Conformal maps

Published

18 October 2026

Draft

These notes are still being edited: the text may contain typos, inaccuracies, and places that have not been fully worked out locally. If you find something, please let the author know (contact info on the course page).

1 Introduction

Let \(D \subset \overline{\mathbb {C}}\).

Definition 1. A map \(f\) of a domain \(D\) is called conformal if

\(f\) is a bijection between \(D\) and \(f(D)\);
\(f\), \(f^{-1}\) are regular (in a suitable local parameter).

What does “in a suitable local parameter” mean? Regularity in the definition above means regularity in a neighborhood of every point of the domain of definition \(D \subset \overline{\mathbb {C}}\). Let \(z_0 \in D\), \(f(z_0) = w_0\). 4 cases:

\(z_0 \neq \infty\), \(w_0 \neq \infty\). Then let \(\zeta = z\), \(\omega = w\);
\(z_0 = \infty\), \(w_0 \neq \infty\). Then let \(\zeta = \frac{1}{z}\), \(\omega = w\);
\(z_0 \neq \infty\), \(w_0 = \infty\). Then let \(\zeta = z\), \(\omega = \frac{1}{w}\);
\(z_0 = \infty\), \(w_0 = \infty\). Then let \(\zeta = \frac{1}{z}\), \(\omega = \frac{1}{w}\);

Regularity of \(f\) in a neighborhood of \(z_0\) means regularity of \(\omega = \omega (\zeta )\) in a neighborhood of \(\zeta_0\) as an ordinary complex function (\(\zeta\) and \(\omega\) are chosen so that \(\zeta_0\) and \(\omega_0\) are always finite points).

That is, this is a map that, in the local parameter, is an ordinary regular map, and the derivative at every point (in local coordinates) exists and is nonzero.

Definition 2. A map \(f\) of a domain \(D\) is called conformal if

\(f\) is univalent (i.e. it is a bijection between \(D\) and \(f(D)\))
\(f\) is regular at every point \(z \in D\), except possibly one pole of order \(1\). Moreover, if \(z_0\) is a point of regularity, then \(f'(z_0) \neq 0\).

Definition 3. A map \(f\) of a domain \(D\) is called conformal if

\(f\) is univalent (i.e. it is a bijection between \(D\) and \(f(D)\))
\(f\) is regular in a punctured neighborhood of every point of \(D\);

Example 1

In which domains is \(\exp\) a conformal map?

Solution

Checking univalence: \(z_1 \neq z_2\), \(e^{z_{1}} \stackrel{?}{=}e^{z_2}\) \(\iff\) \(z_1 - z_2 = 2\pi i k\), \(k\in \mathbb {Z}\). Hence, the domain \(D\) must not contain points differing by \(2\pi i k\), \(k \neq 0\).
\(e^{z}\) is an entire function, with no poles or removable singularities. \(\infty \not\in D\);
\(0 \neq \left(e^{z}\right)' = e^z\), \(z \in \mathbb {C}\).

Any horizontal strip of width less than \(2\pi\) will do.

Conformality is an equivalence relation on the set of all domains on the Riemann sphere \(\overline{\mathbb {C}}\). Into which equivalence classes does this relation split the simply connected domains?

Theorem 1 (Riemann) Simply connected domains split into \(3\) equivalence classes:

\(\left\{ \overline{\mathbb {C}}\right\}\) (i.e. the Riemann sphere is conformal only to itself);
\(\left\{ \overline{\mathbb {C}} \setminus \left\{ z_0\right\} \; : \; z_0 \in \overline{\mathbb {C}}\right\}\);
All the remaining simply connected domains, i.e. simply connected domains whose boundary consists of at least 2 points.

