An "elementary" approach to complex exponents?

Solution 1:

Short version: If we try to extend the functions $x\mapsto a^x$ to complex exponents in the same way we extended them to reals (essentially, by requiring that $a^{m+n}=a^ma^n$ should still hold), we are left with a lot of ambiguity: there is a large family of plausible extensions. The problem is to pick out the particular extension we want. I'll show how we can narrow down the options considerably using reasonable criteria, but the final step is exactly the normalization of a logarithm (i.e., the choice of base) which, I believe, cannot be justified without appeal to calculus (which would open the door to power series definitions, etc., which you forbade). This perhaps explains why all proofs of Euler's formula make crucial use of calculus in one way or another.

Long version:

As I said in my comment, the way we extend the definition from $\mathbb N$ to $\mathbb Z$, and then to $\mathbb Q$, is by asking that the law $a^{m+n}=a^ma^n$ continue to hold on the larger domains; we extend to $\mathbb R$ by asking that the function be continuous. So in my view, the "elementary" definition of exponentiation for reals is this:

Elementary definition of real exponentiation. For any $a\in\mathbb{R}^+$, there is exactly one function $\varphi\colon\mathbb R\to\mathbb R$ satisfying these conditions:

1. $\varphi(1) = a$;
2. $\varphi(x+y)=\varphi(x)\varphi(y)$ for all $x,y\in\mathbb{R}$; and
3. $\varphi$ is continuous.

We write $\varphi(x) = a^x$.

(I consider only positive bases, for simplicity.)

Let's see what functions $\varphi\colon\mathbb C\to\mathbb C$ satisfy these conditions. First note that $$ \varphi(x+iy) = \varphi(x)\varphi(iy) = a^x \varphi(iy) \text{ ,} $$ so really it's a matter of choosing the function $y\mapsto\varphi(iy)$. To describe the candidates for this function, I need the following functional characterization of the trigonometric functions.

Proposition 1. Suppose $C,S\colon\mathbb R\to\mathbb R$ satisfy these conditions:

1. $C$ and $S$ are continuous;
2. $C(u-v) = C(u)C(v)+S(u)S(v)$ for all $u,v\in\mathbb R$;
3. $S(u-v) = S(u)C(v)-C(u)S(v)$ for all $u,v\in\mathbb R$;
4. $C$ and $S$ are not both identically zero.

Then there exists $\lambda\in\mathbb R$ such that

$$ C(u) = \cos(\lambda u) \quad\text{and}\quad S(u) = \sin(\lambda u) \text{ .} $$

The proof is long, so I defer it to an appendix below. Let's continue with identifying the candidate functions $y\mapsto\varphi(iy)$.

Proposition 2. Suppose $\psi\colon\mathbb R\to\mathbb C$ satisfies these conditions:

1. $\psi(0)=1$;
2. $\psi(u+v) = \psi(u)\psi(v)$ for all $u,v\in\mathbb R$; and
3. $\psi$ is continuous.

Then there exist $\lambda\in\mathbb R$ and $\mu\in\mathbb R^+$ such that

$$ \psi(u) = \mu^u(\cos(\lambda u) + i\sin(\lambda u)) \text{ .} $$

Proof. Note that $|\psi(u+v)| = |\psi(u)\psi(v)| = |\psi(u)|\,|\psi(v)|$. By the definition of real exponentiation, $$ |\psi(u)| = |\psi(1)|^u $$ for all $u\in\mathbb R$; let $\mu=|\psi(1)|$. In particular, $|\psi(u)|$ is never zero, so we can define $$ C(u) = \text{Re}\ \frac{\psi(u)}{|\psi(u)|} \quad\text{and}\quad S(u) = \text{Im}\ \frac{\psi(u)}{|\psi(u)|} \text{ .} $$ These functions are continuous, because $\psi$ is, and they are not both identically zero, because $\psi$ is not. Furthermore, \begin{align*} C(u-v) &= \text{Re}\ \frac{\psi(u-v)}{|\psi(u-v)|} \\ &= \text{Re}\ \left(\frac{\psi(u)}{|\psi(u)|}\right) \left(\frac{\psi(v)}{|\psi(v)|}\right)^{-1} \\ &= \text{Re}\ (C(u)+iS(u))(C(v)-iS(v)) \\ &= C(u)C(v)+S(u)S(v) \end{align*} and similarly for $S(u-v)$. By Proposition 1, there exists $\lambda\in\mathbb R$ such that $$ \psi(u) = |\psi(u)|(C(u)+iS(u)) = \mu^u(\cos(\lambda u)+i\sin(\lambda u)) $$ as desired.

