TITLE: Construction of the complex exponential, extending real identities to complex, and proving $e^{ix} = \cos x+i\sin x$ from those identities
QUESTION [8 upvotes]: Let's say that we have already defined $f(x)=e^x$ on $\mathbb R$ as the solution to the equation $f'(x) = f(x)$ with $f(0)=1$, and let's say that we've proved the following three properties:

$f(x) = \sum\limits_{n=0}^\infty \frac{x^n}{n!}$
$f(x+y)=f(x)f(y)$
$f(x) = \lim\limits_{n\to\infty} (1+\frac xn)^n$

Now we want to extend this function to the entire complex plane analytically, and so (using the identity theorem) the continuation is $f(z) = \sum\limits_{n=0}^\infty \frac{z^n}{n!}$.

First question: I know that on $\mathbb C$, properties $2$, $3$, and $f'(z)=f(z)$ still hold. Is this a surprise, or coincidental? That is, in general is it true that if we have some formulas $F_1, \ldots, F_n$ (like the identities above, or things like continued fractions, etc) involving $g: \mathbb R\to \mathbb R$, will those formulas $F_1,\ldots, F_n$ hold on $\mathbb C$ as well if we analytically extend $g$ to the complex plane?

Now let's say that we have all these properties, and we want to use property $2$ to prove $e^{ix}=\cos x+i\sin x$. Well, following the lead of this 3b1b video at @18:50: https://www.youtube.com/watch?v=mvmuCPvRoWQ, (maybe start watching at around minute @18:30), Grant says that it "would be reasonable" to think that pure vertical shifts would result in pure rotations (i.e. exponentiating a pure imaginary would result in a number on the unit circle). Yes, this is reasonable, but how do we prove it? It seems that property $2$ alone (along with the fact that $f(x+i0)=e^x$ for all $x\in \mathbb R$) is not enough to nail down exactly the complex exponential. So:

What's the easiest step we need to take fully justify that pure vertical slides correspond to pure rotations? Note that I'm asking for a step starting from the "group-theoretic" framework Grant laid out in the video above; that is, I'm NOT asking for just any proof of $e^{ix}=\cos x+i\sin x$ using heavy calculus (like Taylor series, or differential equations).

P.S. Are there results like the Bohr-Mollerup theorem for $e^z$? Like is it true that any (continuous/differentiable?) function defined by $f(x+y)=f(x)f(y)$ (+ other conditions?) MUST be $e^z$?

REPLY [2 votes]: I want to add another perspective on a part of your question: property 2 isn't just a matter of the identity theorem, and it's not actually a matter of analysis at all. In fact, the identity is a formal identity in the power series ring $R[[x]]$ over any commutative ring $R$ for the function defined in (1). To see this, note that the coefficient of $x^iy^j$ in the product $f(x)f(y)$ is just $\frac{1}{i!}\frac{1}{j!}$; OTOH, the coefficient in the series $f(x+y)$, expanding each term $(x+y)^n$ using the binomial identity, is ${i+j\choose i}\cdot\frac1{(i+j)!}$. The series (1) is also the formal solution of the differential equation $f'(x)=f(x)$ with constant coefficient 1.
On the other hand, the identity (3) is not formal; see e.g. the discussion on Wikipedia's page on formal power series. For this identity you actually need a topology on the ring $R$ itself which satisfies $\lim_{n\to\infty}{n\choose i}\ /\ n^i = \frac1{i!}$.