2 Fractional-linear maps (Möbius maps)

2.1 Definition

Let \(a, b \in \mathbb {C}\), \(a \neq 0\). Consider the map

\[ z \mapsto w(z) = \begin{cases} az + b, & z \neq \infty \\ \infty , & z = \infty \end{cases} \]

This map \(\overline{\mathbb {C}} \to \overline{\mathbb {C}}\) is called linear. For brevity we write this map as \(w(z) = az + b\), with the understanding that \(z\) may also be \(\infty\) (in which case \(w(\infty )= \infty\)).

Fractional-linear maps are a generalization of linear maps. Let \(a,b,c,d \in \mathbb {C}\), with \(ad - bc \neq 0\). A fractional-linear map (or Möbius map) is a map \(\overline{\mathbb {C}} \to \overline{\mathbb {C}}\) which, in the case \(c = 0\), equals \(\frac{a}{d}z +\frac{b}{d}\) and is given as above, and in the case \(c \neq 0\) is given by

\[ z \mapsto w(z) = \begin{cases} \frac{a z+b}{c z+d}, & z \neq \infty , z \neq -\frac{d}{c} \\ \infty , & z = -\frac{d}{c} \\ \frac{a}{c}, & z = \infty \end{cases} \]

The condition \(a d-b c \neq 0\) means that \(w \not\equiv \operatorname {const}\). Again, for brevity it is customary to write \(w(z) = \frac{a z+b}{c z+d}\), where \(z\) may be substituted by any point of \(\overline{\mathbb {C}}\).

2.2 Properties

2.2.1 Conformality

Lemma 1 (A Möbius map is conformal) The map \(w(z)=\frac{a z+b}{c z+d}\), regarded as a map \(\overline{\mathbb {C}} \to \overline{\mathbb {C}}\), is conformal (on the whole of \(\overline{\mathbb {C}}\)).

In particular, a Möbius map preserves angles between curves.

2.2.2 Möbius maps form a group

It is easy to check that the composition of 2 Möbius maps is again a Möbius map.

Lemma 2 The collection of all fractional-linear maps forms a group under composition (superposition).

Let \(ad - bc \neq 0\). Then the matrix \(\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)\) is nonsingular. To each such matrix corresponds a Möbius map:

\[ \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right) \mathrel {\leadsto } w = w(z) \]

where \(w(z) = \frac{a z+b}{c z+d}\). It turns out (easy to check) that matrix multiplication corresponds to the composition of 2 Möbius maps: if \(\left(\begin{smallmatrix} a_1 & b_1 \\ c_1 & d_{1} \end{smallmatrix}\right)\), \(\left(\begin{smallmatrix} a_2 & b_2 \\ c_2 & d_{2} \end{smallmatrix}\right)\) are two nonsingular matrices, then

\[ \begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \cdot \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} \mathrel {\leadsto } w_1 \circ w_2 \]

where \(w_{1,2}(z) = \frac{a_{1,2}z + b_{1,2}}{c_{1,2}z + d_{1,2}}\)

Lemma 3 (Homomorphism of \(\operatorname {GL}_2(\mathbb {C})\) onto the group of Möbius maps) The map described above is a homomorphism of the group of nonsingular \(2\times 2\) matrices over the complex numbers (usually denoted \(\operatorname {GL}_2(\mathbb {C})\)) onto the group of Möbius maps under composition.

What use is this? If we want to find the inverse map to the Möbius map \(w(z) = \frac{a z+b}{c z+d}\), we just need to see which map corresponds to the matrix \(\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)^{-1}\). Namely,

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \quad \mathrel {\leadsto } \quad w^{-1}(z) = \frac{\frac{1}{ad - bc}dz - \frac{1}{ad - bc} b }{-\frac{1}{ad - bc} cz + \frac{1}{ad - bc} a} = \frac{dz - b}{-cz + a} \]

