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I would like to construct a (conformal) map similar to the exp conformal map, but for which the vertical lines in the source space are evenly separated in the target space. That is, in this simulator: https://www.geogebra.org/m/qCSzkRsP (have to change the function to exp(z)), I would like the red lines to be evenly spaced in the target space.

In the end, what I really want is the inverse transform of that (to be able to "unfold" an 2D object), so it will likely be a modified log conformal map. But I find it easier to imagine it with the exp conformal map first.

Note: I do care about the fact that there is a notch around the origin in the target space.

matthieu
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1 Answers1

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There's no such conformal map.

Suppose we had such a map, $f(x, y) = \rho(x, y) e^{i \alpha(x, y)}$. The condition that vertical lines in the domain are mapped to circles centered at the origin is $$\rho_y = 0 ,$$ and given that it is satisfied, the condition that evenly spaced vertical lines in the domain are mapped to evenly spaced circles is then $$\rho_x = C,$$ for some constant $C$, that is, $\rho_{xx} = \rho_{xy} = 0$.

In these coordinates, the Cauchy-Riemann equations are $$(\log \rho)_x = \alpha_y, \qquad (\log \rho)_y = -\alpha_x.$$

Substituting our conditions gives that $$ \alpha_x = 0, \qquad \alpha_y = \frac{C}{\rho} . $$ But these conditions imply that $\rho$, $\alpha$ are constant. (Why?) Thus, $f(x, y) = \rho e^{i \alpha}$ is constant, hence not conformal.

Dropping the conformality hypothesis In the comments o.p. mentioned they were also in interested a function that had the desired spacing property even without the conformality requirement. In that case, applying our conditions on $\rho$ imposes precisely that $\rho$ is an affine function of $x$ alone, that is, is of the function $$f : (x,y) \mapsto (A x + B) e^{ig(x,y)} .$$ Here, $A$ is the distance between the concentric rings that are the images of vertical lines $1$ unit apart, and $B$ is the radius of the circle to which the imaginary axis is mapped. In particular, we probably want $A \neq 0$, lest our function map all of $\Bbb C$ into the circle of radius $B$ centered at the origin). The Jacobian determinant of $f$ as above is $$\det \operatorname{Jac}(f) = A (A x + B) g_y,$$ so if $A \neq 0$ and $g_y$ vanishes nowhere, the map is invertible everywhere except on the vertical line $x = -\frac{B}{A}$, which is mapped to the origin. The natural choice $B = 0$, $g(x,y) = y$ yields the simple map $$(x,y) \mapsto x e^{i y},$$ perhaps restricted to the right half-plane, $\{x > 0\}$.

Remark 2 If we only impose the condition $\rho_y = 0$, an analysis similar to ours shows that the only conformal maps that send vertical lines in the domain to circles centered at the origin are (locally) the maps $z \mapsto A \exp(b z)$.

Travis Willse
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  • Thank you for your answer! How about a non-conformal mapping then? I don't care as much for it to be strictly conformal. – matthieu Apr 18 '23 at 16:51
  • You're welcome. Looking at our conditions we see that we have a lot of freedom: any function $(x, y) \mapsto (A x + B) e^{i g(x, y)}$ should suffice, where $A$ is the distance between the concentric rings that are the images of vertical lines $1$ unit apart, so we probably want $A \neq 0$ (otherwise our function maps all of $\Bbb C$ into the circle of radius $B$ centered at the origin). – Travis Willse Apr 18 '23 at 18:48
  • The Jacobian of the map is $A (A x + B) g_y$, so if $g_y$ is nonvanishing, the map is invertible everywhere except on the vertical line $x = -\frac{B}{A}$, which is mapped to the origin. A natural choice is $B = 0$, $g(x, y) = y$, giving $(x, y) \mapsto x e^{iy}$, perhaps restricted to the right half-plane, ${x > 0}$. – Travis Willse Apr 18 '23 at 18:50
  • If these comments are useful to you, feel free to edit the original question and then comment here, and I'll update my answer accordingly. – Travis Willse Apr 19 '23 at 00:52
  • Yes, that's what I was looking for ! Thanks a lot :) – matthieu Apr 24 '23 at 16:06