Geometric interpretation of inverse complex function? Function $f\colon\mathbb{R}\to\mathbb{R}$ and its inverse $f^{-1}$ are symmetric over line $y=x$. It's easy to imagine inverse of real function, we just have to "flip" the plot over $y=x$.
But what about complex functions? How to imagine the inverse of function $g\colon\mathbb{C}\to\mathbb{C}$? After looking at plots of $g(z)=\exp(z)$ and $g^{-1}(x)=\ln(x)$ I don't see anything like symmetry or something.
Clarifying, if I have a function $f\colon \mathbb{C}\mapsto\mathbb{C}$ and its plot(s) (any type: real, imaginary, absolute etc. but it must be easy to draw), how to imagine its inverse?
 A: $\newcommand{\Cpx}{\mathbf{C}}$If $g:\Cpx \to \Cpx$ is a function, you can reflect the graph $w = g(z)$ across the (complex) line $w = z$ to get $z = g(w)$, or $w = g^{-1}(z)$. (Of course, $g^{-1}$ is generally multiple-valued, particularly if $g$ is entire.)
Oh right, there's that small obstacle of living in a three-dimensional universe, where the graph of $g$ doesn't fit....
The underlying issue is we don't have a method (analogous to graphing a real-valued function of one variable) for visualizing $g:\Cpx \to \Cpx$. I didn't look at your plots, but presumably the value $w = g(z)$ is encoded as a color and placed as a pixel at $z$. If you try a similar technique for visualizing real-valued functions of one variable, you'll similarly find inverses are not easy to understand.
Despite these reasons why you can't expect an answer to your question, you might try this: If $g(x + iy) = u(x + iy) + iv(x + iy)$ (with $x$, $y$, $u$ and $v$ real-valued), the parametric surface
$$
G(x, y) = (x, y, u, v)
$$
represents the graph of $g$. Pick pairs of components, say $x$-$u$ and $y$-$v$, rotate the corresponding coordinate planes, and project away the fourth coordinate:
$$
G_{t}(x, y) = (x\cos t - u\sin t, y\cos t - v\sin t, u\cos t + x\sin t).
$$
If you make an animation loop with $t$ as time parameter, you'll see the locus $w = g(z)$ "rotate into" something like the locus $z = g(w)$, modulo loss of information from projecting away the fourth coordinate.
A: In the first place: Don't flip the plot, because it mixes up the names of the variables.  When studying $f$ and $f^{-1}$ at the same time you see the graph of $f^{-1}$ in the original $f$-plot when you tilt your head $90^\circ$ sideways.
For functions $f:\>{\mathbb C}\to{\mathbb C}$ and their inverses $f^{-1}$ there is an analogous mischievous symmetry; but you need a four-dimensional vizualization for it: The graph $$\Gamma(f):=\{(z,w)\>|\>z\in{\rm dom\,}(f), \ w=f(z)\}$$ is a subset of ${\mathbb C}^2\sim{\mathbb R}^4$. If you reflect this graph at the diagonal $D:=\{(z,z)\>|\>z\in{\mathbb C}\}$ of ${\mathbb C}^2$ using the map
$$\iota:\quad (z,w)\mapsto (w,z)$$
then $\iota\bigl(\Gamma(f)\bigr)$ can be viewed as graph of $f^{-1}$, but written in the wrong variables.
A: It is fairly hard to visualize the symmetry, because as was written in the other two answers the objects in the Complex case are in general four-dimensional. The symmetry can be displayed however, by dropping one dimension.
To augment the two answers given and help you visualize it, here is the case for the two maps: $\exp$, the principal branch of $\log$ and the identity map, as the surfaces: $(x,y,\Re(\exp))$, $(x,y,\Re(\log))$ and $(x,y,\Re(z))$, with all three surfaces constrained in the upper half plane, so as to outline the symmetry, which is now visible in the stationary image, on the plane $\Im(z)=0$.

