1Complex differentiation
IB Complex Analysis
1.2 Conformal mappings
The course schedules has a weird part where we are supposed to talk about
conformal mappings for a lecture, but not use it anywhere else. We have to put
them somewhere, and we might as well do it now. However, this section will be
slightly disconnected from the rest of the lectures.
Definition
(Conformal function)
.
Let
f
:
U → C
be a function holomorphic at
w ∈ U. If f
0
(w) 6= 0, we say f is conformal at w.
What exactly does
f
0
(
w
)
6
= 0 tell us? In the real case, if we know a function
f
: (
a, b
)
→ R
is continuous differentiable, and
f
0
(
c
)
6
= 0, then
f
is locally
increasing or decreasing at
c
, and hence has a local inverse. This is also true in
the case of complex functions.
We write
f
=
u
+
iv
, then viewed as a map
R
2
→ R
2
, the Jacobian matrix is
given by
Df =
u
x
u
y
v
x
v
y
.
Then
det(Df) = u
x
v
y
− u
y
v
x
= u
2
x
+ u
2
y
.
Using the formula for the complex derivative in terms of the partials, this shows
that if
f
0
(
w
)
6
= 0, then
det
(
Df|
w
)
6
= 0. Hence, by the inverse function theorem
(viewing
f
as a function
R
2
→ R
2
),
f
is locally invertible at
w
(technically, we
need
f
to be continuously differentiable, instead of just differentiable, but we
will later show that
f
in fact must be infinitely differentiable). Moreover, by
the same proof as in real analysis, the local inverse to a holomorphic function is
holomorphic (and conformal).
But being conformal is more than just being locally invertible. An important
property of conformal mappings is that they preserve angles. To give a precise
statement of this, we need to specify how “angles” work.
The idea is to look at tangent vectors of paths. Let
γ
1
, γ
2
: [
−
1
,
1]
→ U
be
continuously differentiable paths that intersect when
t
= 0 at
w
=
γ
1
(0) =
γ
2
(0).
Moreover, assume γ
0
i
(0) 6= 0.
Then we can compare the angles between the paths by looking at the difference
in arguments of the tangents at w. In particular, we define
angle(γ
1
, γ
2
) = arg(γ
0
1
(0)) − arg(γ
0
2
(0)).
Let
f
:
U → C
and
w ∈ U
. Suppose
f
is conformal at
w
. Then
f
maps our two
paths to
f ◦ γ
i
: [
−
1
,
1]
→ C
. These two paths now intersect at
f
(
w
). Then the
angle between them is
angle(f ◦ γ
1
, f ◦ γ
2
) = arg((f ◦ γ
1
)
0
(0)) − arg((f ◦ γ
2
)
0
(0))
= arg
(f ◦ γ
1
)
0
(0)
(f ◦ γ
2
)
0
(0)
= arg
γ
0
1
(0)
γ
0
2
(0)
= angle(γ
1
, γ
2
),
using the chain rule and the fact that f
0
(w) 6= 0. So angles are preserved.
What else can we do with conformal maps? It turns out we can use it to
solve Laplace’s equation.
We will later prove that if
f
:
U → C
is holomorphic on an open set
U
, then
f
0
: U → C is also holomorphic. Hence f is infinitely differentiable.
In particular, if we write
f
=
u
+
iv
, then using the formula for
f
0
in terms
of the partials, we know
u
and
v
are also infinitely differentiable. Differentiating
the Cauchy–Riemann equations, we get
u
xx
= v
yx
= −u
yy
.
In other words,
u
xx
+ u
yy
= 0,
We get similar results for
v
instead. Hence
Re
(
f
) and
Im
(
f
) satisfy the Laplace
equation and are hence harmonic (by definition).
Definition
(Conformal equivalence)
.
If
U
and
V
are open subsets of
C
and
f : U → V is a conformal bijection, then it is a conformal equivalence.
Note that in general, a bijective continuous map need not have a continuous
inverse. However, if we are given further that it is conformal, then the inverse
mapping theorem tells us there is a local conformal inverse, and if the function
is bijective, these patch together to give a global conformal inverse.
The idea is that if we are required to solve the 2D Laplace’s equation on
a funny domain
U
subject to some boundary conditions, we can try to find a
conformal equivalence
f
between
U
and some other nice domain
V
. We can
then solve Laplace’s equation on
V
subject to the boundary conditions carried
forward by
f
, which is hopefully easier. Afterwards, we pack this solution into
the real part of a conformal function
g
, and then
g ◦f
:
U → C
is a holomorphic
function on
U
whose real part satisfies the boundary conditions we want. So we
have found a solution to Laplace’s equation.