This map is not an isomorphism for \(\operatorname {GL}_2(\mathbb {C})\), since the kernel is nontrivial. For example, the identity matrix \(\left(\begin{smallmatrix} 1 & 0 \\ 0 & 1\end{smallmatrix}\right)\) maps to the identity map \(w(z) = \frac{1z + 0}{0z + 1} = z\). But so does, for example, \(\left(\begin{smallmatrix} 2 & 0 \\ 0 & 2\end{smallmatrix}\right)\): in this case \(w(z) = \frac{2z + 0}{0z+ 2} =z\); the kernel consists of all scalar matrices \(\lambda E\), \(\lambda \neq 0\). If we restrict to matrices with unit determinant, the kernel shrinks but remains nontrivial: \(\operatorname {det}(-E) = 1\), and the corresponding Möbius map is the identity. That is, the kernel narrows to \(\left\{ \pm E\right\}\).

Lemma 4 (The group of Möbius maps is isomorphic to \(\operatorname {PSL}_2(\mathbb {C})\)) The map described above induces an isomorphism between the quotient group \(\operatorname {PSL}_2(\mathbb {C}) = \operatorname {SL}_2(\mathbb {C})/\left\{ \pm E\right\}\) (matrices with unit determinant, taken up to sign) and the group of Möbius maps under composition.

2.2.3 The circle-preserving property

Lemma 5 Under a fractional-linear map, the image of any circle on the Riemann sphere \(\overline{\mathbb {C}}\) is again a circle on the Riemann sphere.

Recall that a circle on the Riemann sphere (under stereographic projection) corresponds to a circle in the plane \(\mathbb {C}\) if, on the sphere, the circle did not pass through the top point, i.e. through \(\infty\). If it does pass through \(\infty\), then on the plane it corresponds to a line.

2.2.4 A Möbius map preserves symmetry of points

Let \(z_1, z_2 \in \overline{\mathbb {C}}\), and let \(\gamma\) be a line or a circle. We need to define symmetry of points with respect to \(\gamma\).

Let \(\gamma\) be a line. Finite points \(z_1, z_2 \in \mathbb {C}\) are called symmetric with respect to \(\gamma\) if they are symmetric in the ordinary geometric sense. That is, the segment \([z_1, z_2]\) crosses the line \(\gamma\) perpendicularly, with the distance from \(\gamma\) to \(z_1\) and to \(z_2\) equal.

Further, \(\infty\) is symmetric only to \(\infty\). It is convenient to adopt this convention, since every point on a line is symmetric to itself (with respect to that line), and \(\infty\) lies on the line (once we pass to the Riemann sphere).
Let \(\gamma\) be a circle. Points \(z_1\), \(z_2\) are symmetric with respect to \(\gamma\) if they are related by inversion. Namely, they lie on the same ray from the center \(z_0\) of the circle \(\gamma\), with \(\left|z_1 - z_0\right| \cdot \left|z_2 - z_0\right| = R^2\), where \(R\) is the radius of \(\gamma\).

In particular, the center \(z_0\) itself is symmetric to \(\infty\) with respect to \(\gamma\).

Lemma 6 (The symmetry-preservation property) Under a fractional-linear map, a pair of points symmetric with respect to a circle or a line maps to a pair of points symmetric with respect to the image of that circle or line.

Example 2

Given the fractional-linear function \[ w = f(z) = \frac{1 + iz}{1 - iz}. \] Find the image of the upper half-plane \(D = \left\{ z \; : \; \operatorname {Im}z > 0\right\}\).

Solution

Let \(\gamma = \partial D = \left\{ z \; : \; \operatorname {Im} z = 0\right\}\). \(\gamma\) is the real axis. By the properties of conformal maps, the boundary of \(f(D)\) is the set \(f(\gamma )\). Elementary observations:

\[ 0 \mapsto f(0) = 1, \qquad 1 \mapsto \frac{1 + i}{1-i} = \frac{2i}{2} = i, \qquad \infty \mapsto \lim _{z \to \infty }f(z) = \lim _{z \to \infty }\frac{\frac{1}{z} + i}{\frac{1}{z} - i} = \frac{i}{-i} = -1 \]

Further, the point that sends the denominator to zero maps to \(\infty\), i.e. the point \(-i\). In total, \(-i \mapsto \infty\).

