Conformal Mappings: Maps That Preserve Angles

1. What "conformal" means

Let the map \( w = f(z) = f(x + iy) = u + iv \) send the point \(z_0\) to \(f(z_0)\), and let \(c_1\), \(c_2\) be any two curves starting at \(z_0\). Their images \(c_1'\), \(c_2'\) are curves starting at \(f(z_0)\). The angle between two curves at a point is the angle between their tangents there.

1° Conformal mapping The mapping \( w = f(z) \) is conformal at the point \(z_0\) if, for any two curves \(c_1, c_2\) through \(z_0\), the angle between \(c_1\) and \(c_2\) is equal to the angle between their images \(c_1'\) and \(c_2'\), both in size and in orientation (sense of rotation).

Conformality is a local property: it is about angles at a point, i.e. about infinitesimally small figures. Large figures can be heavily distorted (a conformal map can turn a strip into a disk), but tiny shapes keep their form: a tiny square becomes a tiny (rotated, rescaled) square, never a parallelogram.

Figure 1. The map \(w = z^2\) applied to a grid of horizontal and vertical lines. The images are two families of parabolas, and wherever the original lines met at right angles, the image parabolas also meet at right angles. Only at \(z = 0\), where \(f'(0) = 0\), does this fail.

2. The theorem on conformal mapping

2° Theorem If \(f(z)\) is regular in a domain \(G\) and \( f'(z) \ne 0 \) in \(G\), then the mapping \( w = f(z) \) is conformal in \(G\).

Why. Let a curve through \(z_0\) be \( z = z(t) \), with \( z(t_0) = z_0 \) and tangent vector \( z'(t_0) \ne 0 \). Its image \( w(t) = f(z(t)) \) has, by the chain rule, the tangent vector

\[ w'(t_0) = f'(z_0)\,z'(t_0). \]

Multiplying by the fixed complex number \( f'(z_0) = |f'(z_0)|e^{i\alpha} \) rotates the tangent by \( \alpha = \arg f'(z_0) \) and stretches it by \( |f'(z_0)| \). Since this happens to every curve through \(z_0\) with the same \(\alpha\), the tangents of \(c_1\) and \(c_2\) are both rotated by the same angle, and the angle between them is unchanged, including its orientation.

Two quantities describe the local behaviour of the map:

\( \arg f'(z_0) \): the angle of rotation at \(z_0\);
\( |f'(z_0)| \): the linear stretching factor at \(z_0\) (all directions stretch equally; areas are multiplied by \(|f'(z_0)|^2\)).

Where \(f'(z_0) = 0\) The condition \( f'(z) \ne 0 \) is essential. If \( f'(z_0) = 0 \) and \( f^{(k)}(z_0) \) is the first nonzero derivative, then near \(z_0\), \( f(z) - f(z_0) \approx c(z - z_0)^k \), and angles at \(z_0\) are multiplied by \(k\). For \(w = z^2\) at \(0\), \(k = 2\): the right angle between the positive axes becomes a straight angle.

Example 1 · Non-examples \(w = \bar z\) preserves the size of angles but reverses their orientation (it is a reflection), so it is not conformal in the sense of the definition. The map \( w = x + 2iy \) (stretching the \(y\)-direction) changes angles: the line \(y = x\) (at \(45^\circ\)) goes to \(v = 2u\) (at \(63.4^\circ\)). Neither is analytic.

3. Bilinear transformations

3° Bilinear mapping The mapping \[ w = \frac{\alpha z + \beta}{\gamma z + \delta} \qquad (\alpha\delta - \beta\gamma \ne 0) \] is called bilinear (also: linear fractional, or Möbius transformation).

The condition \( \alpha\delta - \beta\gamma \ne 0 \) excludes the degenerate case where \(w\) is constant. Its derivative is

\[ w' = \frac{\alpha\delta - \beta\gamma}{(\gamma z + \delta)^2} \ne 0, \]

so a bilinear map is conformal everywhere it is defined. In the extended plane it is a bijection: the point \( z = -\frac\delta\gamma \) goes to \(w = \infty\), and \(z = \infty\) goes to \( w = \frac\alpha\gamma \) (for \(\gamma \ne 0\)). The inverse is again bilinear: \( z = \frac{\delta w - \beta}{-\gamma w + \alpha} \).

Building blocks

For \(\gamma \ne 0\), every bilinear map is a composition of simple steps:

\[ w = \frac\alpha\gamma + \frac{\beta\gamma - \alpha\delta}{\gamma}\cdot\frac{1}{\gamma z + \delta}: \]

\( z \mapsto \gamma z + \delta \): rotation, scaling and translation;
\( \zeta \mapsto \frac1\zeta \): inversion (reflection in the unit circle followed by reflection in the real axis);
again a rotation, scaling and translation.

