# Conformal mapping

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Conformal mapping
```(November 23, 2014)
Conformal mapping
Paul Garrett [email protected]
http://www.math.umn.edu/egarrett/
[This document is
http://www.math.umn.edu/˜garrett/m/complex/07 conformal mapping.pdf]
1.
2.
3.
4.
5.
6.
Conformal (angle-preserving) maps
Lines and circles and linear fractional transformations
Elementary examples
f 0 (zo ) = 0 implies local non-injectivity
Automorphisms of the disk and of H
Schwarz’ lemma
1. Conformal (angle-preserving) maps
A complex-valued function f on a non-empty open U ⊂ C is conformal if it preserves angles, in the sense
that, for any two smooth parametrized curves γ : [a, b] → C, δ : [c, d] → C with γ(a) = zo = δ(c), the
angle between γ 0 (a) and δ 0 (c) is equal to the angle between (f ◦ γ)0 (a) and (f ◦ δ)0 (c). The function f is
orientation-preserving if the directed angle is preserved.
Explicitly, for non-zero z, w ∈ C,
directed angle from z to w
That is,
eiθ =
=
θ ∈ R, such that
w
z
= eiθ
|w|
|z|
w .z
|w| |z|
[1.0.1] Claim: A holomorphic function f is conformal and orientation-preserving at points zo where
f 0 (zo ) 6= 0.
Proof: This is a direct computation, using the chain rule, noting that f 0 is the complex derivative of f ,
while γ 0 is the real derivative. With γ(a) = zo = δ(c),
(f ◦ γ)0 (a) = f 0 (γ(a)) · γ 0 (a) = f 0 (zo ) · γ 0 (a)
so
(f ◦ δ)0 (c) = f 0 (δ(a)) · δ 0 (a) = f 0 (zo ) · δ 0 (a)
(f ◦ γ)0 (a) . (f ◦ δ)0 (a)
f 0 (zo ) γ 0 (a) . f 0 (zo ) δ 0 (a)
γ 0 (a) . δ 0 (a)
=
=
|(f ◦ γ)0 (a)| |(f ◦ δ)0 (a)|
|f 0 (zo ) γ 0 (a)| |f 0 (zo ) δ 0 (a)|
|γ 0 (a)| |δ 0 (a)|
showing that directed angles are preserved.
///
[1.0.2] Remark: Holomorphic f on U with non-vanishing derivative maps the mutually orthogonal families
of lines Re (z) = x = const and Im (z) = y = const to mutually orthogonal curves.
For example, letting f (z) = z 2 on the upper half-plane, the lines x → x + iyo become parabolas
x → x2 − 2ixyo − yo2 opening horizontally, and the lines y → xo + iy become parabolas y → −y 2 + 2xo y + x2o
opening vertically.
√
Letting f (z) = z on C − [0, +∞) amounts to looking at inverse images in the previous example, giving
mutually orthogonal lines Re (z 2 ) = const and Im (z 2 ) = const, namely, hyperbolas x2 − y 2 = const and
hyperbolas xy = const.
1
Paul Garrett: Conformal mapping (November 23, 2014)
2. Lines and circles and linear fractional transformations
[2.0.1] Theorem: The collection of lines and circles in C ∪ {∞} is stabilized by linear fractional
transformations, and is acted upon transitively by them.
a b
Proof: Clearly affine maps
(z) = (az + b)/d, with ad 6= 0, which are combinations of translations,
0 d
rotations, and dilations, preserve lines and circles. Given this, a group-theoretic result greatly simplifies
things:
[2.0.2] Claim: (Bruhat
decomposition) Let P be the upper-triangular matrices in GL2 (C), and the Weyl
element wo =
0
1
1
. Then
0
GL2 (C) = P t P wo P
Proof: A group element g =
a
c
b
d
a
c
b
d
(disjoint union)
is in P if c = 0, so consider c 6= 0. Then
1
0
− dc
1
0
1
1
0
=
∗
0
∗
∗
∈ P
That is, for g 6∈ P , gP wo ∩ P 6= φ. That is, gP ∩ P wo 6= φ, so gP ∩ P wo P 6= φ, and g ∈ P wo P .
