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:
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I just don't see the lack of differentiability in this. Any clarification would be very appreciated.
On
A nice way to visual complex functions is thinking is as used here: https://www.youtube.com/watch?v=shEk8sz1oOw&ab_channel=Numberphile where one draws the function as a mapping of points on one complex plane to another.
What it means for a complex function $f$ to be differentiable at a point $a_0$ is that it is approximated by $a(z-z_0) + b$ locally (which moves $z_0$ to $0$, then is a dilation/rotation around $0$ then a translation by $b$), where $a = f'(z_0)$ and $b = f(z_0)$. However, when you zoom in you find that $z \mapsto \overline{z}$ always is a reflection so it cannot be approximated like that. This is captured in the Cauchy-Reimann equations and you will not see by inspecting the real and imaginary components separately since it looks like: $$\operatorname{Re}: (a,b)\mapsto a$$ and $$\operatorname{Im}: (a,b)\mapsto -b.$$ You cannot look at these actions separately. The Cauchy-Reimann equations would tell you how these functions interact.
(See Stein and Shakarchi's Complex Analysis for more:) A function $f: \mathbb{C} \to \mathbb{C}$ is complex differentiable at a point $z_0$ if the limit exists: $$\lim_{|z| \to 0}\frac{f(z_0 + z) - f(z_0)}{z}.$$ It then turns out that this is equivalent to writing $f(z) = u(z) + iv(z)$, where $u$ is the real part of $f$ and $v$ is the imaginary part of $f$, with the requirement that $u$ and $v$ satisfy the Cauchy-Reimann equations: $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ and $$ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$
In fact, if you have a real harmonic function on $\mathbb{C}$: $u: \mathbb{C} \to \mathbb{R}$ with $\Delta u = \frac{\partial}{\partial x} u + \frac{\partial}{\partial y} u = 0$ and this function is either the real of imaginary part of a complex function $f: \mathbb{C} \to \mathbb{C}$, then using the Cauchy-Riemann equations you can construct the other part of $f$ uniquely (ignoring a constant $+ C$).
So, for instance, if we look at the real part of your example $f(x + iy) = x + v(x + iy)$. Then we expect that $v$ satisfy: $$1 = \frac{\partial v}{\partial y}$$ and $$0 = -\frac{\partial v}{\partial x}.$$ This means that $v(x+yi) = y + C$ for some constant $C \in \mathbb{C}$. So, this gives that $f(x+yi) = x + yi$, i.e. $f(z) = z$.
If we were to do the same analysis to the imaginary part of the conjugation function, we would write $f(x + iy) = u(x+iy) - iy$, then we would get exactly $u(x + iy) = -x$, so $f(z) = -z$. The Cauchy-Reimann equations do not allow this reflecting business...
On
When it comes to the real numbers, you can imagine the derivative of a function $f$ at a point $x$ by imagining moving $x$ back and forth a little bit. When you do this, if $f$ is differentiable, then $f(x)$ will also move back and forth a little bit. If the derivative is 2, then $f(x)$ will wiggle in the same direction that $x$ wiggles in, but twice as much. If the derivative is -1, it will wiggle the same amount in the opposite direction. And if the derivative is 0, $f(x)$ will mostly stay put.
If $f$ is not differentiable at $x$, however, then something stranger will happen: perhaps as you wiggle $x$, $f(x)$ will wiggle to the left but it won't wiggle to the right, or something like that. Anything besides the "normal, expected" behavior indicates that $f$ is not differentiable at $x$.
For functions $f$ on the complex numbers, you can imagine the derivative at $z$ by imagining swirling $z$ in tiny little circles instead. When you do this, if $f$ is differentiable, then $f(z)$ will also swirl in little circles. If the derivative is 2, then $f(z)$ will swirl at the same frequency, in the same phase, but the circles will be twice as big. If the derivative is -1, then $f(z)$ will swirl in the same frequency, with the same size circles, but it will be 180 degrees out of phase with $z$. If the derivative is $i$, then $f(z)$ will swirl 90 degrees out of phase with $z$.
Now, what if $f$ is the complex conjugate function? When you swirl $z$ around in little circles, $f(z)$ will also swirl in little circles, but in the wrong direction! The argument of the derivative corresponds to the phase difference between the two "swirls", but since $f(z)$ moves in the wrong direction, the derivative is undefined. On the real line, it looks like the circles are in phase, and so the derivative should be 1. On the imaginary line, it looks like the circles are 180 degrees out of phase, and so the derivative should be -1. Since there's no agreement here, the derivative just ends up being undefined.
The only time the complex derivative is defined is when, when you swirl $z$ in little circles, $f(z)$ swirls in circles in the same direction and frequency (or just stays put). This explains why the following functions are not differentiable:
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.