 1)$

where $ (x^2 + 1)$ denotes the set of polynomial multiples of $ x^2 + 1$, which is also an ideal. Now $ x$ plays the part of the "imaginary unit" since $ x^2 \equiv - 1$.

Recall that if $ m$ is prime, $ \mathbb{Z}/m \mathbb{Z}$ is an integral domain (in fact, a field). Otherwise, zero divisors exist. The same idea holds over polynomials and is generalized by the notion of a prime ideal: in other words, because $ x^2 + 1$ is irreducible, no two polynomials in $ \mathbb{R}[x]/(x^2 + 1)$ that are not divisible by $ x^2 + 1$ can multiply to a multiple of $ x^2 + 1$. In fact, we can say more: the Euclidean algorithm on polynomials allows us to divide, so that $ \mathbb{R}[x]/(x^2 + 1)$ is a field (in analogy to how the Euclidean algorithm on integers allows us to divide modulo a prime).

This is an extremely abstract, if solid, description of the complex numbers (there is nothing "imaginary" about a quotient ring). It does not really tell us anything about them - their topology, their geometric properties, their group actions, anything. It places the complex numbers within a generic algebraic framework; a priori, there is no reason for us to expect that $ \mathbb{R}[x]/(x^3 - x + 1)$, for example, is not equally interesting. Let us turn, then, to a more geometric understanding.


Formulation 2: The complex numbers are the natural algebraic model for 2D rotation.

Our basic intuition for this formulation of the complex numbers is an Argand diagram, but we will not need to refer to it explicitly. This formulation of the complex numbers is one I find explained poorly in most introductions, so I will devote a lot of space to it.

I mentioned in my discussion of groups that geometry is a natural setting for the appearance of groups. In the case of Euclidean geometry, we are interested in the isometry group of (for example, $ \mathbb{R}^2$), the group of transformations that preserve the Euclidean metric. This is an extremely high-concept way to think about (and organize) geometry; the name for this point of view is the Erlangen program.

What we are interested in is the subgroup of rotations about the origin. It's easy to verify intuitively that the set of rotations is a group: the inverse of a rotation by $ \theta$ is a rotation by $ - \theta$ and so forth. The group elements consist of the set of transformations that we will label "rotation by $ \theta, 0 \le \theta < 2 \pi$" and the group operation will be composition of rotations, which amounts to addition of the angles to which they correspond $ \bmod 2 \pi$ (so that in fact this group is abelian). We have talked about groups alone, but this group has a group action on $ \mathbb{R}^2$ (which is why we are interested in it in the first place!). We would like to talk about this group in terms of its interpretation as transformation of points $ (x, y)$ to other points $ (x', y')$. The most important fact about rotation (and it forms the entire basis of our discussion here) is that rotation is a linear transformation.

It's fairly clear that rotation respects scalar multiplication, so the only (possibly unintuitive, depending on how much geometric intuition you have) property we need to verify is that the rotation of a sum of points is the sum of a rotation of points. But the parallelogram law settles the matter (intuitively). The point is that if rotation is a linear transformation it is determined by its action on the basis vectors $ (1, 0), (0, 1)$. A rotation by a counterclockwise angle of $ \theta$ takes $ (1, 0)$ to $ (\cos \theta, \sin \theta)$ (where, for all intents and purposes, this can be taken to be a definition of the sine and cosine functions) and $ (0, 1)$ to $ ( - \sin \theta, \cos \theta)$. It therefore follows that since

$ (x, y) = x(1, 0) + y(0, 1)$

a rotation by $ \theta$ takes $ (x, y)$ to $ (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)$. Furthermore, we immediately obtain the matrix representation

$ \left[ \begin{array}{cc} \cos \theta & - \sin \theta \\
\sin \theta & \cos \theta \end{array} \right] \left[ \begin{array}{c} x \\
y \end{array} \right] = \left[ \begin{array}{c} x \cos \theta - y \sin \theta \\
x \sin \theta + y \cos \theta \end{array} \right]$

which, among other things, reduces the angle addition formulas to a matter of matrix multiplication. Before I develop the next point immediately, I'd like to note that linear transformations can not only be composed (which gives a group structure) but can be added (i.e. their results can be added), giving a (commutative) ring structure. In general, this is the endomorphism ring of an object; here it is a matrix ring, in fact the ring $ \mathcal{M}_2(\mathbb{R})$ of $ 2 \times 2$ matrices with entries in $ \mathbb{R}$. The rotations span a (commutative) subring. What is it?