///
Invoking the Bruhat decomposition, since g ∈ P preserves lines and circles, it suffices to prove that the Weyl
element wo preserves lines and circles. The real and imaginary parts of wo (z) = 1/z are easily observed:
1
z
=
z
|z|2
so
Re (1/z) =
x
|z|2
Im (1/z) = −
y
|z|2
It is convenient that wo2 gives the identity map, so the images and inverse images of sets under wo are the
same things. Given a line ax + by = c with real a, b, c, the image of the line is
a
x
y
−b 2 = c
|z|2
|z|
or
ax − by = cx2 + cy 2
giving the circle
b
a
(x − )2 + (y + )2 =
c
c
a 2
b 2
+
c
c
The image of a circle |z − zo | = r is
1
− zo = r
z
or |1 − zo · z| = r · |z| or
1 − 2 Re (zo · z) + |zo |2 · |z|2 = r2 · |z|2
In the special case that |zo | = r, this is a line. Otherwise, it is a circle.
2
Paul Garrett: Conformal mapping (November 23, 2014)
Three points on a circle determine the circle completely. A line in C can be viewed determined by two
points on it in C, and inevitably passing through ∞. The group of linear fractional transformation actions
is triply transitive in the sense that it can map any triple of distinct points to any other, so is transitive on
circles-and-lines.
///
[2.0.3] Remark: Beware: except for affine maps
a
0
b
d
(z) = (az + b)/d, with ad 6= 0, linear fractional
transformations do not respect centers of circles.
3. Elementary examples
A sector is an open set of the form
{z = reiθ : r > 0, a < θ < b, with |b − a| < 2π}
Two sectors of the special form
{z = reiθ : r > 0, 0 < θ < b0 , with b0 < 2π}
{z = reiθ : r > 0, 0 < θ < b, with b < 2π}
can be mapped holomorphically to each other by a power map
b0
b0
z −→ z b = e b ·log z
with a continuous logarithm defined on the first sector by
Z
z/eiπ
log z = iπ +
1
dw
w
integrating along a straight line segment. In polar coordinates, this is simply
b0
b0
reiθ −→ r b eiθ b
Exhibiting the map as a holomorphic map shows that it preserves angles.
Sectors with edges elsewhere than the positive real axis can be rotated, by map z → µ · z with |µ| = 1, to
put either edge on the positive real axis. Thus, the problem of mapping one sector to another reduces to
that simpler case, by pre-composing and post-composing with rotations:
{z = reiθ : r > 0, a < θ < b, with |b − a| < 2π}
can be mapped to another sector
{z = reiθ : r > 0, a0 < θ < b0 , with |b0 − a0 | < 2π}
by
0 −a0
0 −a0
z bb−a
z bb−a
z
ia0
z −→ ia −→
−→ e · ia
e
eia
e
Again, there is an unambiguous choice of continuous αth power on such a sector, by
z α = eα·log z
3
Paul Garrett: Conformal mapping (November 23, 2014)
where log z is defined on C with any ray {reiθo : r > 0} removed, with this ray not lying in the given sector.
Again, such a logarithm can be defined by
Z
z/ei(θo +π)
log z = i(θo + π) +
1
dw
w
integrating along a straight line segment from 1 to z/ei (θo +π). That is, all sectors are conformally equivalent.
In other words, while z → z α with real α 6= 0 is conformal at every point other than z = 0, it alters angles
at 0 by multiplying angles by α.
A bigon is an open subset of CP1 bounded by two arcs, by which we for present purposes we mean either
straight line segments, possibly infinite in one or both directions, and/or arcs of circles with angle measure
< 2π.
The non-degenerate cases are specified by two vertices z1 , z2 ∈ PC1 , and two distinct circles-or-lines passing
through both z1 , z2 . This configuration cuts PC into four connected components, each of which is a bigon.
In the degenerate case that the two points are identical, CP1 is cut into only three connected components.
[3.0.1] Claim: All bigons in the non-degenerate case are conformally equivalent, and are conformally
equivalent to the upper half-plane.
Proof: Given vertices z1 6= z2 , map z1 → 0 and z2 → ∞ by a linear fractional transformation.
Since linear fractional transformations preserve lines and circles, the boundary of the image consists of
two straight lines from 0 to ∞, so the image is a sector. Rotate the sector until is is of the form
{reiθ : r > 0, 0 < θ < b, with b < 2π}, and the apply a power map to obtain the upper half-plane H.
Thus, any non-degenerate bigon is conformally equivalent to the upper half-plane, so they are equivalent to
each other.
///
[3.0.2] Claim: All degenerate bigons are conformally equivalent, and are conformally equivalent to the strip
0 < Re (z) < 1.
Proof: Map the single vertex to ∞. The two bounded arcs, mapped to arcs passing through ∞, must
become straight lines with no other intersections, thus, parallel. Translate, rotate and dilate to obtain the
lines Re (z) = 0 and Re (z) = 1.
