Complex analytic functions show rigid behavior while realvalued smooth functions are flexible. Why is this the case?
9 Answers
Well, realvalued analytic functions are just as rigid as their complexvalued counterparts. The true question is why complex smooth (or complex differentiable) functions are automatically complex analytic, whilst real smooth (or real differentiable) functions need not be real analytic.
As Qiaochu says, one answer is elliptic regularity: complex differentiable functions obey a nontrivial equation (the CauchyRiemann equation) which implies a integral representation (the Cauchy integral formula) which then implies analyticity (Taylor expansion of the Cauchy kernel); the ellipticity of the CauchyRiemann equation is what gives the analyticity of its fundamental solution, the Cauchy kernel. Real differentiable functions obey no such equation.
Another approach is via Cauchy's theorem. In both the real and complex setting, differentiability implies that the integral over a closed (or more precisely, exact) contour is zero. But in the real case this conclusion has trivial content because all closed contours are degenerate in one (topological) dimension. In the complex case we have nontrivial closed contours, and this makes all the difference.
EDIT: Actually, the above two answers are basically equivalent; the latter is basically the integral form of the former (Morera's theorem). Also, to be truly nitpicky, "differentiable" should be "continuously differentiable" in the above discussion.

7$\begingroup$ Dear Terry, since I like nitpicking too, here are two results: a) if a function is only supposed differentiable in a domain of C, then its derivative is automatically continuous and the function is analytic. b) if a function is continuous in a domain in C and if its integrals over closed contours are zero, then the function is differentiable and a) applies (Morera). Since you know this infinitely better than I, I wonder what you meant in your last sentence . $\endgroup$ Commented Nov 2, 2009 at 20:34

3$\begingroup$ Ah, right, one doesn't need continuous differentiability to prove Cauchy's theorem. (One does need continuous differentiability for the real variable Stokes' theorem, which is why I was unnecessarily conservative, but of course one can do better in the complex differentiable case, thanks to elliptic regularity (or the slick BolzanoWeierstrass type proof by contradiction for Cauchy's theorem).) $\endgroup$ Commented Nov 2, 2009 at 21:40

8$\begingroup$ Historical note: I think Goursat was first to remove the continuous differentiability hypothesis from Cauchy's theorem. $\endgroup$– timurCommented Jan 5, 2012 at 22:54

1$\begingroup$ Yes, and Goursat was born the year after Cauchy died, so it was a long time coming! $\endgroup$ Commented Apr 14, 2012 at 17:50
One answer is that property of being complex analytic is equivalent to being a solution to a differential equation (namely the CauchyRiemann equations) whereas there is no analagous formulation for being a smooth function. Once you have this formulation it should be immediately clear that complex analytic functions are rigid because solutions to differential equations are rigid. The whole idea of boundary value problems and initial value problems is predicated on the fact that knowing a solution to differential equation in a small area determines its values everywhere, which is precisely the type of rigidity that complex analytic functions have.
So you might ask why do complex analytic functions satisfy the CauchyRiemann equations. Well every real smooth function from R^2 to R^2 has a derivative, which is a 2x2 matrix. Requiring the function to be complex differentiable is the same as requiring that matrix to be a "complex number", i.e. a matrix of the following form: first row = [a, b], second row = [b, a]. Well this condition is precisely the CauchyRiemann equations.

3$\begingroup$ I would add that it is a complex number only in the onedimensional (complex) situation. The CR equations actually say that the derivative map is complex linear and it is only in dimension one that the space of complex linear maps from C to C is canonically C itself. In higher dimensions, this is not the case. $\endgroup$ Commented Nov 4, 2009 at 3:24
The book Visual Complex Analysis gives a good explanation: locally, analytic functions are rotations and dilations. Disks go to disks. A smooth function of two real variables may map disks to ellipses. That is, a real valued function can distort disks in a way that analytic functions cannot.
To expand on Dinakar's comment about boundary value problems, the physical intuition one should have here is that the real and complex parts of a complex differentiable functions are harmonic functions (this is just a restatement of the CR equations). An important way harmonic functions arise is as solutions of the steadystate heat equation, so one can think of the values of a harmonic function on a contour as temperatures and the values of a harmonic function in the interior of the contour as the steadystate distribution of temperature determined by the distribution of temperature on the contour. When the contour is a circle one can compute this distribution by convolving with the Poisson kernel; this is a special case of Cauchy's integral formula. The Poisson kernel itself is the canonical example of a "good kernel," and it finds application in Fourier analysis for that reason. One generally expects convolution by a good kernel to have nice properties.
In turn, one reason why diffusion should have anything to do with complex differentiability is that in both cases one wants path integrals to be homotopy invariant, in the first case because path integrals should give the difference in potential and in the second case because this is the correct generalization of the fundamental theorem of calculus.
Here is a related discussion from combinatorics and more: http://gilkalai.wordpress.com/2009/06/29/testyourintuition6/#comment2057 A very nice explanation by John Baez (in two installments) can be found here
http://golem.ph.utexas.edu/category/2006/10/knowledge_of_the_reasoned_fact.html#c005571
and here
http://golem.ph.utexas.edu/category/2006/10/knowledge_of_the_reasoned_fact.html#c005623

$\begingroup$ I see that the second of these also explains elliptic regularity (from the comment under the OP). $\endgroup$ Commented Apr 14, 2012 at 21:09
Consider the definition of derivative at a point a, which is formally the same for real functions of a real variable and for complex functions of a complex variable:
f'(a) = lim_{h > 0}(f(a+h)f(a))/h.
In the real case, h is real, and approaches 0 along the real line, from the right and from the left. In the complex case, h is complex, and approaches 0 from any possible direction. This makes it more difficult for the limit to exist, and thus for complex functions of a complex variable to have a derivative. When such a function is viewed as a pair of real functions of two real variables, that is, when we write
f(x + i*y) = u(x,y) + i*v(x,y), u and v real valued,
the existence of the above limit translates into the CauchyRiemannn equations:
u_x = v_y, u_y = v_x .
This again shows that a function of the form u(x,y) + i*v(x,y) is smooth in the complex sense only if it satisfies the rigidity imposed by the CauchyRiemann equations.
A complex function is analytic if and only if locally it can be represented by a power series. This means that (at least locally) an analytic function is determined by countable data (namely, the Taylor coefficients of its local expansion). This is not true for smooth functions on the real line. In my thinking the rigidity of analytic functions stems from this countable determinacy.
Another instance of this countable determinacy is of course the identity theorem.

$\begingroup$ Yes, but I think this is just a rephrasing of the question. Why, conceptually, is analytic equivalent to complex differentiable? $\endgroup$ Commented Nov 2, 2009 at 17:59

8$\begingroup$ A smooth (or even just continuous) function on the real line can also be determined by just countable data: its value on rationals. $\endgroup$ Commented Apr 14, 2012 at 17:44
Simple answer: because there are so many realvalued smooth functions, and so few complex analytic functions.
(Which is just a smug way of rephrasing Armin Straub's answer. The key is that analyticity is such a strong requirement as opposed to smoothness.)