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4. Week 4: Conformal Mappings

≪ 3. Week 3: Complex Derivatives | Table of Contents | 5. Week 5: Estimating and Sometimes Computing Path Integrals ≫

Today’s focus will be on finding conformal mappings between two arbitrarily chosen (but usually geometrically significant) subsets of the complex plane. Let’s first recall the definition of a conformal mapping.

This seems like a rather arbitrary condition, but conformal mappings preserve important “small-scale” geometric properties of their domains. One particularly important property is that conformal mappings preserve harmonicity, and harmonicity shows up all the time in applied mathematics and physics. Frequently, one will attempt to solve a PDE (with relations to harmonicity, e.g. the heat equation) on a strange domain (say, the 2D interior of a house or rocket). Solving this equation is doable on a simple domain like the open disc, and so one uses conformal maps to transform the domain into a simpler, more manageable scenario, solves the equation using known techniques, then undoes the conformal map at the end.

The point being, understanding conformal maps, when functions are conformal, and how to find conformal maps between arbitrary domains is of interest in many applications, and that’s what we’ll be discussing today.

You have all learned in class one sufficient condition for being conformal:

It turns out that this is also a necessary condition, i.e. conformal maps on the complex plane are exactly the holomorphic functions with nonvanishing derivatives. One can understand this in terms of the Jacobian matrix of a holomorphic map and its geometric relationship to the Cauchy-Riemann equations. (That said, the notion of a “conformal map” extends well beyond the setting of a single complex variable, from where the equivalence with holomorphy is weakened.)

Finding Conformal Maps

A fundamental type of conformal map is a Möbius transform, or a fractional-linear transformation. These are functions of the form \[f(z) = \frac{az+b}{cz+d},\] where the matrix \(\begin{pmatrix} a&b\\c&d \end{pmatrix}\) has nonzero determinant. (What goes wrong if the determinant is zero?)

In lecture, you have described some algebraic properties of these transformations, such as rules for composing two such transformations and “fitting” a transformation to three prescribed points. But Möbius transformations enjoy a very rich set of geometric properties: if \(f\) is a Möbius transform, then the image of a circle or line under \(f\) is also a circle or a line, for instance.

Suppose has a problem of the following setup: you are given two domains \(D_0\) and \(D_1\), which are either half-planes, the interiors of circles, or the exteriors of circles (or a mix of any of the three). One can *always* find a bijective Möbius transform \(f: D_0\to D_1\), and the process is as follows:

1. Identify the “boundaries” of \(D_0\) and \(D_1\), which will either be a circle or a line. Pick any points \(a_0\) and \(a_1\) on the boundaries of \(D_0\) and \(D_1\), respectively.
2. Identify a pair of “symmetric points” for each, one point on the inside and one point on the outside. For circles, one point is the centre of the circle and the other is the point at \(\infty\). For half-planes, pick two points that are reflections of each other across the border. Let \(b_0, c_0\) be the points on the inside and outside, respectively; let \(b_1, c_1\) be defined analogously.
3. Find \(f: \mathbb{C}\to \mathbb{C}\) by determining the unique Möbius transformation taking \(a_0\mapsto a_1\), \(b_0\mapsto b_1\), and \(c_0\mapsto c_1\).

Note that this mapping is not unique, and generally there are infinitely many conformal mappings between any pair of domains. The non-uniqueness comes from the freedom in choosing the boundary points \(a_0\) and \(a_1\), as well as the freedom in choosing “symmetry points”. (In fact, there are more ways to choose symmetry points for circles, but this is easist for me to remember.)

Solution

Let’s follow the procedure above. The boundaries of \(\mathbb{D}\) and \(\mathbb{H} \) are the unit circle and the real axis, respectively. We may therefore pick \(a_0=1\) and \(a_1 = 0\).

Next, the pair of symmetric points for \(\mathbb{D}\) are \(b_0 = 0\) and \(c_0 = \infty\). I have the freedom to pick any pair of reflected points for \(\mathbb{H}\), so I’ll pick \(b_1 = i\) and \(c_1 = -i\). Thus, I need to find any Möbius transform taking \(1\mapsto 0\), \(0\mapsto i\) and \(\infty\mapsto -i\). The map \[f(z) = \frac{-i z + i}{ z + 1 }\] will do.

Half-planes and interiors/exteriors of circles aren’t the only things that need to be conformally mapped to each other. Most often, we will be encountering strips, which are infinitely long rectangles, and sectors, which are infinitely long pie slices (which includes slit-planes).

For this, recall the geometric action of the exponential map, which we discussed two weeks ago: the exponential takes a horizontal strip onto a sector, and the logarithm takes sectors back to horizontal strips. In particular, when our horizontal strips have height (\pi ), the exponential turns them into a half-plane (which is a type of sector), and we can then chain together a Möbius transform to turn it into a circle if necessary.

Thus, we have all the tools to build conformal equivalences between sectors, strips, and circles.

Suppose \(D_0\) and \(D_1\) are one of the above. Generally, conformal equivalences between the three happen in the following steps:

1. “Normalise” \(D_0\) and \(D_1\) if it’s a rectangular strip or sector: * By multiplying by a complex number of modulus \(1\), we can rotate any rectangular strip to be horizontal. By translating, we can make the strip lie above the real axis. By rescaling, we can make the strip have any height we want (usually \(\pi \), so we can turn it into a half-plane). * By translating, we can move any sector so that its “pointed tip” is at the origin. By mapping \(z\mapsto z ^{\alpha }\), we can either fan out or compress the sector to cover any angle we want (typically \(\pi \), so it’s a half-plane). By rotating, we can force one end of the sector to lie on the real axis.
2. Perform a conformal mapping between the normalised \(D_0\) and \(D_1\) using Möbius transforms.
3. Undo the normalisations of \(D_1\).

By chaining all three steps together, and using the fact that compositions of conformal mappings are themselves conformal, we can find conformal equivalences between virtually anything!