///
[3.0.3] Remark: An open disk can be considered as a non-degenerate bigon in many ways, as can a
half-plane. The Cayley map
z+i
iz + 1
z −→
maps i → ∞, −i → 0, and 1 → 1, so maps the unit circle to the real line, because linear fractional
transformations preserve circles-and-lines. Thus, it maps the connected components of the complement of
the circle to the connected components of the complement of the real line: since 0 → i, the interior of the
circle is mapped to the upper half-plane.
4
Paul Garrett: Conformal mapping (November 23, 2014)
4. f 0(z) = 0 implies local non-injectivity
The following generally-useful corollary of the argument principle could have been proven earlier, but is
perhaps of interest here as a sort of converse to the holomorphic inverse-function theorem:
[4.0.1] Claim: A non-constant holomorphic function f near zo vanishing to order exactly k at zo is (k + 1)to-1 in every sufficiently small punctured disk at zo , in the following precise sense. Given a sufficiently small
r > 0, there is a neighborhood U of zo such that, for all z1 in U , the value f (z1 ) is hit k + 1 times inside
|z − zo | = r.
Proof: The idea is that, on one hand, since f (z) − f (zo ) vanishes to order k at zo , the argument principle
would give
k+1 ≤
1
2πi
Z
γ
(f (w) − f (zo ))0 dw
1
=
f (w) − f (zo )
2πi
Z
γ
f 0 (w) dw
f (w) − f (zo )
where γ is the counterclockwise path around |z − zo | = r. On the other hand,
Z
1
f 0 (w) dw
z −→
2πi γ f (w) − f (z)
should be a holomorphic function of z on some neighborhood U of zo . An integer-valued holomorphic
function is constant. Thus, it should be every value f (z1 ) for z1 ∈ U is hit at least k + 1 times inside the
circle |z − zo | = r.
Two details are missing. First, we want to be sure that the denominator f (w) − f (zo ) does not vanish on the
circle |z − zo | = r. We claim that we can shrink r if necessary so that f (w) 6= f (zo ) on that circle: indeed, if
there were a sequence of points z1 , z2 , . . . approaching zo with f (zj ) = f (zo ), then by the identity principle
f is constant. Second, we want the denominator f (w) − f (z) to not vanish for all w on the circle and for
all z sufficiently near to zo . Indeed, the set of images I = {f (z) : |z − zo | = r} is a continuous image of a
compact set, so is compact, so is closed. Since f (zo ) 6∈ I, there is some neighborhood N of f (zo ) disjoint
from I. The inverse image f −1 (N ) is open, by continuity, and contains zo , so there is a neighborhood U of
zo such that f (U ) ∩ I = φ. Shrink U to be an open disk at zo , so it is connected.
///
5. Automorphisms of the disk and of H
First, we demonstrate some explicit groups of linear transformations stabilizing the upper half-plane, or
stabilizing the unit disk, in both cases large enough to act transitively. Then we invoke Schwarz’ lemma
(from the following section) to see that these groups are all the holomorphic automorphisms of these regions.
[5.0.1] Claim: The linear fractional transformations arising from
SL2 (R) = {two-by-two real matrices with determinant = 1}
Im (z)
a b
stabilize the upper half-plane H, and act transitively on it. In particular, Im
(z) =
.
c d
|cz + d|2
Proof: First, SL2 (R) is a subgroup of GL2 (R), including inverses. Directly compute the effect on imaginary
parts:
2iIm
=
a
c
b
d
(x + iy) = 2iIm
az + b
az + b az + b
(az + b)(cz + d) − (az + b)(cz + d)
=
−
=
cz + d
cz + d cz + d
|cz + d|2
(aczz + bcz + adz + bd) − (aczz + bcz + adz + bd)
z−z
2iIm (z)
=
=
=
|cz + d|2
|cz + d|2
|cz + d|2
|cz + d|2
5
Paul Garrett: Conformal mapping (November 23, 2014)
This shows that SL2 (R) stabilizes the upper half-plane. To show transitivity, observe that for x + iy ∈ H
1
0
x
1
√
y
0
0
(i) = x + iy
√1
y
so the point i ∈ H can be mapped to any other.
///
[5.0.2] Remark: The special orthogonal group
cos θ
SO2 (R) = {
− sin θ
sin θ
cos θ
: θ ∈ R} ⊂ SL2 (R)
is the stabilizer subgroup in SL2 (R) of the point i ∈ H.
Let g → g ∗ be conjugate transpose:
−1
0
and put S =
0
1
a
c
b
d
∗
=
a
b
c
d
[5.0.3] Claim: The subgroup
SU (1, 1) = {g =
a
c
b
d
∈ GL2 (C) : g ∗ Sg = S, det g = 1} ⊂ GL2 (C)
stabilizes the open unit disk and acts transitively on it.
