Conformal mapping

A conformal map is a holomorphic function with non-zero derivative, and it preserves angles between curves. This makes conformal maps the natural coordinate changes of two-dimensional physics: they carry solutions of Laplace’s equation (electrostatics, fluid flow, heat conduction) from a complicated domain to a simple one where the problem is easy, and then pull the answer back. The Riemann mapping theorem guarantees that, up to a Möbius transformation, every simply connected domain looks like the unit disk.

Authoritative sources

L. Ahlfors, Complex Analysis, Ch. 4

J. B. Conway, Functions of One Complex Variable I, Ch. 6

Conformal (angle-preserving) maps

Definition. A holomorphic function $f$ is conformal (or angle-preserving) at $z_0$ if $f'(z_0) \neq 0$ . At such a point, $f$ maps the tangent direction of any curve through $z_0$ to the tangent direction of its image, preserving both the magnitude and the orientation of the angle between curves.

Details

Why a non-zero derivative preserves angles

Write $f'(z_0) = re^{i\theta}$ . For a curve $\gamma$ through $z_0$ with tangent direction $\gamma'(t_0)$ , the image curve $f \circ \gamma$ has tangent

$(f \circ \gamma)'(t_0) = f'(z_0)\, \gamma'(t_0) = re^{i\theta}\, \gamma'(t_0).$

Multiplication by $re^{i\theta}$ scales every tangent vector by $r$ and rotates it by the same angle $\theta$ . Since the same scale and rotation is applied to every direction, the angle between any two curves is unchanged. This is exactly the content of the Cauchy–Riemann equations: the derivative is a similarity (uniform scaling plus rotation), not a general linear map.

Why the non-zero condition is essential

If $f'(z_0) = 0$ , the first-order term vanishes and the map is governed by the first non-zero higher derivative $f^{(k)}(z_0) (z-z_0)^k / k!$ . Such a map multiplies angles at $z_0$ by the integer $k$ : for example $f(z) = z^2$ doubles angles at the origin. So $z^2$ is conformal everywhere except at $0$ .

Worked example

The map $f(z) = e^z = e^x e^{iy}$ has $f'(z) = e^z \neq 0$ everywhere, so it is conformal on all of $\mathbb{C}$ . It sends vertical lines $x = c$ to circles $|w| = e^c$ and horizontal lines $y = c$ to rays $\arg w = c$ , so it maps the horizontal strip $0 < y < \pi$ conformally onto the upper half-plane. Grid lines that met at right angles still meet at right angles.

Example problem

Problem. Show that $f(z) = z^2$ is conformal at every $z \neq 0$ but not at $z = 0$ .

Solution

$f'(z) = 2z$ , which is non-zero for $z \neq 0$ , so $f$ is conformal there. At $z = 0$ , $f'(0) = 0$ . To see the angle behavior, a curve with tangent direction $\phi$ at $0$ is sent to direction $2\phi$ (since $z^2$ squares the argument). Two curves meeting at angle $\theta$ meet at angle $2\theta$ in the image, so angles are not preserved. ∎

Möbius transformations

Definition. A Möbius (or linear fractional) transformation is a map of the form

$f(z) = \frac{az + b}{cz + d}, \qquad ad - bc \neq 0,$

defined on the Riemann sphere $\widehat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ . Möbius transformations are conformal automorphisms of the Riemann sphere and map generalized circles (circles and lines) to generalized circles.

Details

Why they are conformal and circle-preserving

On $\mathbb{C}$ (away from the pole $z = -d/c$ ), $f'(z) = (ad - bc)/(cz+d)^2 \neq 0$ , so Möbius maps are conformal. They are built from translations, scalings, and the inversion $z \mapsto 1/z$ , each of which sends circles and lines to circles and lines; a composition does too. A line is a “circle through $\infty$ ”, and Möbius maps permute the points of the sphere, so they carry the family of generalized circles to itself.

The three building blocks

Translation $z \mapsto z + b$ and scaling $z \mapsto a z$ : obvious circle-preservers.
Inversion $z \mapsto 1/z$ : a circle not through $0$ maps to a circle; a circle through $0$ maps to a line; a line not through $0$ maps to a circle through $0$ ; a line through $0$ maps to itself.

Worked example: mapping a half-plane to a disk

The map $f(z) = \frac{z - i}{z + i}$ sends the upper half-plane $\operatorname{Im} z > 0$ conformally onto the unit disk. Check the boundary: the real axis $z = x$ maps to $f(x) = (x-i)/(x+i)$ , whose modulus is $|x-i|/|x+i| = 1$ , so the boundary goes to $|w| = 1$ . The point $z = i$ (inside) maps to $0$ (inside). Hence the interior maps to the disk.

Example problem

Problem. Find a Möbius transformation that maps the unit disk to the upper half-plane and sends $z = 1$ to $\infty$ .

