Why are differentiable complex functions infinitely differentiable?

Solution 1:

As Akhil mentions, the keyword is elliptic regularity. Since I don't know anything about this, let me just say some low-level things and maybe they'll make sense to you.

A differentiable function $f : \mathbb{R} \to \mathbb{R}$ can be thought of as a function which behaves locally like a linear function $f(x) = ax + b$. So, very roughly, it is a collection of tiny vectors which fit together. These tiny vectors can, however, fit together in a very erratic manner. That's because since you only have to fit one vector to the two vectors that are its neighbors, there is a lot of room for bad behavior.

A differentiable function $f : \mathbb{C} \to \mathbb{C}$ has to satisfy a much more stringent requirement: locally, it has to behave like a linear function $f(z) = az + b$ where $z, a, b$ are complex, which is a rotation (and scale, and translation). So, very roughly, it is a collection of tiny rotations which fit together. Now one rotation has a continuum of neighbors to worry about, and it becomes much harder for erratic behavior to persist.

Solution 2:

The existence of a complex derivative means that locally a function can only rotate and expand. That is, in the limit, disks are mapped to disks. This rigidity is what makes a complex differentiable function infinitely differentiable, and even more, analytic.

For a complex derivative to exist, it has to exist and have the same value for all ways the "h" term can go to zero in $\frac {(f(z+h) - f(z))}{h}$. In particular, h could approach 0 along any radial path, and that's why an analytic function must map disks to disks in the limits.

By contrast, an infinitely differentiable function of two real variables could map a disk to an ellipse, stretching more in one direction than another. An analytic function can't do that.

A smooth function of two variables could also flip a disk over, such as $f(x, y) = (x, -y)$. An analytic function can't do that either. That's why complex conjugation is not an analytic function.

Solution 3:

When one uses the complex plane to represent the set of complex numbers ${\bf C}$,

$z=x+iy$

looks so similar to the point $(x,y)$ in ${\bf R}^2$.

However, there is a difference between them which is not that obvious. The linear transformation in ${\bf R}^2$, can be represented by a $2\times 2$ matrix as long as one chooses a basis in ${\bf R}^2$, and conversely, any $2\times 2$ matrix can define a linear transformation by using the matrix multiplication $A(x,y)^{T}$.

On the other hand, the linear transformation on $\bf C$ is different. Let $f:{\bf C}\to{\bf C}$ where $f(z)=pz$, $p \in{\bf C}$. If one writes $p=a+ib$ and $z=x+iy$, this transformation can be written as

$$ \begin{bmatrix} x\\ y \end{bmatrix}\to \begin{bmatrix} a &-b\\ b &a \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} $$

when one sees it as in the complex plane. Hence, not all matrices can define a linear transformation $f:\bf C\to C$.

The derivative, which can be regarded as a "linear transformation", is also different for $f:{\bf R}^2\to {\bf R}^2$ and $f:\bf C\to C$. In the real case

$$ f \left( \begin{bmatrix} x\\ y \end{bmatrix} \right) = \begin{bmatrix} f_1(x,y)\\ f_2(x,y) \end{bmatrix} $$

$f_1$ and $f_2$ are "independent" for the sake of $f$ being differentiable.

While in the complex case $f_1$ and $f_2$ have to satisfy the Cauchy-Riemann equations.

The relationship between $f:{\bf R}^2\to{\bf R}^2$ and $f:{\bf C}\to{\bf C}$ is also discussed here.

Solution 4:

The proofs I have seen derive this as a corollary of Cauchy's integral formula. Look at the difference quotient as an integral, play around with it, and you get that it converges to what you'd get if you differentiated under the integral sign.

Note that since harmonic functions also satisfy a similar integral equation, they are also infinitely differentiable in the same way (this also follows since they are real and imaginary parts of holomorphic functions).