Geometrical Interpretation of Cauchy Riemann equations? Differentiation has an obvious geometric interpretation, and the Cauchy Riemann equations are closely linked with differentiation.
Do the Cauchy Riemann equations have a geometric interpretation?

The version of the Cauchy-Riemann equations that I prefer is (assuming we are using $x,y$ as coordinates for the real and imaginary part on the domain)
$$
i\partial_x f = \partial_y f
$$
This has a direct geometrical interpretation: the partial derivative with respect to the real part rotated by $\pi/2$ (which is what multiplying by $i$ does to a complex number) is the partial derivative with respect to the imaginary part.
The geometric interpretation of the complex derivative itself
$$
f(z+h) = f(z) + f'(z)h + o(h)
$$
is that $f$ can be approximated at the first order near $z$ by an affine transformation with a linear part represented by complex multiplication. If $f'(z)=re^{i\theta}$, then $h\mapsto f'(z)h$ is a rotation by $\theta$ followed (or preceded) by a uniform scaling of factor $r$. 
This is much more restrictive than real differentiability for functions $\mathbb R^2\to\mathbb R^2$, which allows approximation by arbitrary affine transformations. For example the mapping $c: (x, y) \mapsto (x, -y)$ has itself as a real differential everywhere (because it's a linear transformation). Geometrically $c$ is a reflection, which changes the orientation of the plane (it sends the base $(1, 0),\ (0, 1)$ which is clockwise or anticlockwise oriented depending on how you draw the axes, to $(1, 0),\ (0, -1)$ which has the opposite orientation). When thought of as a transformation of the complex plane, $c$ is just the complex conjugation $z \mapsto \overline z$. From a geometric viewpoint $c$ has no complex derivative anywhere because if such a derivative $c'(z)$ existed the linear transformation $h\mapsto c'(z)h$ would have to be a reflection, which is impossible to obtain by composing uniform scalings with rotations.
The Cauchy-Riemann equations can be seen as a consequence of the relation between the complex derivative and the real differential. Because of the approximation argument given above and the unicity of the real differential, the complex linear transformation $h\mapsto f'(z)\cdot h$ in $\mathbb C$ corresponds in $\mathbb R^2$ to the real differential
$$
  df(z_x, z_y): \begin{bmatrix}h_x\\h_y\end{bmatrix}\mapsto 
  \begin{bmatrix}\partial_x f(z_x, z_y) & \partial_y f(z_x, z_y)\end{bmatrix}
  \begin{bmatrix}h_x\\h_y\end{bmatrix}
$$
where the real partial derivatives $\partial_x f(z_x, z_y)$ and $\partial_y f(z_x, z_y)$ are column vectors and correspond to the complex partial derivatives $\partial_x f(z), \partial_y f(z)$. Taking $h = 1$, i.e. $h_x = 1, h_y = 0$, in $\mathbb R^2$ we have
$$
  \begin{bmatrix}\partial_x f(z_x, z_y) & \partial_y f(z_x, z_y)\end{bmatrix}
  \begin{bmatrix}1 \\ 0\end{bmatrix} = 
  \begin{bmatrix}\partial_x f(z_x, z_y)\end{bmatrix}
$$
which translated back to $\mathbb C$ is $f'(z)\cdot 1 = \partial_x f(z)$. Taking $h = i$ (i.e. $h_x = 0, h_y = 1$) yields $f'(z)\cdot i = \partial_y f(z)$. Together these two identities give the Cauchy-Riemann equations $\partial_y f(z) = if'(z) = i\partial_x f(z)$.