The circle property A bilinear map takes every circle or straight line to a circle or straight line. (A line counts as a "circle through \(\infty\)".)

Translations, rotations and scalings clearly keep circles circles and lines lines; the only thing to check is the inversion \(w = \frac1z\), and that is a short computation. A circle through the pole \(z = -\frac\delta\gamma\) is sent through \(\infty\), so it becomes a line.

Example 2 · Inversion of a circle through \(0\) Under \( w = \frac1z \), the circle \( |z - 1| = 1 \) (which passes through \(0\)) becomes the vertical line \( \operatorname{Re}w = \frac12 \). Check: \( z = 1 + e^{it} \) gives \( \frac{1}{1 + e^{it}} = \frac{e^{-it/2}}{2\cos\frac t2} = \frac12 - \frac i2\tan\frac t2 \), which indeed has real part \(\frac12\).

A bilinear map has three free complex parameters (the four coefficients matter only up to a common factor). Correspondingly: for any three distinct points \(z_1, z_2, z_3\) and any three distinct points \(w_1, w_2, w_3\), there is exactly one bilinear map sending \(z_k\) to \(w_k\). The tool for finding it is the cross ratio.

4. The cross ratio

Invariance of the cross ratio For four points \(z_1, z_2, z_3, z_4\), the cross ratio \[ (z_1, z_2, z_3, z_4) = \frac{(z_4 - z_1)(z_2 - z_3)}{(z_2 - z_1)(z_4 - z_3)} \] is invariant under bilinear mappings: if \( w_k = \frac{\alpha z_k + \beta}{\gamma z_k + \delta} \), then \( (w_1, w_2, w_3, w_4) = (z_1, z_2, z_3, z_4) \).

Why: for a bilinear map, \( w_j - w_k = \frac{(\alpha\delta - \beta\gamma)(z_j - z_k)}{(\gamma z_j + \delta)(\gamma z_k + \delta)} \). Each point appears once in the numerator and once in the denominator of the cross ratio, so all the factors \( \gamma z_k + \delta \) and \( \alpha\delta - \beta\gamma \) cancel. (If one of the points is \(\infty\), the factors containing it are simply dropped.)

Example 3 · A map through three given points Find the bilinear map with \( 0 \mapsto -1 \), \( 1 \mapsto 0 \), \( \infty \mapsto 1 \). Write the invariance with \(z_4 = z\), \(w_4 = w\): \[ \frac{(w + 1)(0 - 1)}{(0 + 1)(w - 1)} = \frac{(z - 0)(1 - \infty)}{(1 - 0)(z - \infty)}. \] On the right, the factors with \(\infty\) cancel, leaving \(z\). So \( -\frac{w + 1}{w - 1} = z \), i.e. \[ w = \frac{z - 1}{z + 1}. \] Check: \( 0 \mapsto -1 \), \( 1 \mapsto 0 \), \( \infty \mapsto 1 \). ✓ This map sends the imaginary axis to the unit circle and the right half-plane \( \operatorname{Re}z > 0 \) onto the unit disk.

5. The upper half-plane onto the unit disk

4° Mapping the upper half-plane onto the unit disk For any \(z_0\) with \( \operatorname{Im}z_0 > 0 \) and any real \(\theta_0\), the bilinear mapping \[ w = e^{i\theta_0}\,\frac{z - z_0}{z - \bar z_0} \] maps the upper half-plane \( \operatorname{Im}z > 0 \) onto the interior of the unit disk \( |w| < 1 \). The point \(z_0\) goes to the centre \(w = 0\), and the real axis goes to the unit circle.

Typo in the book The denominator must be \( z - \bar z_0 \) (with the conjugate). As printed, \( e^{i\theta_0}\frac{z - z_0}{z - z_0} = e^{i\theta_0} \) would be a constant.

Why it works: \( |w| = \frac{|z - z_0|}{|z - \bar z_0|} \) is the ratio of the distances from \(z\) to \(z_0\) and to its mirror image \(\bar z_0\) in the real axis.

On the real axis, a point is equally far from \(z_0\) and \(\bar z_0\): \( |w| = 1 \).
In the upper half-plane, a point is closer to \(z_0\) (on the same side) than to \(\bar z_0\): \( |w| < 1 \).
\(z_0\) itself gives \(w = 0\), and the factor \(e^{i\theta_0}\) only rotates the disk.

The two free parameters have a clear meaning: \(z_0\) is the point that becomes the centre, and \(\theta_0\) fixes the rotation (it is the argument of the image of \(z = \infty\): \( w(\infty) = e^{i\theta_0} \)). In fact, every conformal one-to-one map of the upper half-plane onto the unit disk has this form.