Let us find \(f(\gamma )\), i.e. the image of the line \(\operatorname {Im} z = 0\). The image of any line under a Möbius map is a line (if a point on the line maps to infinity) or a circle (otherwise). In our case \(-i \mapsto \infty\), and \(-i \not\in \gamma\) does not lie on the line in question. Hence, \(\infty \not\in f(\gamma )\), so \(f(\gamma )\) is indeed a circle. How do we find this circle?

Method 1, using 3 points. \(0,1,\infty\) lay on \(\gamma\) (in \(\overline{\mathbb {C}}\)), so \(f(\gamma )\) will pass through \(1, i, -1\). The unique line or circle passing through these points is the circle \(\left\{ \left|z\right| = 1\right\}\). Hence, that is \(f(\gamma )\).
Method 2, via intersection angles and 2 points. We need to understand where the real axis goes. One can note that substituting into \(f(z) = \frac{1 + iz}{1 - iz}\) points from the line \(\operatorname {Re}z = 0\), i.e. points of the form \(z = iy\), yields real numbers. Hence, the imaginary axis maps to the real axis. The imaginary and real axes intersect at a right angle, so their images must intersect at a right angle too. Moreover, we know that \(-1, 1 \in f(\gamma )\). Hence \(f(\gamma )\) is a line or circle passing through \(-1,1\) that intersects the real axis at right angles. The only option is the circle \(\left|z\right| = 1\).
Method 3, using the center and a point on the circle. We have found that \(f(\gamma )\) is a circle. How do we find the point \(w_0\), the center of this circle? \(w_0\) is symmetric to \(\infty\) with respect to \(f(\gamma )\), so \(z_0 := f^{-1}(w_0)\) is symmetric to \(f^{-1}(\infty ) = -i\) with respect to \(\gamma\). I.e. \(z_0 = i\), \(w_0 = f(z_0) = 0\). Further, we know that \(1 \in f(\gamma )\). \(f(\gamma )\) is a circle centered at \(0\) and passing through \(1\). The only option: \(f(\gamma ) = \left\{ \left|z\right| = 1\right\}\).

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: the conformal map D -> f(D).

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/330/010_conformal_map_D_fig.tex

We have found the boundary of the image. It remains to determine on which side of this boundary the image lies. We have already said that \(-i \mapsto \infty\). Moreover \(-i \not\in D\), so \(\infty \not\in f(D)\), so \(f(D) = \left\{ \left|z\right| < 1\right\}\).

Answer

\(f(D) = \left\{ \left|z\right| < 1\right\}\).

Example 3

Given the Möbius map \[ w = f(z) = \frac{2iz}{z + 3} \] Find the image of \(D = \left\{ \left|z - 1\right| < 2\right\}\).

Solution

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: D -> f(D), the image is the unit disk.

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/330/020_conformal_map_unit_disk_fig.tex

\(-3 \mapsto \infty\). \(-3 \not\in \gamma = \partial D = \left\{ \left|z-1\right| = 2\right\}\). Hence, the image of the boundary \(f(\gamma )\) is a circle.

\(-1 \mapsto \frac{-2i}{2} = -i\), \(3 \mapsto \frac{6i}{6} = i\). Further, the points \(-3\) and \(0\) are symmetric with respect to \(\gamma\) (they lie on the same ray from the center \(1\), and \(\left|-3-1\right|\cdot \left|0-1\right| = 4 = R^2\)), so their images \(\infty\) and \(f(0) = 0\) are symmetric with respect to \(f(\gamma )\), i.e. \(0\) is the center of the circle \(f(\gamma )\). It passes through \(\pm i\), i.e. \(f(\gamma ) = \left\{ \left|w\right| = 1\right\}\).