In fact, precisely the matrices of the form $ \left[ \begin{array}{cc} a & - b \\
b & a \end{array} \right]$. Verify that the matrices of this form are closed under both addition and multiplication; moreover, any such matrix can be uniquely written as the product of a scalar multiplication by $ \sqrt {a^2 + b^2}$ and a rotation matrix. In other words, the subring of the endomorphism ring spanned by the group of rotations in $ \mathbb{R}^2$ is the ring of rotations-and-scalings.

To gain some intuition of why this is true geometrically (although it is easy to verify algebraically), take a point $ P$ and apply two transformations, both of which are rotations-and-scalings (say, by angles $ \alpha, \beta$ and scale factors $ a, b$) to get points $ A, B$, and then add them to get the point $ A + B$. The points $ A, B, P, A + B$ form a quadrilateral with known angles and side lengths; in particular, the angle between $ A + B$ and $ P$ is fixed and the ratio of the length of the diagonal $ A + B$ and the length of the vector $ P$ (and here I am carelessly interpreting points as vectors and vectors as points) is also fixed. Hence the set of rotations-and-scalings is closed under addition.

The ring of rotations-and-scalings is in fact a field; the inverse of a rotation by $ \theta$ and scaling by $ k$ is a rotation by $ - \theta$ and a scaling by $ \frac {1}{k}$, which is undefined only when $ k = 0$ - but this is the zero transformation. On the other hand, we can regard this ring as a real vector space (of matrices!) and then it clearly has basis

$ \left[ \begin{array}{cc} 1 & 0 \\
0 & 1 \end{array} \right], \left[ \begin{array}{cc} 0 & - 1 \\
1 & 0 \end{array} \right]$,

the first of which is the identity transformation and the second of which is rotation by $ 90^{\circ}$. (This choice of basis is the most natural given the presentation we gave above because we chose the basis $ (1, 0), (0, 1)$ for $ \mathbb{R}^2$.) We get the complex numbers by identifying the second basis element with $ i$ and the first with $ 1$.

Again, there is nothing "imaginary" about rotation. The rotation by $ 90^{\circ}$ is a natural candidate for the "square root of minus one" because two rotations by $ 90^{\circ}$ is a rotation by $ 180^{\circ}$ - the negative of the identity matrix. Now, we have identified the complex numbers with matrices rather than vectors as the Argand diagram would suggest. But the vector $ \left < a, b \right >$ corresponding to the complex number $ a + bi$ is just the image of $ \left < 1, 0 \right >$ under the rotation matrix $ \left[ \begin{array}{cc} a & - b \\
b & a \end{array} \right]$. The map that takes a linear transformation to its image on some fixed vector is a type of evaluation character and is in fact a homomorphism of algebras.

The matrix exponential naturally leads to Euler's formula here. In fact, letting $ \mathbf{J} = \left[ \begin{array}{cc} 0 & - 1 \\
1 & 0 \end{array} \right]$ (the engineer's convention for $ \imath$ so as to distinguish it from the identity, and not to be confused with the split-complex $ j$), we have exactly

$ e^{\mathbf{J} \theta} = \mathbf{I} \cos \theta + \mathbf{J} \sin \theta = \left[ \begin{array}{cc} \cos \theta & - \sin \theta \\
\sin \theta & \cos \theta \end{array} \right]$

which is now no longer mysterious but a direct consequence of the fact that the solution to the differential equation

$ \mathbf{P}' = \mathbf{JP}$

is a curve whose tangents are perpendicular, at every point, to the line from the origin to that point; in other words, a circle. Some of the familiar properties of the complex numbers are therefore beginning to emerge, as well as their intrinsic ties to rotation. In fact, this is a great intuition for understanding the Cauchy-Riemann equations. As MathWorld states, a function is complex-differentiable if and only if its Jacobian is a rotation-and-scaling matrix. Thus complex-differentiable functions (with nonzero derivative) are conformal maps - that is, locally they behave approximately like rotations. This is the complex analogue of the notion that a real differentiable function is locally approximately linear.