Example 4 · The Cayley transform With \(z_0 = i\) and \(\theta_0 = 0\): \[ w = \frac{z - i}{z + i}. \] Real points go around the unit circle: \( -1 \mapsto i \), \( 0 \mapsto -1 \), \( 1 \mapsto -i \), \( \pm\infty \mapsto 1 \). As \(x\) runs from \(-\infty\) to \(\infty\), \(w\) goes once around the circle, starting and ending at \(1\). The point \(i\) goes to the centre.

Figure 2. The Cayley transform \(w = \frac{z - i}{z + i}\). Left: the upper half-plane with points \(A, \dots, F\) on the real axis, the point \(P = i\), and a grid of horizontal lines (blue) and vertical lines (red). Right: their images. The real axis becomes the unit circle, \(P\) goes to the centre, horizontal lines become circles touching the unit circle at \(1\), and vertical lines become arcs through \(1\), still meeting the blue circles at right angles.

This map is the workhorse for problems on the half-plane: steady temperatures, potentials and flows are easy to compute in a disk, and the Cayley transform carries the solution over.

6. Summary

Concept	Statement
Conformal at \(z_0\)	angles between curves preserved in size and orientation
Theorem	\(f\) regular and \( f' \ne 0 \) in \(G\) \(\Rightarrow\) conformal in \(G\)
Local behaviour	rotation by \(\arg f'(z_0)\), stretching by \(\|f'(z_0)\|\)
Bilinear map	\( w = \frac{\alpha z + \beta}{\gamma z + \delta} \), \( \alpha\delta - \beta\gamma \ne 0 \); circles/lines \(\to\) circles/lines
Cross ratio	\( \frac{(z_4 - z_1)(z_2 - z_3)}{(z_2 - z_1)(z_4 - z_3)} \) is invariant; 3 points determine the map
Half-plane \(\to\) disk	\( w = e^{i\theta_0}\frac{z - z_0}{z - \bar z_0} \), \( \operatorname{Im}z_0 > 0 \)

7. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Complex Numbers.

Problem 1: Find the angle of rotation and the stretching factor of \( w = e^z \) at \( z_0 = 1 + \frac{\pi}{2}i \).

\( f'(z_0) = e^{z_0} = e\cdot e^{i\pi/2} = ie \): rotation by \(90^\circ\), stretching by \(e \approx 2.72\).

Problem 2: Where is \( w = z^3 - 3z \) not conformal?

\( w' = 3z^2 - 3 = 0 \) at \( z = \pm1 \). There angles are doubled (since \(w'' = 6z \ne 0\)).

Problem 3: Find the image of the real axis under \( w = \frac{z - i}{z + i} \) and of the imaginary axis.

The real axis goes to the unit circle (Section 5). The imaginary axis \( z = iy \): \( w = \frac{y - 1}{y + 1} \), real; as \(y\) runs over \(\mathbb R\) it covers the whole real axis. So the imaginary axis maps onto the real axis (a line through \(\infty\), the image of \(z = -i\)).

Problem 4: Find the bilinear map with \( -1 \mapsto 0 \), \( 0 \mapsto i \), \( 1 \mapsto \infty \).

Zero at \(-1\), pole at \(1\): \( w = k\frac{z + 1}{z - 1} \). From \(0 \mapsto i\): \( -k = i \), \(k = -i\). So \( w = -i\frac{z + 1}{z - 1} = i\frac{1 + z}{1 - z} \).

Problem 5: Verify that the cross ratio of \(1, i, -1, -i\) is real. What does that say?

\( \frac{(-i - 1)(i + 1)}{(i - 1)(-i + 1)} = \frac{-(1 + i)^2}{-(1 - i)^2} = \frac{2i}{-2i} = -1 \). Four points have a real cross ratio exactly when they lie on one circle or line; here, the unit circle.

Problem 6: Map the upper half-plane onto the unit disk so that \(z_0 = 2i\) goes to the centre and \(\infty\) goes to \(w = -1\).

\( w = e^{i\theta_0}\frac{z - 2i}{z + 2i} \), and \( w(\infty) = e^{i\theta_0} = -1 \), so \( w = -\frac{z - 2i}{z + 2i} = \frac{2i - z}{z + 2i} \).

Final thoughts

Conformality is the geometric face of complex differentiability: since \(f'(z_0)\) acts on every small displacement as one complex number (one rotation, one scaling), an analytic map with \(f' \ne 0\) preserves angles. Among conformal maps, the bilinear transformations are the most manageable. They are built from translations, rotations, scalings and one inversion; they send circles and lines to circles and lines; three points determine them, via the invariant cross ratio. The map \( w = e^{i\theta_0}\frac{z - z_0}{z - \bar z_0} \) shows what they can do: an infinite half-plane becomes a bounded disk, with all angles intact.