Finally, \(-3 \not\in D\), so \(\infty = f(-3) \not\in f(D)\): the image is the interior of the disk, \(f(D) = \left\{ \left|w\right| < 1\right\}\).

Answer

\(f(D) = \left\{ \left|z\right| < 1\right\}\).

Problem 27.4

Find the image of the half-plane \(\left\{ z: \operatorname {Re} z < 1\right\}\) under the following maps:

\(w = \frac{z-3+i}{z+1+i}\).

2.2.5 A Möbius map preserves the cross-ratio of 4 points

Let \(z_1, z_2, z_3, z_4 \in \overline{\mathbb {C}}\). The cross-ratio of these four numbers is defined as

\[ (z_1, z_2, z_3, z_4 ) \quad := \quad \frac{z_3 - z_1}{z_3 - z_2} \; \colon \; \frac{z_4 - z_1}{z_4 - z_2} = \frac{(z_3 - z_1) (z_4 - z_2)}{(z_3 - z_2) (z_4 - z_1)} \]

In particular, if \(z_1 = \infty\), then

\[ (\infty , z_2, z_3, z_4 ) = \frac{(z_3 - \infty ) (z_4 - z_2)}{(z_3 - z_2) (z_4 - \infty )} = \lim _{z_1 \to \infty } \frac{(z_3 - z_1) (z_4 - z_2)}{(z_3 - z_2) (z_4 - z_1)} = \frac{z_4 - z_2}{z_3 - z_2} \]

Lemma 7 Let \(w = w(z)\) be a Möbius map, \(w_i := w(z_i)\). Then \[ (z_1, z_2, z_3, z_4 ) = (w_1, w_2, w_3, w_4 ) \]

2.3 Examples of fractional-linear maps

2.3.1 3 points to 3 points

There exists a unique fractional-linear map under which three distinct points \(z_{1}, z_{2}, z_{3}\) of \(\overline{\mathbb {C}}\) map respectively to three distinct points \(w_{1}, w_{2}, w_{3}\) of \(\overline{\mathbb {C}}\). This map can be found from the equality of cross-ratios: \((z_1, z_2, z_3, z) = (w_1, w_2, w_3, w)\). Namely,

\[ \frac{w_3-w_{1}}{w_3-w_{2}} \colon \frac{w-w_{1}}{w-w_{2}}=\frac{z_3-z_{1}}{z_3-z_{2}} \colon \frac{z-z_{1}}{z-z_{2}} . \]

2.3.2 Upper half-plane to the unit disk

Every conformal map from the upper half-plane \(\{ z: \operatorname {Im} z>0\}\) onto the disk \(\{ w:|w|<1\}\) has the form

\[ w(z)=\frac{z-z_{0}}{z-\bar{z}_{0}} \cdot e^{i \alpha } \]

where \(\operatorname {Im} z_{0}>0, \quad \alpha \in \mathbb {R}\).

2.3.3 Unit disk to the upper half-plane

We seek the inverse of the previous map:

\[ we^{-i\alpha } (z - \overline{z_0}) = z - z_0 \iff z = \frac{we^{-i\alpha }\overline{z_0} - z_0}{we^{-i\alpha } - 1} = \frac{we^{-2i\operatorname {arg}z_0} - e^{i\alpha }}{w - e^{i\alpha }}z_0 = \frac{we^{-2i\beta } - e^{i\alpha }}{w - e^{i\alpha }}r \cdot e^{i\beta } \]

So, every conformal map from the unit disk \(\{ w:|w|<1\}\) to the upper half-plane \(\{ z: \operatorname {Im} z>0\}\) has the form

\[ w(z)=\frac{z e^{-2i\beta } - e^{i\alpha }}{z - e^{i\alpha }}r \cdot e^{i\beta } \]

where \(r > 0, \beta \in (0,\pi ), \quad \alpha \in \mathbb {R}\).