Still, there are many more interesting transformations to study besides rotations, so we have not quite resolved the question of why the complex numbers are so fascinating. The next formulation is perhaps the first really suggestive one.


Formulation 3: The complex numbers are an algebraic closure of the real numbers.

This formulation is perhaps the first one a student encounters but is, in my opinion, the most mysterious. The fundamental theorem of algebra is a surprisingly difficult beast to tame. Its statement does not even require the notion of a complex number: it is equivalent to the statement that every real polynomial can be written as the product of irreducible polynomials of at most second degree. Nevertheless, as the Wikipedia article puts it,

"In spite of its name, there is no known purely algebraic proof of the theorem, and many mathematicians believe that such a proof does not exist."

I find the complex-analytic proofs the most enlightening (perhaps because, as a number theorist, I routinely marvel at the explicative power of complex analysis). One of the proofs I learned recently, the second proof in the Wiki article, makes use of

Liouville's Theorem: Every bounded entire function is constant.

Liouville's Theorem is just one of many examples in complex analysis of an extremely rigid statement about complex-differentiable functions that simply does not hold in real analysis. Many examples of bounded functions analytic everywhere (such as the sine and cosine) abound. What happens, without going into too much detail, when we move into the complex numbers? $ \sin$ and $ \cos$ can perhaps be used as a case study: while they are bounded on the real line, the identities

$ i \sin ix = \text{sinh } x = \frac {e^x - e^{ - x}}{2}$
$ \cos ix = \text{cosh } x = \frac {e^x + e^{ - x}}{2}$

demonstrate that $ \sin z, \cos z$ are clearly unbounded on the purely imaginary line. Rather than go into a lot of detail about the proof of Liouville's Theorem, and keeping in mind that my experience with complex analysis is extremely limited, I propose the following (possibly flawed)

Intuition about the complex numbers: The complex numbers provide more "wiggle room" than the real numbers.

In less vague terms, perhaps Liouville's Theorem is true because the Fourier coefficients of a bounded non-constant analytic function on the real line force it to diverge once it has the appropriate imaginary inputs, and perhaps one should think of the Fundamental Theorem of Algebra as being a statement about polynomials having enough "wiggle room" to always have roots. (This is perhaps a trivial rephrasing.)

In any case, we now see why $ \mathbb{R}[x]/(x^3 - x + 1)$ is uninteresting: over the reals, any polynomial of higher than second degree is reducible, so any such construction involving polynomials of degree $ 3$ or higher does not produce an integral domain. We must therefore contend ourselves with $ \mathbb{R}[x]/(ax^2 + bx + c)$ where $ ax^2 + bx + c$ is irreducible (over the reals, which means it has no real roots) - but it's a straightforward exercise to show that these fields are all isomorphic to $ \mathbb{R}[x]/(x^2 + 1)$.

In some sense, therefore, the complex numbers are an "inevitable" (that is, special) extension of the real numbers. The power of complex analysis may also convince us that the complex numbers are extremely interesting. The next formulation makes the first intuition precise.


Formulation 4: The complex numbers are only commutative normed division algebra besides the reals.

Corollary: The complex numbers are the unique (up to isomorphism) algebraic closure of the reals.

This is a consequence of Hurwitz' s theorem and generalizes the observation we made about quotient rings and is perhaps the most suggestive one yet. It confirms that the choice of the complex numbers as an object of study was not an accident or indeed much of a choice at all (given the properties we'd like a nice extension of the reals to have). Indeed, the complex norm is crucial to defining notions of limit and convergence - in other words, to perform complex analysis.

This places Formulation 2 in an interesting light. The strongest condition above is really the notion of a norm, and on the complex numbers that notion of norm is Euclidean distance, so the role of the complex numbers as rotations becomes a straightforward consequence of the fact that they preserve distance, which is something of a "low-dimensional" coincidence: the orthogonal matrices are much more complicated in higher dimensions. It is an interesting geometric fact, on the other hand, that certain aspects of high-dimensional geometry are easier than low-dimensional geometry for certain abstruse reasons I don't understand well, but perhaps the complex numbers are exactly at the tipping point between too low-dimensional and too high-dimensional. In these respects complex analysis deserves its special place over the more general multivariable calculus (from which it certainly inherits traits, but retains a unique flavor).


