It is quite easy to see why $f(z)=\bar{z}$ isn't complex differentiable.
$\frac{\partial{u}}{\partial{x}}=1\neq-1=\frac{\partial{v}}{\partial{y}}$
But I struggle to see why this is the case.
Visualization of the conjugate function in my head is simply flipping the complex plane upside down. I don't see how this creates any discontinuities.
Am I having the wrong visualization or have I misunderstood complex differentiability?
Best Answer
Of course it doesn't satisfy one of the possible definitions (Cauchy-Riemann equations, the specific limit does not exist), but in my opinion that doesn't give you much intuition - at least of the geometric kind. Hopefully the following will compensate for that.
TL;DR:
If $f\colon \mathbb C \to \mathbb C$ is complex-differentiable, then it is differentiable as an $\mathbb R^2 \to \mathbb R^2$ function, and it locally almost keeps angles and circles. To see this, look at one equivalent version of the complex-differentiability: $$f(z_0 + re^{it}) = f(z_0) + re^{it}f'(z_0) + \omicron(r), $$ where $z_0 \in \mathbb C, 0 \leq r, 0 \leq t \leq 2\pi$. In fact, as a demonstration have a picture of the image of two circles around $1 + i$ by $\operatorname{id_{\mathbb C}}^2$.
The black,$\{z_0 + re^{it} \mid t \in [0,2\pi]\}$ ones are our original circles ,
the reds, $\,\,\color{red}{f(\{z_0 + re^{it} \mid t \in [0,2\pi]\})}$ are the blacks' images by $f$, and
the blues, $\color{blue}{\{f(z_0) + re^{it}f'(z_0) \mid t \in [0,2\pi] \}}$ are the first order approximation of the red ones.
If we pick $z_0 = 1 + i, r=0.1$ and $0.2$, then we get, in order:
As you can see in the $r=0.1$ case the red and blue circles are almost equal. As $r \to 0$, the difference becomes smaller and smaller. That's what I meant by locally almost keeping circles.
Here's another picture, displaying the image of a point from each of the two black circles. The black and blue circles represent the same as before, the green circles are due to the $\omicron(r)$ part.
Look for the points $\color{red}{a}, \color{blue}{a'}, \color{red}{f(a)}$ and $\color{purple}{b},\color{blue}{b'}, \color{purple}{f(b)}$.
In fact, in the $r=0.1$ case the green circle is so small, you can barely see any of it (this is why the red and blue circles from a moment ago were almost equal).
As you can see, the image of a circle is essentially the sum of multiple circles, in this case, two; however, if $f$ has lots of non-zero derivatives, then there'll be lots of circles, and each "new" circle will have smaller radius, but higher frequency than the previous one (assuming $r < 1$).
If you know about Fourier-series, then this concept might sound familiar. By the way, if you haven't already, read this answer.
Anyway, this is not so surprising: if you consider the Taylor series of a holomorphic $f$ at a point, and plug in a point from a circle around that point, you'll get $$ f(z_0 + re^{it}) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}r^ne^{int}.$$ Here the $n$th term will correspond to the circle $\left\{\frac{f^{(n)}(z_0)}{n!}r^ne^{int} \mid t \in [0,2\pi]\right\}$. OK, back to the original question.
Hopefully I've convinced you that the $\omicron(r)$ part is not significant. (Of course only so if the first derivative is non-zero, but in the other cases we can say something similar too.)
Let's say a word about locally almost keeping angles. Remember $$f(z_0 + re^{it}) = f(z_0) + re^{it}f'(z_0) + \omicron(r).$$ Picking two numbers, $z_0 + re^{it}, z_0 + re^{is}$ the angle between them that is kept locally is $(t-s)$. Their images are $$ f(z_0) + re^{it}f'(z_0) + \omicron(r), \text{and } f(z_0) + re^{is}f'(z_0) + \omicron(r). $$ As you can see, the local (i.e. looking from the point $f(z_0)$) angle between them is again, approximately $(t-s)$.
Now let's really get back to your original question. For $f(z) = \bar z$, assuming differentiability at $z_0$, the following would be true for all $t\in [0,2\pi]$: \begin{align} f(z_0 + re^{it}) &= f(z_0) + re^{it}f'(z_0) + \omicron(r) \\ \iff re^{-it} &= re^{it}f'(z_0) + \omicron(r). \end{align} This is absurd. Of course we could also note that conjugation locally flips angles. Going back to our points $z_0 + re^{it}, z_0 + re^{is}$, the angle is still $(t-s)$. Their images now are $$\bar{z_0} + re^{-it}, \text{ and } \bar{z_0} + re^{-is}, $$ meaning that the appropriate angle is $-(t-s)$. Thus again, conjugation can not be complex-differentiable.