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Bourbaki shows in a very natural way that every continuous group isomorphism of the additive reals to the positive multiplicative reals is determined by its value at $1$, and in fact, that every such isomorphism is of the form $f_a(x)=a^x$ for $a>0$ and $a\neq 1$. We get the standard real exponential (where $a=e$) when we notice that for any $f_a$, $(f_a)'=g(a)f_a$ where $g$ is a continuous group isomorphism from the positive multiplicative reals to the additive reals. By the intermediate value theorem, there exists some positive real $e$ such that $g(e)=1$ (by our earlier classification of continuous group homomorphisms, we notice that $g$ is in fact the natural log).

Notice that every deduction above follows from a natural question. We never need to guess anything to proceed.

Is there any natural way like the above to derive the complex exponential? The only way I've seen it derived is as follows:

Derive the real exponential by some method (inverse function to the natural log, which is the integral of $1/t$ on the interval $[1,x)$, Bourbaki's method, or some other derivation), then show that it is analytic with infinite radius of convergence (where it converges uniformly and absolutely), which means that it is equal to its Taylor series at 0, which means that we can, by a general result of complex analysis, extend it to an entire function on the complex plane.

This derivation doesn't seem natural to me in the same sense as Bourbaki's derivation of the real exponential, since it requires that we notice some analytic properties of the function, instead of relying on its unique algebraic and topological properties.

Does anyone know of a derivation similar to Bourbaki's for the complex exponential?

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Is it fair to say that anything relying on getting $\cos\theta+i\sin\theta=e^{i\theta}$ from the power series is too analytic/not what you're looking for? –  Isaac Jul 29 '10 at 11:12
    
Yes. The case of representing exp by its taylor series is going to be the easiest problem of this kind (since we have an algebro-topological description (up to change of base) of the map that doesn't depend on any sort of analytic intuition. I can't see how we can define sine and cosine on the complex plane in any useful way without depending on Euler's theorem (or repeating the logic I applied to extend exp above), which follows from the characterization of the imaginary exponential as rotation. –  user126 Jul 29 '10 at 11:21
    
I don't think you need to be able to define sine and cosine other than on the real line to get to Euler's theorem, but if you have Euler's theorem, then you can connect the imaginary exponential to rotations (using real sine and cosine to perform the rotations), then get to polar-form complex as $re^{i\theta}$, and on to $e^{a+bi}$ from there... or some such. It was really just a vague idea, not fully-formed. –  Isaac Jul 29 '10 at 11:46
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The problem is not a problem of defining maps. It is the problem of deriving these things naturally without Taylor's theorem. –  user126 Jul 29 '10 at 11:54
    
You can obtain it simply by integrating a certain function (don't remember exactly which offhand) - would that count? –  BlueRaja - Danny Pflughoeft Jul 29 '10 at 13:11

6 Answers 6

up vote 13 down vote accepted

I think essentially the same characterization holds. The complex exponential is the unique Lie group homomorphism from $\mathbb{C}$ to $\mathbb{C}^*$ such that the (real) derivative at the identity is the identity matrix.

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Great! This was what I was looking for. –  user126 Jul 29 '10 at 17:04

So, what's unnatural about the complex differential equation... f : C -> C satisfying f'(z) = f(z) and f(0)=1 ?

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I guess that works, but it's not really pretty like the Bourbaki construction, which involves only the topological and algebraic structure up to base change. However, I guess this is a satisfying enough answer should no other answers be given. –  user126 Jul 29 '10 at 14:21

Also, if you are OK with power series, the Prologue to Walter Rudin's Real and Complex Analysis seems like exactly what you want. It's a beautiful development of exp, as well as sin, cos, e, and even π, all quite organically.

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Let

f(x) = cos(x) + i*sin(x)

Then

df/dx = -sin(x) + i*cos(x)
      = i*f(x)

So that

∫(1/f(x)) df = ∫i dx
ln(f(x)) = ix  + C
f(x) = e^(ix + C) = cos(x) + i*sin(x)

Since f(0) = 1, C = 0, so

e^(ix) = cos(x) + i*sin(x)

(I will update this with LaTeX when that functionality becomes available)

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What is your definition of sin and cos? –  BBischof Jul 29 '10 at 15:14
    
@BBischof: I'm not sure I understand your question. Though f(x) is a complex function, x is real, and no complex-analysis is used in this proof (other than the trivial ∫i dx = ix + C) - so the usual high-school definitions of sin/cos apply. –  BlueRaja - Danny Pflughoeft Jul 29 '10 at 16:27
    
@Downvoter: Why the downvote? This derivation is correct. –  BlueRaja - Danny Pflughoeft Jul 29 '10 at 18:20
    
Whether or not something is correct doesn't have anything to do with whether or not it answers the question. –  user126 Jul 29 '10 at 22:09
    
@97832123: ...I specifically asked you if this answer would be alright before posting it. –  BlueRaja - Danny Pflughoeft Jul 29 '10 at 22:24

Some Assumptions

I will assume that you are ok with power series being used, just not Taylor's theorem. I will also assume you will allow us to observe a solution to a DE since you used it in your derivation.

Defn A series of the form $\sum_{n=0}^{\infty}c_n\left(z-z_0\right)^n$ for $c_n,z,z_0\in\mathbb{C}$ is called a power series.

Thm There is some $R\in[0,\infty]$ such that the power series above converges absolutely for all $z\in\mathbb{C}$ with $\mid z-z_0\mid < R$ and uniformly in $D\left(z_0,\rho\right)$ for all $\rho < R$. Further, the terms are unbounded for all $z$ with $\mid z-z_0\mid > R$.

pf Use the geometric series' convergence.

