Is there a minimum set of properties that uniquely defines complex exponentiation?

Question

Starting with real numbers, the four basic operations, and limits, we can define exponentiation using the properties:
$a^0 = 1$ (base case for our purposes)
$a^1 = a$ (base case)
$a^{b+c} = a^ba^c$ (defines positive integer exponents)
$a^{-b} = 1/a^b$ (defines the rest of the integers)
$a^{n/m} = (a^{1/m})^n$ (defines rational exponents including $(-1)^{1/2}=i$ and necessitates a principal branch. $a$ is now a complex number)
$a^x = \lim_{c\to x}a^c$ (defines real exponents and necessitates continuous principal branch so $(-1)^{1/3}=\frac12+i\frac{\sqrt{3}}2$. $0^0$ is still defined by base case)
$a^i =\ ???$
$a^z =\ ???$

In order to extend the definition to include $a^z$ for any two complex numbers $(a, z)$, I still need some more properties. If I could prove even a single value of $a^z$ for a defined $a$ and non-real $z$ without assuming anything beyond the properties above (meaning we can't define e using complex exponents), I'd have everything I need, but I haven't been able to find a way to do so. Something like $e^i$ or $(-1)^i$ would be sufficient, but I'm unwilling to simply define $e^{ix} = \cos(x)+i\cdot \sin(x)$. The binomial theorem seems to be the crux of proving that $\frac{d}{dz}e^z=e^z$, but it's not obvious that it extends to complex exponents like it extends to rationals and it's not a natural-feeling property to include in the extension when compared to all the other properties.

Answer 1

Here's what I figured out. We need the property $$\lim_{z\to 0}\frac{a^z-1}{z} = \lim_{x\to 0}\frac{a^x-1}{x}$$ where $z$ is a complex number and $x$ is a real number. Since a can't be $0$ on either side we don't really need to explicitly include that exception. This property essentially sets the derivative of $a^z$ at $z=0$ equal to $\ln(a)$, making the function holomorphic. You can also simplify this to just setting the derivative of $e^z$ at $0$ equal to $1$ rather than using the arbitrary complex base $a$. After that the only thing left is to define which principal branch to use.

Answer 2

$a^i = \cos \omega + i\sin \omega$ where $\omega = \ln a$

$\cos$ and $\sin$ are defined in radians.

But what if $\cos$ and $\sin$ were defined in degrees?

There exists some $a$ such that $a^i = \cos 1^\circ + i\sin 1^\circ$

More abstractly $a$ defines some angle which could make the "unit angle." And $a^i = \cos_a 1 + i\sin_a 1$ with transformed base units for our trigonometric functions.

$a^{iy} = \cos_a y + i\sin_a y$
and $a^z = a^{x+ iy} = a^xa^{iy}$

Answer 3

I've been playing around with this for years and haven't gotten very far, in part because I'm trying to avoid using calculus.

Suppose $e^{ix}$ is some complex number for real x, Euler's Constant e and $i$ the imaginary unit.

$$e^{ix}=a+bi$$ $$e^{-ix}=\frac{a-bi}{a^2+b^2}$$

So with minimum assumptions, we have that inverting the complex exponent gives us a real multiple of the complex conjugate.

Now $x$ was arbitrary, it could have been $-x$. For reciprocation and argument negation to be compatible, we need $a^2+b^2=1$. So the modulus is 1.

So for reals $x$ and $t$, we have $e^{ix}=\cos{t}+i\sin{t}$, but we for some $t$ given x.

Were you to assume $x=f(t)$ and $x(0)=1$, you can get $x=t$ using calculus from here. Then you'd be done. But that's cheating!

Suppose $e^{ix}=\cos{u}+i\sin{u}$ and $e^{iy}=\cos{v}+i\sin{v}$.

Then with some trig identities, $e^{i(x+y)}=\cos{(u+v)}+i\sin{(u+v)}$.

DeMoivre's Theorem gives some meaning to the associated exponentiation/multiplying the argument by a real.

It looks like several plain algebraic properties of complex numbers and real exponentiation give some structure to the notion of complex exponentiation, but I'm not sure how to hammer down Euler's Formula without calculus.