Why are differentiable complex functions infinitely differentiable? When I studied complex analysis, I could never understand how once-differentiable complex functions could be possibly be infinitely differentiable. After all, this doesn't hold for functions from $\mathbb R ^2$ to $\mathbb R ^2$. Can anyone explain what is different about complex numbers?

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.