Lemma Inside the disk of convergence $\sum_{n=1}^{\infty}nc_n\left(z-z_0\right)^{n-1}$ is the derivative of the power series.

Construction

For power functions $y(z)=\sum c_nz^n$ can we find a unique solution to $y'(z)=y(z)$ in $\mathbb{C}$? We can observe that this implies $nc_n=c_{n-1}$. Hence

Defn Let $E\left(z\right):=\sum_{n=0}^{\infty}\frac{1}{n!}z^n$.

Thm 1) $E'=E$

2) $E\left(z_1+z_2\right)=E\left(z_1\right)E\left(z_2\right)$

3)$E_{\mid_{\mathbb{R}}}$ is strictly increasing and $E(\mathbb{R})=(0,\infty)$

4) $x\mapsto E(ix)$ sends $\mathbb{R}$ onto $\mathbb{T}$.

5) There is some real $\pi>0$ such that $E\left(\frac{\pi}{2}i\right)=i$ and for all $z_1, z_2\in\mathbb{C}$, then $E\left(z_1\right)=E\left(z_2\right)$ iff $\frac{z_1-z_2}{2\pi i}\in\mathbb{Z}$

6) $E\left(\mathbb{C}\right)=\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.

Note that all of these can be shown purely at the level of power series using no complex analysis. Further the proofs are not hard, if you want more details here let me know.

Corollary $E$ is a homomorphism of the additive group $\mathbb{C}$ onto the multiplicative group with $\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.

Application

If you care about loops(which I think you do!) lets observe that $\gamma(t)=e^{it}$ for $t\in[0,2\pi]$, is $\mathbb{T}$ (Note here that $\pi$ is simply the real number we found before, not some existential thing!). We could now push this to get winding numbers.

Or you could use this definition of exponential, and the power series definition of inverse to get a (branchless) logarithm. In particular, you can show that the derivative of that fella is $\frac{1}{x}$. So we don't have to define it that way. :)

Comments

I agree with you that the definitions of complex exponentials feel contrived and the logarithm is even worse. The branched logarithms is the only part of Palka that I dislike as a complex book. These definitions I can stomach, as they require no Dues ex Machina.

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My Tex Code is correct, I don't know why it looks so crappy. –  BBischof Jul 29 '10 at 15:07
    
I fixed it, apparently \dfrac doesn't work. –  BBischof Jul 29 '10 at 15:13
    
Also, you can do something similar for Sin and Cos, just start with the power series and show all the relevant properties. –  BBischof Jul 29 '10 at 15:14
    
This misses the point again. I realize that you can just "start with the power series", but that's pulling magical expressions out of thin air. –  user126 Jul 29 '10 at 17:01
    
What expression? The coefficients? They follow from the DE. –  BBischof Jul 29 '10 at 17:07

This is more a commentary @Charles Staats answer than a direct answer to the OP. In fact Charles' contribution generated the supplementary question: what about holomorphy? I think we can provide an alternative definition for the complex exponential:

Definition $\exp$ is the unique holomorphic group homomorphism $\mathbb{C}\to \mathbb{C}^\star$ such that its derivative at the origin is $1$.

It is quite easy to show such a homomorphism is indeed unique. It is less easy to prove its existence if starting from scratch. One way I have found illuminating is the following (based on the Italian classic Analisi matematica by Giovanni Prodi, chapter 6: "Funzioni esponenziali e circolari")

We can show that every continuous group homomorphisms $\mathbb{R}\to \mathbb{C}^\star$ (the latter is the complex multiplicative group) is automatically differentiable (Prodi, §39.1 - sorry, I don't know an English reference. I'm quite sure this is well known, though). Also, it is uniquely determined by its derivative at the origin. This furnishes a (IMHO) very effective definition for real exponential and trigonometric functions:

  • $x \mapsto e^x$ is the unique continuous group homomorphism $\mathbb{R}\to \mathbb{C}^\star$ s.t. its derivative at the origin is $1$;
  • $x \mapsto \cos x+i \sin x$ is the unique continuous group homomorphism $\mathbb{R}\to \mathbb{C}^\star$ s.t. its derivative at the origin is $i$.

The images of such homomorphisms are the subgroups $\mathbb{R}^+$ and $\mathbb{S}^1$ respectively. Turns out that their direct product $\mathbb{R}^+\times \mathbb{S}^1$ is the whole of $\mathbb{C}^\star$, and this is why the mapping

$$f(a+ib)=e^a(\cos b + i \sin b)$$

is a surjective homomorphism $\mathbb{C} \to \mathbb{C}^\star$. We can check directly that it is holomorphic at the origin with complex derivative $1$ and so we can rightfully call it complex exponential.


A substantial difference between this definition and Charles' one is that the adjective holomorphic is really necessary here. To wit, first note that every continuous group homomorphism $f \colon \mathbb{C} \to \mathbb{C}^\star$ is automatically (real) differentiable: if we define two mappings $\mathbb{R} \to \mathbb{C}^\star$ by the equations

$$f_1(a)=f(a+i0),\ f_2(b)=f(0+ib),$$

we get two continuous group homomorphisms $\mathbb{R}\to \mathbb{C}^\star$ and appealing to the previous differentiability result we easily show that $f_1, f_2$ are $C^\infty$ mappings. And so the same holds for $f$: however, it needs not be holomorphic: take for example

$$f(a+ib)=e^{2a}(\cos b + i \sin b).$$

Ok, I'm finished. Just wanted to share my thoughts with the community.

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Thanks for your thoughts. The adjective "holomorphic" is also necessary here because without it, the derivative at the origin is a 2x2 matrix rather than a single complex number. (Incidentally, the @name functionality seems to be useless in the body of a question.) –  Charles Staats Jun 9 '11 at 19:08

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