Satisfying Cauchy Riemann equations at discontinuity What does it mean for a function to satisfy Cauchy-Riemann equations at a point discontinuity? The Cauchy-Riemann equations are about the partial derivatives of a function satisfying particular condition(s), but how would one calculate the partial derivatives at all at that point? Do you just take the limit of the partial derivative as one approaches that point?
 A: Sastisfation of the Cauchy-Riemann equations alone is not sufficient to garantee the existence of $f'(z_0)$. For example: 
$f(z) = \begin{cases} \frac{\overline{z}^2}{z} &, z\neq 0 \\ 0 &,z=0 \end{cases}$
It satisfies the Cauchy-Riemann equations but $f'(0)$ does not exist. 
Now, if $f$ is defined in a certain neighborhood of $z_0$ and its first order derivative exists everywhere in that same neighborhood and the Cauchy-Riemann equations hold then $f'(z_0)$ exists. (see Chapter 2 - sec. 21 Churchill - Complex Variables and Applications).
I also suggest you that read a bit more about the concept of analytic function here.
A: A function can be discontinuous while still allowing the existence of partial derivatives. Only total derivatives cannot exist at points of discontinuity. For instance, consider this very simple function:
$$f:\mathbb C\to\mathbb C\\
x+\mathrm i y\mapsto\begin{cases}1&\textrm{if }x=0\textrm{ or }y=0\\0&\textrm{otherwise}\end{cases}$$
Geometrically, the graph of this function is essentially a kind of cross embossed onto the plane: The function is $1$ on the real and imaginary axes, but $0$ everywhere else. This function is obviously discontinuous on that cross, since every point of that cross is essentially freefloating above the plane, with no connection to the plane. However, this function still admits both partial derivatives at the origin: Approaching the origin along the real axis, we always stay on the cross, so the function is essentially constant for this purpose. Its partial derivative with respect to $x$ is then $0$. The same for the imaginary axis: Along this axis, it's also constant, so the partial derivative with respect to $y$ is also $0$. So at the origin, which lies on both axes, the partial derivatives along both axes exist and are $0$, satisfying the Cauchy-Riemann equations. The function is not complex differentiable at the origin, though, since complex differentiability relies on being able to approach the point in question from any direction, not just along the two coordinate axes. Just like total differentiability (which automatically holds if the function is complex differentiable). Complex and total differentiability both require continuity. Partial differentiability does not.