Thus we can describe all the $\mathbb C\to\mathbb C$ extensions:

Corollary. Let $a\in\mathbb{R}^+$. Suppose $\varphi\colon\mathbb C\to\mathbb C$ satisfies these conditions:

1. $\varphi(1) = a$;
2. $\varphi(x+y)=\varphi(x)\varphi(y)$ for all $x,y\in\mathbb{R}$; and
3. $\varphi$ is continuous.

Then there exist $\lambda\in\mathbb R$ and $\mu\in\mathbb R^+$ such that

$$ \varphi(x+iy) = a^x \mu^y (\cos(\lambda y) + i\sin(\lambda y)) $$

for all $x,y\in\mathbb R$.

At this point I think it's natural to pick out the case $\mu=1$. For one thing, the curves $\psi(\mathbb R)$ are logarithmic spirals, except in the case $\mu=1$, which is the unit circle, which is qualitatively distinct. For another thing, the exponentiation part of $\psi$ (the $\mu^u$) is behaviour we already have; it's the rotation part (the $\cos$ and $\sin$) which is new. It seems most tidy to have $\varphi(x)$ do the exponentiation and $\varphi(iy)$ do the rotation.

One elegant algebraic criterion which picks out the $\mu=1$ case is to ask that $\varphi$ respect conjugation, i.e., $\varphi(\overline z)=\overline{\varphi(z)}$. This requires the curve $\psi(\mathbb R)$ to be symmetrical under reflection in the real axis, which excludes all the logarithmic spirals. Alternatively, we could require that $\psi(\mathbb R)$ be bounded, or that it be bounded away from zero. I'll state the conjugation version:

Corollary. Let $a\in\mathbb R^+$. Suppose $\varphi\colon\mathbb C\to\mathbb C$ satisfies these conditions:

1. $\varphi(1) = a$;
2. $\varphi(w+z)=\varphi(w)\varphi(z)$ for all $w,z\in\mathbb C$;
3. $\varphi$ is continuous; and
4. $\varphi(\overline z) = \overline{\varphi(z)}$ for all $z\in\mathbb C$.

Then there exists $\lambda\in\mathbb R$ such that, for all $x,y\in\mathbb R$,

$$ \varphi(x+iy) = a^x (\cos(\lambda y) + i\sin(\lambda y)) \text{ .} $$

(Some treatments of complex exponentiation present this conjugation condition as if it were a provable conclusion and not an imposed condition; see this question.)

We still have the ambiguity of $\lambda$, and indeed, nothing said so far gives any clue about what $\lambda$ we should take for a given $a$. We don't even have any connection between the choice of $\lambda$ for different bases. The latter issue can be partly addressed by also assuming the rule $(ab)^z=a^zb^z$. Thus:

Proposition 3. Let $\lambda\colon\mathbb R^+\to\mathbb R$. For $a\in\mathbb R^+$, define

$$ \varphi_a(x+iy) = a^x (\cos(\lambda(a)y) + i\sin(\lambda(a)y)) $$

If these functions satisfy the condition $\varphi_{ab}(z) = \varphi_a(z)\varphi_b(z)$, then $\lambda$ satisfies $\lambda(ab)=\lambda(a)+\lambda(b)$.

The proof is direct, so I omit it.

At this point it's natural to take $\lambda(a) = \log_c(a)$ for some $c$; then we get Euler's formula $c^{i\pi} = -1$. To justify the normalization $c=e$ requires, I believe, ideas from calculus, which pretty much opens the door to the series definition of the exponential, and the continuously compounded interest definition, which you forbade.

Another perhaps interesting issue is that the functional equation $\lambda(ab)=\lambda(a)+\lambda(b)$ apparently doesn't quite characterize the logarithms; on Wikipedia, at least, there is only mention of a characterization which also assumes $\lambda$ is increasing, which seems quite artificial in our context. So there is something more to pin down here, I think.

Appendix: Proof of Proposition 1

(My proof here draws heavily on Robison, "A New Approach to Circular Functions, $\pi$, and $\lim (\sin x)/x$", Math. Mag. 41 (1968), 66–70, jstor, which gives a different functional characterization which is very nice but not quite adapted to my needs here.)

Lemma. $S(0)=0$. Proof: Take $u=v$ in condition 3.

Lemma. $C(u)^2+S(u)^2=C(0)$. Proof: Take $u=v$ in condition 2.

Lemma. $C(0)\ne 0$. Proof: Otherwise $C(u)^2+S(u)^2=0$ for all $u$, so $C$ and $S$ would be identically zero, contrary to condition 4.

