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This is a follow-up to a question I asked a year ago, which was helpfully answered by Anton Petrunin:

Fitting a mesh to a density function

I am trying to come up with a way to make a picture of an equilateral triangular mesh in the plane in which the density of triangles has elliptical level sets. For example, the following picture has a triangular mesh in which the density of triangles has circular level sets, which I obtained by taking the complex map $z\mapsto z^{2}$:

alt text

The picture below is closer to what I was looking for, but it's actually a total hack: I just took the complex map $z\mapsto \sin{z}$ and then physically drew a black ellipse in the center of the picture to suggest some kind of elliptical density:


If we zoom closer we can see that the level sets of the density are clearly not ellipses:

alt text

Does anyone have suggestions for better maps that might produce the kind of picture I'm looking for? I browsed through the "Dictionary of Conformal Mappings" at but haven't been able to find anything better than what I've got.

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Not directly relevant but you might be interested in the article "Conformal and equivalent world maps" by B. H. Brown in the Amer. Math. Monthly. vol. 42. As the title suggests, it is motivated by (two central) problems of mathematical cartography but the main results are purely mathematical. The relevant one is an explicit description of all conformal mappings of the plane which map the rectangular array onto systems of conic sections. (He also solves the same problem for area-preserving mappings). – jbc Mar 24 '13 at 7:00

Well, if $z \mapsto z^2$ gives circles, then $(x,y) \mapsto (x^2-y^2,xy)$ should give you ellipses with the horizontal axis twice as large as the vertical. Note, this map is not conformal.

It therefore "feels" like you cannot find such a map that you are looking for, ellipses are not really conformal objects in some sense, maybe someone else can give a rigorous proof on why there cannot be such a map?

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Ellipses are "conformal objects": the map $z = w + 1/w$ takes the circle $|w| = r$ to an ellipse for $r \ne 1$. – Robert Israel Feb 27 '13 at 0:34
@Robert Israel: Ah, yes, then this map should do the job? – Per Alexandersson Feb 27 '13 at 6:34
Not quite. What you want, I think, is $z = f(w)$ where $|f'(f^{-1}(z))|$ is constant on ellipses. – Robert Israel Feb 27 '13 at 17:28

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