### Useful conformal mappings

This post is to be a list of conformal mappings, so that I can get better at answering questions like “Find a conformal mapping from <this domain> to <this domain>”. The following Mathematica code is rough-and-ready, but it is designed to demonstrate where a given region goes under a given transformation.

```
whereRegionGoes[f_, pred_, xrange_, yrange_] :=
whereRegionGoes[f, pred, xrange, yrange] =
With[{xlist = Join[{x}, xrange], ylist = Join[{y}, yrange]},
ListPlot[
[email protected]
Through[{Re, Im}[
f /@ (#[[1]] + #[[2]] I & /@
Select[Flatten[Table[{x, y}, xlist, ylist], 1],
With[{z = #[[1]] + I #[[2]]}, pred[z]] &])]]]]
```

- Möbius maps - these are of the form \(z \mapsto \dfrac{az+b}{c z+d}\). They keep circles and lines as circles and lines, so they are extremely useful when mapping a disc to a half-plane. A map is defined entirely by how it acts on any three points: there is a unique Möbius map taking any three points to any three points (and hence any circle/line to circle/line). (Some of the following are Möbius maps.)
- To take the unit disc to the upper half plane, \(z \mapsto \dfrac{z-i}{i z-1\)}
- To take the upper half plane to the unit disc, \(z \mapsto \dfrac{z-i}{z+i}\) (the Cayley transform)
- To rotate by 90 degrees about the origin, \(z \mapsto i \)z
- To translate by \(a\), \(z \mapsto a+\)z
- To scale by factor \(a \in \mathbb{R}\) from the origin, \(z \mapsto a \)z
- \(z \mapsto exp(z)\) takes a vertical strip to an annulus - but note that it is not bijective, because its domain is simply connected while its range is not.
- \(z \mapsto exp(z)\) takes a horizontal strip, width \(\pi\) centred on \(\mathbb{R}\) onto the right-half-plane.

## Maps which might not be conformal

These maps are useful but we can only use them when the domain doesn’t include a point where \(f’(z) = 0\) (as that would stop the map from being conformal).

- To “broaden” a wedge symmetric about the real axis pointing rightwards, \(z \mapsto z^\)2
- To take a half-strip \(Re(z) > 0, 0 < Im(z) < \dfrac{\pi}{2}\) to the top-right quadrant: \(z \mapsto \sinh(z\))
- to take a half-strip \(Im(z) > 0, -\frac{\pi}{2} < Re(z) < \frac{\pi}{2}\) to the upper half plane, \(z \mapsto \sin(z\))