Perhaps the next question to ask from here is why, specifically, complex analysis is so powerful. Unfortunately, this is a question I don't have the background to answer. (The answer seems to be related to aspects of the complex numbers I don't have the background to discuss, such as their topology.)


Practice Problem 1: Prove the claim I made in Formulation 3. More precisely, for a polynomial $ p(x)$ of degree greater than or equal to $ 2$, the quotient ring $ \mathbb{R}[x]/(p(x))$ is either not a field or is isomorphic to $ \mathbb{C}$.

Practice Problem 2: Prove that $ \mathbb{R}^3$ cannot be made into a division algebra over the reals.

Comment

4 Comments

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Good insights. It's funny how obvious it's become that you're through with contest math.

Your idea of defining $ \sin$ and $ \cos$ by the rotations is interesting. In my post about different metrics on $ \mathbb{R}^2$, which you commented on, I had also considered the isometries on those metrics, and this would result in a different definition. Although the alternate definition is kind of boring; just $ \sin^{2/ \alpha}$ and $ \cos^{2/ \alpha}$.

by MellowMelon, Jul 23, 2008, 1:29 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Excellent post! If possible, can you post the solution to the first practice problem? I would appreciate it. Thanks for the great posts and keep them coming.

by JZambrano201, Jul 29, 2008, 11:10 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
$ \mathbb{R}[x]/(ax^2+bx+c)$ is isomorphic to the reals of the form $ x + y \left( \frac{-b + i \sqrt{4ac-b^2}}{2a} \right)$, which is precisely the complex numbers (just with a different basis). This is because, just as $ x$ behaves like $ i$ in $ \mathbb{R}[x]/(x^2+1)$, $ x$ here behaves like one of the roots of $ ax^2 + bx + c$ (in that $ ax^2 \equiv -bx - c$).

by t0rajir0u, Jul 31, 2008, 8:38 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Alternatively, you can substitute $ x - t$ for an appropriate $ t$ to eliminate the $ x$ coefficient, and then substitute $ x/c$ to make the constant term $ 1$. However, there is similar messiness as you must show that the first transformation gives a positive constant term if the polynomial is irreducible.

by MellowMelon, Aug 2, 2008, 4:20 AM

Various mathematical thoughts, ranging from problem-solving techniques to attempts to tie together related concepts.

avatar

t0rajir0u
Archives
+ March 2009
+ October 2007
+ May 2007
Shouts
Submit
  • orz $~~~~$

    by clarkculus, Jan 10, 2025, 4:13 PM

  • Insanely auraful

    by centslordm, Jan 1, 2025, 11:17 PM

  • Fly High :(

    by Siddharthmaybe, Oct 22, 2024, 8:34 PM

  • Dang it he is gone :(( /revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive/revive.

    by 799786, Aug 4, 2022, 1:56 PM

  • annoying precision

    by centslordm, May 16, 2021, 7:34 PM

  • rip t0rajir0u

    by OlympusHero, Dec 5, 2020, 9:29 PM

  • Shoutbox bump xD

    by DuoDuoling0, Oct 4, 2020, 2:25 AM

  • dang hes gone :(

    by OlympusHero, Jul 28, 2020, 3:52 AM

  • First shout in July

    by smartguy888, Jul 20, 2020, 3:08 PM

  • https://artofproblemsolving.com/community/c2448

    has more.

    -πφ

    by piphi, Jun 12, 2020, 8:20 PM

  • wait hold up 310,000
    people visited this man?!?!??

    by srisainandan6, May 29, 2020, 5:16 PM

  • first shout in 2020

    by OlympusHero, Apr 4, 2020, 1:15 AM

  • in his latest post he says he moved to wordpress

    by MelonGirl, Nov 16, 2019, 2:43 AM

  • Please revive!

    by AopsUser101, Oct 30, 2019, 7:10 PM

  • first shout in october fj9odiais

    by bulbasaur., Oct 14, 2019, 1:14 AM

128 shouts
Tags
About Owner
  • Posts: 12167
  • Joined: Nov 20, 2005
Blog Stats
  • Blog created: Dec 5, 2006
  • Total entries: 48
  • Total visits: 321832
  • Total comments: 202
Search Blog
a