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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?

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$\newcommand{\Cpx}{\mathbf{C}}$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.

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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.

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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 $\Im(z)=0$.

enter image description hereenter image description here

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  • $\begingroup$ Yes, but you showed me plane $\Im(z)=0$ which is actually real plot. That's only one plane where I can see the symmetry. Everything breaks down for example on plane $\Im(z)=2$ (plot) $\endgroup$ – Kamil Jarosz Aug 29 '15 at 6:50
  • $\begingroup$ The stationary figure shows much more than just symmetry of the real case. Please read the two other answers given until you understand them fully, and then try again, by considering the projection of the two objects on the plane $\Im(z)=0$. $\endgroup$ – Yiannis Galidakis Aug 29 '15 at 8:43
  • $\begingroup$ Indeed, thanks so much! $\endgroup$ – Kamil Jarosz Aug 29 '15 at 8:56

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