Lemma. $C(0)=1$. Proof: $C(0)=C(0)^2+S(0)^2=C(0)^2$, which implies $C(0)=1$ because $C(0)\ne 0$.

Corollary. $C(u)^2+S(u)^2=1$.

Lemma. $C(-u)=C(u)$. Proof. $C(-u) = C(0-u) = C(0)C(u)+S(0)S(u) = C(u)$.

Lemma. $S(-u)=-S(u)$. Proof. Similar.

Corollary. $S(u+v) = S(u)C(v) + C(u)S(v)$.

Corollary. $C(u+v) = C(u)C(v) - S(u)S(v)$.

Corollary. $C(2u) = C(u)^2 - S(u)^2$.

Corollary. $C(2u) = 2C(u)^2 - 1$.

Lemma. If $C$ is not constant then $C$ has a root. Proof. Let $y_0\in\mathbb R$ be such that $C(y_0)\ne 1$; since $C(y_0)^2 = 1 - S(y_0)^2 \le 1$, we have $C(y_0) < 1$. Define $a_n = C(2^ny_0)$ and $L=\inf_n a_n$. Then $$ L = \inf_{n\ge 0} a_n \le \inf_{n\ge 0} a_{n+1} = \inf_{n\ge 0} (2a_n^2-1) = 2\inf_{n\ge 0} a_n^2 - 1 \le 2L^2-1 $$ and so either $L\ge 1$ or $L\le-\frac12$. If $L\ge 1$ then $C(y_0)=a_0\ge 1$, contrary to our choice of $y_0$; so $L\le-\frac12$. In particular, $C$ takes negative values (as well as the positive value $C(0)=1$), so by IVT, $C$ has a root.

In view of this last lemma, we break into cases.

Case 1. $C$ is constant. Then $C$ is constantly $1$, and $S=\pm\sqrt{1-C^2}$ is constantly $0$, so we can take $\lambda=0$ in the proposition.

Case 2. $C$ is not constant. Let $p$ be the smallest positive root of $C$. (Roots exist as just shown; positive roots exist because $C$ is even, as shown above; and a smallest such root exists because $C$ is continuous.) We continue as follows:

Lemma. If $u\in[0,p)$ then $C(u)>0$. Proof. $C(0)>0$, $C$ is continuous, and $p$ is its smallest positive root.

Corollary. If $u\in[0,p)$ then $C(u) = \sqrt{\frac12(1+C(2u))}$.

Corollary. $C(p/2^n) = \cos(\pi/2^{n+1})$ for all $n\in\mathbb N$. Proof. By induction.

Lemma. $S(p)=\pm 1$. Proof. $S(p)^2=1-C(p)^2=1$.

Lemma. $S(u)=S(p)C(p-u)$. Proof. $S(u)=S(p-(p-u))$ and $C(p)=0$.

Corollary. If $u\in(0,p]$ then $S(u)$ has the same sign as $S(p)$.

Corollary. If $u\in(0,p]$ then $S(u) = S(p)\sqrt{1-C(u)^2}$.

Corollary. $S(p/2^n) = S(p)\sin(\pi/2^{n+1})$ for all $n\in\mathbb N$.

Now to complete the proof of the proposition. Take $\lambda = S(p)\pi/2p$. We have already shown that $C(u) = \cos(\lambda u)$ and $S(u) = \sin(\lambda u)$ for $u = p/2^n$ with $n\in\mathbb N$. (Note that $\cos(S(p)v) = \cos v$ and $\sin(S(p)v) = S(p)\sin v$.) Using the addition formulas we can extend these identities to $u = mp/2^n$ with $m,n\in\mathbb N$; continuity then extends them to all $u\ge 0$; the evenness of $C$ and $\cos$ and the oddness of $S$ and $\sin$ then finish the job.

Solution 2:

So here's a good place to start

$$e^{i\theta}$$

Is interpreted as the complex number that is formed if you form a circle of radius 1 in the complex number field. And starting from the point 1 + 0i you move along the circle for an angle $\theta$ to a new number in the complex number field:

$$\sin(\theta) + \cos(\theta)i)$$

Now note that ANY complex number is of the form

$$r e^{i\theta}$$

Where $r$ is the absolute value of the complex number (or) it's distance from the point 0. The value of the complex exponential simply indicates the angle. In other words we have polar coordinates here.

Then taking exponents becomes quite obvious with it simply distributing over both items which can be factored into complex exponential themselves.

If 'factor' your exponentials into a product of numbers of this format then it becomes intuitive what physically is occurring.

Hope that helps :)