Proof: Observe that
−|α|2 + |β|2 =
For g =
a
c
b
d
∗ α
α
S
β
β
∈ U (1, 1),
az + b 2 − 1 · |cz + d|2 = |az + b|2 − |cz + d|2
|gz|2 − 1 · |cz + d|2 = cz + d
∗ ∗
∗ az + b
az + b
z
z
z
z
S
= g
S g
=
g ∗ Sg
cz + d
cz + d
1
1
1
1
∗
z
z
=
S
= |z|2 − 1 < 0
(for z in the unit disk)
1
1
µ 0
2
This proves that U (1, 1) stabilizes the open disk. There are rotations z → µ ·z given by
in SU (1, 1)
0 µ
for |µ| = 1, and these are transitive on each circle of radius 0 ≤ r < 1. The elements
=
cosh t sinh t
sinh t cosh t
are in SU (1, 1), and send
0 −→ tanh t =
e2t − 1
et − e−t
=
et + e−t
e2t + 1
6
(for t ∈ R)
Paul Garrett: Conformal mapping (November 23, 2014)
By the intermediate value theorem, since limt→+∞ tanh t = 1, every real value T in the interval 0 ≤ T < 1.
Thus, SU (1, 1) is transitive on the open unit disk.
///
1 i
[5.0.4] Remark: Conjugation by the Cayley map C =
has the property that C · g · C −1 ∈ SL2 (R)
i 1
for g ∈ SU (1, 1), and vice-versa. In particular, C −1 conjugates rotations of the disk to the group SO2 (R).
[5.0.5] Corollary: (of Schwarz’ lemma) Any holomorphic automorphism of the open unit disk is given by
an element of SU (1, 1). Any holomorphic automorphism of the upper half-plane is given by an element of
SL2 (R).
Proof: Given a holomorphic map f of the open disk to itself, compose with an element of U (1, 1) to adjust
so that f (0) = 0. Certainly |f (z)| ≤ 1 for |z| < 1. By Schwarz’ lemma, |f (z)| ≤ |z| and |f 0 (0)| ≤ 1, and if
equality holds at any point with 0 < |z| < 1 or if |f 0 (0)| = 1, then f (z) = µ · z with |µ| = 1.
By the holomorphic inverse function theorem, (f −1 )0 (0) = 1/f 0 (0). Also, Schwarz’ lemma applies to f −1 .
Thus,
1
= |(f −1 )0 (0)| ≤ 1
1 ≤
|f 0 (0)|
implies equality, and that f (z) = µ · z for some |µ| = 1. This shows that the linear fractional transformations
given by SU (1, 1) are the whole holomorphic automorphism of the open unit disk.
Similarly, for a holomorphic automorphism f of H, let f (i) = z, and let g ∈ SL2 (C) map g(z) = i. Then
h = C −1 ◦ g ◦ f ◦ C
is a holomorphic automorphism of the open unit disk fixing 0, so is a rotation coming from SU (1, 1). Note
that CgC −1 ∈ SO2 (R). Then
f = g ◦ (C ◦ h ◦ C −1 )
expresses f : H → H as a composition of linear fractional transformations g ∈ SL2 (R).
///
[5.0.6] Remark: The action on CP1 given by scalar elements of GL2 (C) is trivial, so the central subgroups
of U (1, 1) and of SL2 (R) act trivially.
6. Schwarz’ lemma
[6.0.1] Theorem: For f holomorphic on |z| < 1, with bound |f (z)| ≤ 1, and with f (0) = 0,
|f (z)| ≤ |z|
and
|f 0 (0)| ≤ 1
Equality |f (zo )| = |zo | holds for some 0 < |zo | < 1 if and only if f (z) = µ · z for some constant µ with |µ| = 1.
Proof: The function F (z) = f (z)/z has a removable singularity at z = 0, and takes value f 0 (0) there. For
each 0 < r < 1,
|f (z)|
1
≤
(on the circle |z| = r)
r
r
By the maximum modulus principle, |F (z)| ≤ 1/r on |z| ≤ r. Thus, for all r0 with 0 < r < r0 < 1,
|F (z)| ≤ 1/r0 on |z| ≤ r, so |F (z)| ≤ 1 on |z| ≤ r. This holds for every 0 < r < 1, so |F (z)| ≤ 1 on |z| < 1.
This already gives f 0 (0) ≤ 1.
|F (z)| =
If |F (zo )| = 1 for some 0 < |zo | < 1, or if |f 0 (0)| = 1, then |F | attains its sup in the interior, so is a constant
µ, and f (z) = µ · z.
///
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