The simplest example: \(\alpha = 0\), \(\beta = \pi /2\), \(r= 1\),

\[ w(z) = \frac{-z - 1}{z-1}i = \frac{1 + z}{1 - z}i \]

2.3.4 Unit disk to the unit disk

Every conformal map from the disk \(\{ z:|z|<1\}\) onto the disk \(\{ w:|w|<1\}\) has the form

\[ w(z)=\frac{z-z_{0}}{1-\bar{z}_{0}z} \cdot e^{i \alpha } \]

where \(\left|z_{0}\right|<1, \quad \alpha \in \mathbb {R}\).

2.3.5 Upper half-plane to the upper half-plane

Every conformal map from the upper half-plane \(\{ z: \operatorname {Im} z>0\}\) onto the upper half-plane \(\{ w: \operatorname {Im} w>0\}\) has the form

\[ w(z)=\frac{a z+b}{c z+d} \]

where \(a, b, c, d \in \mathbb {R}\) and \(a d-b c>0\).

Example 27.7

Find the fractional-linear functions \(w(z)\) satisfying the conditions:

\(w(i) = 0, w(\infty ) = 1, w(-i) = \infty\); Find the image of the half-plane \(\left\{ z: \operatorname {Re} z > 0\right\}\) under the maps given by these functions.

Solution

Equality of cross-ratios: \((i, \infty , -i, z) = (0,1,\infty , w)\). Let us simplify:

\[ \begin{align} (i, \infty , -i, z) & = \frac{-i - i}{-i -\infty } \colon \frac{z - i}{ z - \infty } = \frac{-2i}{-i -\infty } \cdot \frac{z - \infty }{z - i} = \frac{-2i}{z - i} \\ (0,1,\infty , w) & = \frac{\infty - 0}{\infty - 1} \colon \frac{w - 0}{w - 1} = \frac{w-1}{w} = 1 - \frac{1}{w} \end{align} \]

We get \(1 - \frac{1}{w} = \frac{-2i}{z - i}\) and

\[ w = \left(1 - \frac{-2i}{z - i}\right)^{-1} = \left(\frac{z - i - (-2i)}{z - i}\right)^{-1} = \frac{z-i}{z + i} \]

Further, let \(D = \left\{ \operatorname {Re}z > 0\right\}\), then \(\gamma = \partial D = \left\{ \operatorname {Re}z = 0\right\} = \left\{ z = iy, \; y \in \mathbb {R}\right\}\). Substituting:

\[ w(iy) = \frac{iy-i}{iy + i} = \frac{y-1}{y+1} \]

This is always a real number. Hence the imaginary line \(\gamma = \left\{ \operatorname {Re}z = 0\right\}\) maps to the real line \(\left\{ \operatorname {Im}z = 0\right\}\).

Further, \(1 \mapsto \frac{1-i}{1+i} = \frac{(1-i)^2}{2} = -i\). Hence, \(-i \in f(D)\). Hence, \(f(D) = \left\{ \operatorname {Im}z < 0\right\}\).

Answer

\(w = \frac{z-i}{z+i}, \operatorname {Im} w < 0\);

Problem 27.7

Find the fractional-linear functions \(w(z)\) satisfying the conditions:

\(w(i) = 2, w(\infty ) = 1+i, w(-i) = 0\); Find the image of the half-plane \(\left\{ z: \operatorname {Re} z > 0\right\}\) under the maps given by these functions.

Example 27.8

Find a function \(w(z)\) that conformally maps the domain \(D\) onto the domain \(D_{1}\) and satisfies the given conditions:

\(D = \left\{ z: \operatorname {Re} z > -1\right\} , D_{1} = \left\{ w:|w| < 1\right\} , w(0) = 0, \arg w^{\prime }(0) = \pi\).