Solution

We want a map sending the boundary circle $|z| = 1$ to the real axis and $z = 1$ to $\infty$ (so the pole is at $z = 1$ ). Take $f(z) = i \frac{1 + z}{1 - z}$ . The pole is at $z = 1$ . For $z$ on the unit circle, $z = e^{i\theta}$ , one checks $f(z)$ is real: $f(e^{i\theta}) = i \frac{1 + e^{i\theta}}{1 - e^{i\theta}} = i \cdot \frac{e^{i\theta/2}(e^{-i\theta/2} + e^{i\theta/2})}{e^{i\theta/2}(e^{-i\theta/2} - e^{i\theta/2})} = i \cdot \frac{2\cos(\theta/2)}{-2i \sin(\theta/2)} = -\cot(\theta/2) \in \mathbb{R}$ . The point $z = 0$ (inside) maps to $i$ , which is in the upper half-plane, so the disk maps to the upper half-plane. ∎

The Riemann mapping theorem

Theorem (Riemann mapping). Let $\Omega \subsetneq \widehat{\mathbb{C}}$ be a simply connected domain whose boundary contains more than one point. Then there exists a conformal map $f: \Omega \to \mathbb{D}$ (the unit disk). Moreover, $f$ is unique up to post-composition with a rotation of the disk: if $f(0) = 0$ and $f'(0) > 0$ , then $f$ is unique.

Details

What the theorem says, in plain terms

Up to a conformal change of coordinates, every “nice” simply connected domain is the unit disk. The complicated geometry of the domain is entirely absorbed by the map $f$ ; once you are in the disk, problems (e.g. solving Laplace’s equation with given boundary data) can be attacked with the simple disk geometry and then transported back.

Why the hypotheses are needed

Simply connected: the disk is simply connected, and conformal maps preserve the fundamental group. A domain with a hole (e.g. an annulus) cannot be conformally mapped to the disk.
Proper subset of the sphere: the whole sphere and the whole plane are simply connected but not conformally equivalent to the disk (Liouville’s theorem: a bounded holomorphic function on $\mathbb{C}$ is constant, so no conformal map $\mathbb{C} \to \mathbb{D}$ exists).
Boundary with more than one point: excludes pathological domains.

Sketch of existence (extremal length / normal families)

One classical proof uses the family of all conformal maps from $\Omega$ into $\mathbb{D}$ and shows, via a normal-family (Montel) compactness argument, that the one maximizing $|f'(z_0)|$ exists and is onto. Uniqueness up to rotation follows from the Schwarz lemma applied to $f_1 \circ f_2^{-1}$ .

Use in potential theory

To solve $\Delta u = 0$ in $\Omega$ with boundary data $g$ , map $\Omega$ to $\mathbb{D}$ by $f$ , solve the (easier) Dirichlet problem in the disk with data $g \circ f^{-1}$ (via the Poisson integral), and pull back: $u = u_{\mathbb{D}} \circ f$ . Because conformal maps preserve harmonicity (the real part of a holomorphic function is harmonic, and composition with a conformal map preserves the Laplacian up to a positive factor), this gives the solution in $\Omega$ .

Example problem

Problem. Explain why there is no conformal map from the annulus $\{z : 1 < |z| < 2\}$ onto the unit disk.

Solution

The annulus is not simply connected: a loop around the hole $|z| = 1$ cannot be contracted to a point within the annulus, so its fundamental group is $\mathbb{Z}$ . The disk is simply connected (fundamental group trivial). Conformal maps are homeomorphisms, and homeomorphisms preserve the fundamental group. Since $\mathbb{Z} \not\cong \{0\}$ , no conformal (indeed no homeomorphic) map from the annulus onto the disk exists. ∎

Quiz

Example problem

Problem. Find a conformal map from the upper half-plane $\mathbb{H} = \{z : \operatorname{Im} z > 0\}$ onto the first quadrant $\{w : \operatorname{Re} w > 0, \operatorname{Im} w > 0\}$ .

Solution

Compose two elementary conformal maps. First, $z \mapsto z^2$ maps the upper half-plane onto the plane slit along the positive real axis: a point $z = re^{i\theta}$ with $0 < \theta < \pi$ goes to $r^2 e^{2i\theta}$ with $0 < 2\theta < 2\pi$ , i.e. the whole plane minus the positive real axis. Then $w \mapsto \sqrt{w}$ (the branch with $0 < \arg < \pi$ ) is the inverse of $z^2$ on that slit plane and maps it onto the upper half-plane — not quite the quadrant. Instead, use $f(z) = z^{1/2}$ composed differently: take $f(z) = \sqrt{z}$ with the branch giving $0 < \arg f < \pi/2$ when $0 < \arg z < \pi$ . Concretely, $f(z) = z^{1/2}$ (principal branch) sends $z = re^{i\theta}$ , $0 < \theta < \pi$ , to $\sqrt{r}\, e^{i\theta/2}$ with $0 < \theta/2 < \pi/2$ , i.e. the first quadrant. Since $f'(z) = \frac{1}{2} z^{-1/2} \neq 0$ on $\mathbb{H}$ , $f$ is conformal there. ∎

Conformal mapping §

Conformal (angle-preserving) maps §

Why a non-zero derivative preserves angles §

Why the non-zero condition is essential §

Worked example §

Möbius transformations §

Why they are conformal and circle-preserving §

The three building blocks §

Worked example: mapping a half-plane to a disk §

The Riemann mapping theorem §

What the theorem says, in plain terms §

Why the hypotheses are needed §

Sketch of existence (extremal length / normal families) §

Use in potential theory §

Quiz §

Conformal mapping

Conformal (angle-preserving) maps

Why a non-zero derivative preserves angles

Why the non-zero condition is essential

Worked example

Möbius transformations

Why they are conformal and circle-preserving

The three building blocks

Worked example: mapping a half-plane to a disk

The Riemann mapping theorem

What the theorem says, in plain terms

Why the hypotheses are needed

Sketch of existence (extremal length / normal families)

Use in potential theory

Quiz