# One conformal mapping

1. Oct 9, 2007

### jostpuur

When trying to solve one problem (my own, not an exercise), I encountered the need for a conformal mapping between a square [0,1]^2 and a triangle (0,0)-(1,1)-(2,0), so that the side (0,0)-(0,1) of the square gets mapped into a point (0,0), and the three other sides become the sides of the triangle.

I have no clue how to construct such mapping, or if there exists general techniques for it or already known solution. I barely now the definition of the conformal mapping. Any hints appreciated, if somebody happens to something about this business.

2. Oct 9, 2007

### eendavid

I don't know anything about this. I wonder though. A conformal mapping is a mapping that conserves angles, while you ask a mapping that bends some angles 90° to 45° (and that maps curve to a point). That's impossible, not?

3. Oct 9, 2007

### jostpuur

I succeeded in drawing a picture that looked very convincing. It could be that you just didn't try hard enough when trying to see what kind of mapping it's going to be.

Is there any well known existence proofs for conformal mappings? I think I heard about something like that, but I don't know what are the assumptions in those proofs.

4. Oct 9, 2007

### HallsofIvy

There certainly exist a mapping that will do that but not a "conformal mapping". As eendavid said, conformal mapping preserve angles and so cannot convert a right angle, at a corner of the square into a straight angle in the triangle. A conformal mapping can only change a rectangle into another rectangle.

5. Oct 9, 2007

### matt grime

Typically, one really only asks that the map be conformal on the interior, and bijective on the boundary (but you've also broken that rule).

6. Oct 9, 2007

### jostpuur

hmhm.. to me this seems that it could still be bijective and angle preserving in the interior.

Okey, matt was first to mention interior.

The conformal mapping from the interior of a square to a circular sector

$$\{(r\cos\theta,\;r\sin\theta)\in\mathbb{R}^2\;|\; 0<r<R\;\textrm{and}\; 0<\theta<\pi/4\}$$

is trivial. Map horizontal lines to the lines $\theta=$ constant, and vertical lines to the arcs r=constant. The mapping I'm trying to find would look a little bit the same, but a little bit different too.

Last edited: Oct 9, 2007
7. Oct 9, 2007

### matt grime

Different in the sense that you specifically say the map is _not_ to be bijective on the boundary, you mean?

8. Oct 9, 2007

### jostpuur

I mentioned the circular sector example only to make clear what I'm talking about, since I didn't want to start putting pictures on the internet for such small thing. I didn't try to mean much with it. "Different" meant that the sector example doesn't yet solve the problem.