Solution

We know how to map the upper half-plane to the unit disk. So the first step is to turn the domain \(\left\{ \operatorname {Re}z > -1\right\}\) into the upper half-plane.

\[ \left\{ \operatorname {Re}z > -1\right\} \quad \xmapsto {\, w_1(z)=z + 1\, } \quad \left\{ \operatorname {Re}w_1 > 0\right\} \quad \xmapsto {\, w_2(w_1)= w_1 \cdot e^{i \pi /2} = iw_1\, } \quad \left\{ \operatorname {Im}w_2 > 0\right\} \]

Further, the general form of a conformal map from \(\left\{ \operatorname {Im}w_2 > 0\right\}\) to \(\left\{ \left|w\right| < 1\right\}\):

\[ w(w_2) = \frac{w_2 - z_0}{w_2 - \overline{z_0}}e^{i\alpha } \]

where \(\operatorname {Im}z_0 > 0\).

The resulting map:

\[ w(z) = \left(w \circ w_2 \circ w_1 \right)(z) =\frac{i(z+1) - z_0}{i(z+1) - \overline{z_0}}e^{i\alpha } \]

Substituting the conditions:

\[ 0 = w(0) = \frac{i - z_0}{i - \overline{z_0}}e^{i\alpha } \]

Hence \(z_0 = i\) and \(w(z) = \frac{iz}{iz + 2i}e^{i\alpha } = \frac{z}{z + 2}e^{i\alpha }\). Further,

\[ \pi = \operatorname {arg}w'(0) = \operatorname {arg}\left(\frac{d}{dz}\left(\frac{z}{z + 2}e^{i\alpha }\right)\right)\bigg\rvert _{z = 0} = \operatorname {arg}\left(\frac{2}{(0 + 2)^2}e^{i\alpha }\right) = \alpha \]

In total \(\alpha = \pi\) and

\[ w(z) = \frac{z}{z+2}e^{i\pi } = -\frac{z}{z+2} \]

Answer

\(w = -\frac{z}{z+2}\).

Problem 27.8

Find a function \(w(z)\) that conformally maps the domain \(D\) onto the domain \(D_{1}\) and satisfies the given conditions:

\(D = \left\{ z:|z| < 1\right\} , D_{1} = \left\{ w:|w| < 1\right\} , w\left(z_{0}\right) = w_{0}\), \(\arg w^{\prime }\left(z_{0}\right) = \alpha\), \(\left(\left|z_{0}\right| < 1,\left|w_{0}\right| < 1\right)\);
\(D = \left\{ z: \operatorname {Im} z > 0\right\} , D_{1} = \left\{ w:|w| < 1\right\} , w(0) = i, w(-1) = 1\), \(w(\infty ) = -1\);

Example 6

Find the Möbius map \(0 \mapsto 4\), \(1 + i \mapsto 2 + 2i\), \(2i \mapsto 0\).

Solution

One can solve this directly using the equality of cross-ratios.

Or as follows: from \(2i \mapsto 0\) we seek \(f\) in the form \(f(z) = \frac{a(z-2i)}{cz + d}\). The condition \(0 \mapsto 4\): \(\frac{-2ia}{d} = 4\), i.e. \(d = -\frac{ia}{2}\). The condition \(1+i \mapsto 2+2i\):

\[ \frac{a(1+i-2i)}{c(1+i) + d} = 2 + 2i \iff a(1-i) = \underbrace{(2+2i)(1+i)}_{=4i}c + \underbrace{(2+2i)\left(-\tfrac {ia}{2}\right)}_{= a(1-i)} \iff 4ic = 0 \]

I.e. \(c = 0\): the map is linear, \(f(z) = \frac{a}{d}(z - 2i)\), where \(\frac{a}{d} = \frac{a}{-ia/2} = 2i\). In total

\[ f(z) = 2i(z - 2i) = 2iz + 4 \]

Check: \(f(1+i) = 2i + 2i^2 + 4 = 2 + 2i\) \(\checkmark\).

Answer

\(f(z) = 2iz + 4\).