 ### Patrick Stevens

Former mathematics student at the University of Cambridge; now a software engineer.

# 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 /@ (#[] + #[] I & /@

Select[Flatten[Table[{x, y}, xlist, ylist], 1],

With[{z = #[] + I #[]}, 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 $% 0, 0 < Im(z) < \dfrac{\pi}{2} %]]>$ to the top-right quadrant: $z \mapsto \sinh(z)$
• to take a half-strip $% 0, -\frac{\pi}{2} < Re(z) < \frac{\pi}{2} %]]>$ to the upper half plane, $z \mapsto \sin(z)$