Construction of the complex exponential, extending real identities to complex, and proving $e^{ix} = \cos x+i\sin x$ from those identities 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 $e^z$?
 A: The answer to the first question is indeed the identity theorem.
Suppose $f(z)$ has been extended to the complex plane by the infinite series. It is thus an entire function.
For fixed real $y$, the entire functions $g(z)=f(z)f(y)$ and $h(z)=f(z+y)$ coincide in the real axis which is certainly a set having an accumulation point. So they coincide on $\mathbb C$ and we have $f(z)f(y)=f(z+y)$ for complex $z$ and real $y$. In a second step, we fix some complex $x$ and consider $g(z)=f(x)f(z)$ and $h(z)=f(x+z)$. Again they coincide for real $z$ by the result of the first step and, again by the identity theorem coincide on $\mathbb C$. Thus $f(x)f(z)=f(x+z)$ for all complex $x,z$.
For property 3., one proceeds similarly. First, one has to prove that
$$g(z)=\lim_{n\to\infty}\left(1+\frac zn\right)^n$$
converges uniformly on compact subsets of $\mathbb C$. Then by property 3 for real z, $f(z)=g(z)$ for real $z$. The identity theorem again yields that $f(z)=g(z)$ for
all complex $z$. Observe that the convergence for complex $z$ has tobe proved. It does not follow from the identity theorem.
For $f'(z)=f(z)$, it is again the same: It is known that $g(z)=f'(z)$ and $f(z)$ coincide for real $z$. So these holomorphic functions must coincide on $\mathbb C$ by the identity theorem. Of course, it follows also easily from the power series definition of $f$ that $f'(z)=f(z)$ on $\mathbb C$.
Consider now the second question. First, the definition by the series shows that
$$f(z)=\overline{f(\bar z)}\mbox { for complex }z.$$
By the way, this could also be proved by the identity theorem (also requires analyticity of $\overline{f(\overline{z})}$: How do I rigorously show $f(z)$ is analytic if and only if $\overline{f(\bar{z})}$ is? -- EDIT: in the case that $f$ is entire and real on $\mathbb R$, this can be seen as a consequence of the Schwarz reflection principle, since that tells us $\tilde F(\bar z) := \overline{F(z)}$ is a holomorphic extension of $F$, and since $\tilde F = F$ on $\mathbb R$, the identity theorem forces $\tilde F = F$ on all of $\mathbb C$, i.e. $F(z) = \overline{F(\bar z)}$ for all $z\in \mathbb C$.)...
This implies that $f(-it)=\overline{f(it)}$ for all real $t$.
Therefore
$$f(it)\overline{f(it)}=f(it)f(-it)=f(0)=1$$
and hence the modulus $|f(it)|=1$ for all real $t$.
Let us now write
$$f(it)=c(t)+i\,s(t),\ t\in{\mathbb R}$$
with real valued functions $c,s$. Then we already have $c^2(t)+s^2(t)=|f(it)|^2=1$ for
all real $t$ and $c(0)=1$, $s(0)=0$. Therefore the matrix
$$U(t)=\begin{pmatrix}c(t)&s(t)\\-s(t)&c(t)\end{pmatrix}$$
satisfies $U(t)\,U(t)^T=I$ and hence is orthogonal. As it does not have real eigenvalues (unless $s(t)=0$ in which case $c(t)=\pm1$ and therefore $U(t)=\pm I$), it is indeed a rotation.
$$***$$
In a different approach, differentiation gives $\frac d{dt}f(it)=if(it)$ and separation of real and imaginary part give
$$c'=-s,\ s'=c.$$
Hence $c''=-c,\,c(0)=1,c'(0)=0$ and $s''=-s,s(0)=0,s'(0)=1$.
These are some well known properties defining $\sin$ and $\cos$.
One could now deduce that $c$ must have a positive zero, since otherwise, $s$ would be strictly increasing ($s'=c$) and the graph of $c$ must be below a certain straight line with negative slope ($c'=-s$) which leads to a contradiction. If  $p$ is the first such zero, one could show that $c,s$ are $4p$-periodic ($2p$ had been named $\pi$).
Finally, we obtain the angle addition formulae
$$c(x+y)=c(x)c(y)-s(x)s(y), s(x+y)=s(x)c(y)+c(x)s(y)$$
simply from our definition and $f(i(x+y))=f(ix)f(iy)$.
This also implies that $c$ must have a zero $p>0$: Indeed $s'(0)=c(0)=1$ implies that $s(\delta)>0$ for small positive $\delta$. As $c^2+s^2=1$ and $c$ is continuous, we must have $0<c(\delta)<1$ for small positive $\delta$. The angle addition formula implies that
$$c(2x)=c(x)^2-s(x)^2\leq c(x)^2\,\mbox{ for all }x.$$
Hence $c(2^n\delta)\leq c(\delta)^{2^n}<\sqrt{2}/2$ for sufficiently large $n$. Unless $c(2^k\delta)$ is already below $0$ for some $k\leq n$, we conclude
that $s^2(2^n\delta)=1-c^2(2^n\delta)>1/2>c^2(2^n\delta)$ and hence
$c(2^{n+1}\delta)=c^2(2^n\delta)-s^2(2^n\delta)<0$. In any case there exists
some positive integer $n$ such that  $c(2^n\delta)<0$. As $c$ is continuous and $c(0)=1$, we conclude that there exists $p>0$ such that $c(p)=0$. We can assume that $p>0$ is minimal with that property. As $c^2+s^2=1$, we conclude that $s(p)^2=1$. Now the angle addition formulae yield that $s(x)$ is positive as long as $0<x<p$ since $c$ is positive on $[0,p[$. Hence $s(p)=1$.
The angle addition formulae then yield
$$c(x+p)=-s(x),\ s(x+p)=c(x)\mbox{ for all }x.$$
Hence $c(x+2p)=-s(x+p)=-c(x)$ and $s(x+2p)=c(x+p)=-s(p)$. As a consequence, $c$ and $s$ are $4p$-periodic and $4p$ is the minimal period. (We have $4p=2\pi$).
I hope all this sufficiently justifies that $f(it)$, $t$ real, is related to rotations.
It is well known that the only continuous functions $f:{\mathbb R}\to{\mathbb R}$ satisfying $f(x+y)=f(x)f(y)$ for all real $x,y$ are given by $f(x)=\exp(c\,x)$ with a certain constant $c$. See also here.
The constant can be determined using $f'(0)=c$; in the case of the classical exponential we have $c=1$.This is a characterisation of the exponential function like the Bohr-Mollerup Theorem. Other characterisations can be found here.
A: 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 $\lim_{n\to\infty}{n\choose i}\ /\ n^i = \frac1{i!}$.
