Graphical explanation for non-differentiability of $z \mapsto \bar z$ Let $f(z) = \bar z$ where $z \in \mathbb C$ and $\bar z$ is the complex conjugate of $z$. 
I can prove that $f$ is nowhere differentiable, but I can't picture it. With a real-valued function I have a good mental picture of what a lack of differentiability looks like, but I don't have that here.
Plotting the real and imaginary parts with Wolfram Alpha, I get this:
.
I just don't see the lack of differentiability in this. Any clarification would be very appreciated.

If $f(z)$ is complex-differentiable at $z_0$, this means that we have an approximation $f(z) \approx f(z_0) + c(z-z_0)$ for some $c \in \mathbb{C}$.  Recall how complex multiplication works: multiplying by $c$ is the same as applying a rotation and a scaling.  So, locally, a complex-differentiable map looks like applying a rotation and a scaling.  The problem with $f(z) = \bar{z}$ is that it does not look like this - in fact it's performing a reflection.
You might like Tristan Needham's excellent book "Visual Complex Analysis" which describes in detail this intuitive, geometric meaning of complex-differentiation (and other important concepts in complex analysis) with lots of figures.
