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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:

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

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 $e^z$?

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    $\begingroup$ P.S. It seems that $f(z) = \exp(\;\overline{z}\;)$ is continuous and satisfies $f(x+y)=f(x)f(y)$. $\endgroup$
    – GEdgar
    May 14, 2020 at 23:41
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    $\begingroup$ @GEdgar : But it’s not differentiable $\endgroup$
    – MPW
    May 14, 2020 at 23:48
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    $\begingroup$ You may already be aware, and I don't think this quite answers any of your questions, but it's relevant: The identity theorem for holomorphic functions gives you that since the maclaurin series converges on the real line, the analytic continuation is unique. $\endgroup$
    – Mark S.
    May 15, 2020 at 0:05
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    $\begingroup$ @MarkS. yes, this theorem is what I used above to do the extension from the real line to the complex plane. $\endgroup$
    – D.R.
    May 15, 2020 at 0:56
  • $\begingroup$ Property 3 can be used as definition by just letting $x\in\mathbb {C} $ instead of $x\in\mathbb{R} $. And you can use it to establish all properties of complex exponential function. The first part is proving property 2 using 3 and then proving that the limit exist (this uses the fact that limit exists when $x\in\mathbb {R} $). $\endgroup$
    – Paramanand Singh
    May 18, 2020 at 12:04

2 Answers 2

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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.

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  • $\begingroup$ for your derivation of the cos/sin formula, you first said that it's represented by $U(t)$, a rotation matrix, and then you proceed to do some differential equations stuff. Are these meant to be two separate explanations, or are they supposed to be related? I really would like to avoid differential equations if possible... $\endgroup$
    – D.R.
    May 23, 2020 at 20:26
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    $\begingroup$ They are two separate approaches. The first does not show that $c,s$ are periodic functions. In the second approach I indicate how this could be proved from the differential equations (they are natural!). Probably the existence of a zero of $c$ and thus the periodicity can also be proved from the angle addition formulae. $\endgroup$
    – Helmut
    May 23, 2020 at 21:38
  • $\begingroup$ Ok thanks for the response. Lastly, are there any "natural" seeming formulas that can not be extended to the complex plane via the identity theorem? It seems that the identity theorem is absurdly powerful. $\endgroup$
    – D.R.
    May 24, 2020 at 3:07
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    $\begingroup$ @D.R. Yes. For example, most formulas involving Fourier series cannot be extended. The reason is often that the series only converges for (a subset) of the real numbers. See here for an exmple. $\endgroup$
    – Helmut
    May 24, 2020 at 10:36
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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 $\lim_{n\to\infty}{n\choose i}\ /\ n^i = \frac1{i!}$.

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