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I am interested in finding explicit formulae for (better yet characterizing) conformal functions from various domains onto the open unit disc $\mathbb{D}\subset\mathbb{C}$, and in understanding the key ideas necessary to establish such functions.

Specifically, what can $f$ look like when $f:G\to\mathbb{D}$ is conformal and

(1) $G=\{x+iy~|~x,y>0\}$ is the open first quadrant.

(2) $G=\{x+iy~|~x>0,~0<y<1\}$ is an open horizontal strip in the first quadrant.

(3) $G=\{z\in\mathbb{C}~|~\frac{1}{2}<|z|<1\}$ is an annulus.

(4) $G=\mathbb{D}\cap\{|z-\frac{1}{2}|>\frac{1}{2}\}$ is something else (torus?).

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Do you want these to be bijective? By "conformal" some people mean "biholomorphic", and others mean "holomorphic with non-vanishing derivative". –  Dylan Moreland Jul 19 '11 at 5:51
not necessarily, but it would be nice to also know when/if this is possible. The definition of conformal I was introduced to was "angle-preserving" which was shown to require non-vanishing derivative (Are these conditions equivalent for holomorphic functions?). This was what I had in mind, but all discussion is welcome. –  RHP Jul 19 '11 at 5:55
Sure. I must run off, but I'll mention that I often find the Cayley transform to be useful for problems like this. Here it would be helpful for (1). –  Dylan Moreland Jul 19 '11 at 6:01
@RHP: Thanks, great to hear! As for $3$, the point is that the annulus has a hole, while the disk doesn't have one. If you want a biholomorphic map, you need the same number of holes, intuitively. By the way, one can show (not easy!) that if you have an annulus with radii $0 \lt r \lt R$ and you want to map it to another annulus with radii $0 \lt r' \lt R'$ then you must have $\frac{R}{r} = \frac{R'}{r'}$ "the ratio of radii of annuli is a conformal invariant", (this is such a strange sentence that you can't forget it). I like Ahlfors a lot (but it's tough) and ... –  t.b. Jul 19 '11 at 22:44
...for historical remarks (and many other things) I recommend the great books by Remmert Part 1 and Part 2. I worked through them in German, but I guess the English editions aren't too different. Cartan is great and I have some prejudices against Lang, but that's a matter of taste, I guess :) I never looked at Nevanlinna. –  t.b. Jul 19 '11 at 22:48

2 Answers 2

up vote 15 down vote accepted

Since you didn't show too many own thoughts, here are some hints only. By conformal I understand biholomorphic.

  1. First take $f(z) = z^2$ to map the quadrant biholomorphically onto the upper half-plane, then compose with the Cayley transform $\kappa(z) = \frac{z-i}{z+i}$ to get $\kappa(f(z)) = \frac{z^2-i}{z^2+i}$.

  2. Look at $\cos{(z)}$ and modify appropriately.

  3. Impossible, since $G$ is not simply connected.

  4. Map the region $G$ to the strip between two parallel lines using a Möbius transformation sending $1$ to infinity (e.g. using the inverse Cayley transformation). Then use the exponential function.

This should be enough to figure the solutions out.

For the precise relationship between "conformal" and "analytic", as well as for explanations on how to find such maps, I refer you to Ahlfors or (probably—I never really read it) Needham or any decent text on complex analysis treating conformal mapping.

The characterization of biholomorphisms between simply connected regions is essentially the content of the Riemann mapping theorem.

Sometimes biholomorhic mappings between polygonal regions and the unit disk can be computed via the Schwarz-Christoffel formula, but usually it leads to elliptic integrals that can't be solved explicitly in elementary terms.


Since the solution of 4. is a bit trickier, here's a rather detailed outline:

First note that $G$ is the region enclosed between the circles $\{|z| = 1\}$ and $\{|z - \frac{1}{2}| = \frac{1}{2}\}$. Applying the Möbius transformation (= the inverse Cayley transform) $\kappa^{-1}(z) = i\frac{1+z}{1-z}$ sends $G$ to the horizontal strip $\{0 \lt \operatorname{Im}{z} \lt 1\}$. To see this, look at this picture from Wikipedia illustrating the Cayley transform:

Cayley transform

Finally, the exponential function $g(z) = \exp{(\pi z)}$ sends this strip to the upper half plane. Composing this with the Cayley transform we get the biholomorphic map $h = \kappa \circ g \circ \kappa^{-1}: G \to \mathbb{D}$.

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Here is a sketch:

For (1) we want to map the open first quadrant onto the unit disk. What we can do is first map the open first quadrant onto the upper half plane then the upper half plane onto the unit disk. the mapping $z \mapsto z^2$ maps the first quadrant to the upper half plane. Then the upper half plane can be mapped to the unit disk by the mapping $z \mapsto \frac{z-i}{z+i}$. What remains to do is to compose.

For (2) $z \mapsto cosh(\pi z)$ maps the open half strip of width 1 to the upper half plane. You can now compose with the mapping in (1) which maps the upper half plane to the unit disk.

EDIT: As noted by Theo, these two last examples are false.

For (3) I am not so sure. $z \mapsto lnz$ maps an annulus onto a rectangle. A rectangle in the plane is simply connected so by the Riemann Mapping Theorem one can find a unique conformal mapping between the rectangle and the unit disk. However I don't know which one.

For (4) It sound like the region described in the interior of a parabola. In which the case the mapping onto the unit disk would be $tan^2 \frac{\pi}{4} \sqrt{\frac{z}{p}} $, $p$ being one fourth of the height of the segment on the $y$-axis formed by the intersection of the parabola with the $y$-axis. I am really unsure about this last one. Maybe someone else could help.

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Your solutions to 3) and 4) are wrong. In 3) the map is not everywhere defined, in 4) you have the difference between two circles (precisely the region inside the unit disk but outside the circle with radius $1/2$ around $1/2$. –  t.b. Jul 19 '11 at 7:48
Thanks! I'll edit accordingly. –  user786 Jul 19 '11 at 7:53

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