What is it that makes holomorphic functions so rigid?

Question

This semester I took a complex analysis class and, as far as I've seen, holomorphic functions on the complex plane have very "powerful" properties; For example, the identifying theorem, or the Casorati-Weierstrass theorem, Morera's Theorem, or even Gauss' mean value theorem.

Thing is, most of the theorems have very "light" conditions but lead to "heavy" results. One comes to the conclusion that the holomorphy of a function is itself a very strong property.

But what is it that makes it that way? In real analysis, one can say some things about differentiable functions, but not that much as one can say about complex-differentiable functions.

EDIT: As stated in the answers, complex-differentiability implies complex-analyticity and that is a big thing indeed. Actually, the proof of this theorem uses Cauchy's integral formula and the uniform convergence of a series on a circle. I'm already familiar with this things though and I was kinda hoping for an answer orientated around the fact that the field $\mathbb{C}$ is algebraicly closed. Does this algebraic property of the plane play a crucial role in the implicitation "differentiability $\rightarrow$ analyticity" ?

Answer 1

Most of those striking theorems (identity theorem, Liouville theorem, Cauchy inequality...) are rather direct consequences of the fact that holomorphic functions are analytic.

An analytic function is one that can be locally written on the form $f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$ near every $z_0$ in its domain of definition. The true magic is that being holomorphic (i.e. being complex differentiable) implies being analytic (something far from obvious, that usually takes a good chunk of a first course on complex analysis).

But if you admit that non-obvious fact, it should not be very surprising that analytic functions possess all kinds of strong properties (for one thing, all those coefficients $a_n$, hence all of the function, can be read with just local information near $z_0$).

Note that real analytic functions also satisfy those very strong properties, but of course as we know real differentiability does not imply real analyticity.

EDIT

I have no deep insight as to why complex differentiability implies analyticity whereas real differentiability does not. However, I can say that the fact that $\mathbb C$ is algebraically closed is not a sufficient reason. Indeed, there are algebraically closed and complete fields denoted by $\mathbb{C}_p$ (called $p$-adic numbers), on which this is not true. In fact, $\mathbb{C}_p$ is isomorphic as a field to $\mathbb{C}$ (though they are not homeomorphic). So this is more topological than algebraic in nature.

Answer 2

We get the feeling of surprise in the first encounter with complex variable theory because we are trained for years in the very "loose" world of real functions. However, even though the general theorems in calculus (e.g. mean value theorem) apply to all differentiable functions, all of the actual examples that we work with (differentiating and integrating) are analytic. Even when we define a function to be one expression in one interval, and a different one in another, it still is perfectly analytic (hence rigid) in each domain separately.

So, instead of thinking of functions of a complex variable as surprisingly rigid, it may be better to think of real non-analytic functions as being "too loose" or too general to be of much use in applications. That perhaps was the more dominant view up to (and including) Weierstrass's time.

Here's another twist. Even though a general continuous function on $\mathbb R$ appears to be too loose, one often(?) can extend it to a holomorphic function in a neighborhood just above the real line, so it is "as rigid as holomorphic functions". (You can't extend to both sides, unless the function was analytic to begin with.) In fact, Weierstrass's famous example of continuous but nowhere differentiable function was first discovered as the boundary value of a holomorphic function on the open unit disk.

(BTW, by taking the boundary value of a holomorphic function, one may also obtain generalized functions such as Schwartz's distributions.)

For another piece of "evidence" that holomorphic functions (of a complex variable) are "as many as" functions of a real variable, there is Paley-Wiener type theorem: smooth compactly-supported functions on $\mathbb R$ are in one to one correspondence, under Fourier transform, with entire functions with special growth condition.

Roughly speaking, then,

{holomorphic functions on a complex domain} $\approx$ {continuous (or generalized) functions on a real interval}.

With this dictum in mind, there must be a lot more holomorphic functions out there, more than one might expect from a first course in complex analysis. This partially accounts for the ubiquity of holomorphic functions in many disparate areas of mathematics, perhaps most prominently in number theory (L-functions and modular forms come to mind).

If you really want to compare the situation of real differentiable functions vs holomorphic functions, the "light" or "mild" condition (of being holomorphic) is actually very strong because in two dimensions there's a lot more "room" (or directions) for the condition to hold, instead of just the two directions in the real world.

What is actually "more loose" is the general (say, continuous) functions on the complex domain, or even $C^1$ functions in the multi-variable sense (i.e., as $\mathbb R^2\to \mathbb R^2$). The derivative at a point has four degrees of freedom (being a $2\times 2$ matrix), whereas a holomorphic function has two (rotation and dilation only).

Answer 3

Let $U$ be an open subset of $\mathbb{C}$ and $f\colon U\rightarrow\mathbb{C}$ be a map. Let us identify $\mathbb{R}^2$ and $\mathbb{C}$ through the following isometry $(x,y)\mapsto x+iy$, then $f$ is holomorphic at $z_0\in U$ if and only if $f$ is differentiable and $\mathrm{d}_{z_0}f$ is either zero or a direct similarity, namely there exists $(a,b)\in\mathbb{R}^2\setminus\{(0,0)\}$ such that: $$\mathrm{d}_{z_0}f=\begin{pmatrix}a&-b\\b&a\end{pmatrix}.$$ In fact, $f'(z_0)=a+ib$. From there, on can derives two facts:

• If $f$ is holomorphic at $z_0$, then $f$ satisfies the Cauchy-Riemann equation: $$\frac{\partial f}{\partial x}(z_0)=-i\frac{\partial f}{\partial y}(z_0).$$ This is a pretty strong condition that implies that $\textrm{Re}(f)$ and $\textrm{Im}(f)$ are harmonic maps.

• If $f$ is holomorphic at $z_0$ and its derivative is nonzero, then $f$ is a conformal mapping around $z_0$, namely if $\gamma_1$ and $\gamma_2$ are two smooth curves crossing at $z_0$ with angle $\theta$, then $f\circ\gamma_1$ and $f\circ\gamma_2$ cross at $f(z_0)$ with angle $\theta$. One can write this relation as: $$\frac{\langle{\gamma_1}'(0),{\gamma_2}'(0)\rangle}{\|{\gamma_1}'(0)\|\|{\gamma_2}'(0)\|}=\frac{\langle(f\circ\gamma_1)'(0),(f\circ\gamma_2)'(0)\rangle}{\|(f\circ\gamma_1)'(0)\|\|(f\circ\gamma_2)'(0)\|}.$$

I believe that this two consequences can give some insight on how strong being holomorphic is. The first point is actually equivalent to the holomorphy.