You will have a chance to try that on the first example sheet. Instead, we
will focus on finding conformal equivalences between different regions, since
examiners like these questions.
Example.
Any M¨obius map
A
(
z
) =
az+b
cz+d
(with
ad−bc 6
= 0) defines a conformal
equivalence
C ∪ {∞} → C ∪ {∞}
in the obvious sense.
A
0
(
z
)
6
= 0 follows from
the chain rule and the invertibility of A(z).
In particular, the M¨obius group of the disk
M¨ob(D) = {f ∈ M¨obius group : f(D) = D}
=
λ
z − a
¯az −1
∈ M¨ob : |a| < 1, |λ| = 1
is a group of conformal equivalences of the disk. You will prove that the M¨obius
group of the disk is indeed of this form in the first example sheet, and that these
are all conformal equivalences of the disk on example sheet 2.
Example.
The map
z 7→ z
n
for
n ≥
2 is holomorphic everywhere and conformal
except at z = 0. This gives a conformal equivalence
n
z ∈ C
∗
: 0 < arg(z) <
π
n
o
↔ H,
where we adopt the following notation:
Notation. We write C
∗
= C \ {0} and
H = {z ∈ C : Im(z) > 0}
is the upper half plane.
Example.
Note that
z ∈ H
if and only if
z
is closer to
i
than to
−i
. In other
words,
|z − i| < |z + i|,
or
z − i
z + i
< 1.
So
z 7→
z−i
z+i
defines a conformal equivalence
H → D
, the unit disk. We know
this is conformal since it is a special case of the M¨obius map.
Example. Consider the map
z 7→ w =
1
2
z +
1
z
,
assuming z ∈ C
∗
. This can also be written as
w + 1
w −1
= 1 +
2
w −1
= 1 +
4
z +
1
z
− 2
= 1 +
4z
z
2
− 2z + 1
=
z + 1
z − 1
2
.
So this is just squaring in some funny coordinates given by
z+1
z−1
. This map is
holomorphic (except at z = 0). Also, we have
f
0
(z) = 1 −
z
2
+ 1
2z
2
.
So f is conformal except at ±1.
Recall that the first thing we learnt about M¨obius maps is that they take
lines and circles to lines and circles. This does something different. We write
z = re
iθ
. Then if we write z 7→ w = u + iv, we have
u =
1
2
r +
1
r
cos θ
v =
1
2
r −
1
r
sin θ
Fixing the radius and argument fixed respectively, we see that a circle of radius
ρ is mapped to the ellipse
u
2
1
4
ρ +
1
ρ
2
+
v
2
1
4
ρ −
1
ρ
2
= 1,
while the half-line arg(z) = µ is mapped to the hyperbola
u
2
cos
2
µ
−
v
2
sin
2
µ
= 1.
We can do something more interesting. Consider a off-centered circle, chosen to
pass through the point −1 and −i. Then the image looks like this:
−1
−i
f
f(−1)
f(−i)
Note that we have a singularity at
f
(
−
1) =
−
1. This is exactly the point where
f is not conformal, and is no longer required to preserve angles.
This is a crude model of an aerofoil, and the transformation is known as the
Joukowsky transform.
In applied mathematics, this is used to model fluid flow over a wing in terms
of the analytically simpler flow across a circular section.
We interlude with a little trick. Often, there is no simple way to describe
regions in space. However, if the region is bounded by circular arcs, there is a
trick that can be useful.
Suppose we have a circular arc between α and β.
α
z
β
φ
θ
µ
Along this arc,
µ
=
θ −φ
=
arg
(
z − α
)
− arg
(
z − β
) is constant, by elementary
geometry. Thus, for each fixed µ, the equation
arg(z − α) − arg(z − β) = µ
determines an arc through the points α, β.
To obtain a region bounded by two arcs, we find the two
µ
−
and
µ
+
that
describe the boundary arcs. Then a point lie between the two arcs if and only if
its µ is in between µ
−
and µ
+
, i.e. the region is
z : arg
z − α
z − β
∈ [µ
−
, µ
+
]
.
This says the point has to lie in some arc between those given by µ
−
and µ
+
.
For example, the following region:
−1 1
i
can be given by
U =
z : arg
z − 1
z + 1
∈
h
π
2
, π
i
.
Thus for instance the map
z 7→ −
z − 1
z + 1
2
is a conformal equivalence from
U
to
H
. This is since if
z ∈ U
, then
z−1
z+1
has
argument in
π
2
, π
, and can have arbitrary magnitude since
z
can be made
as close to
−
1 as you wish. Squaring doubles the angle and gives the lower
half-plane, and multiplying by −1 gives the upper half plane.
z 7→
z−1
z+1
z 7→ z
2
z 7→ −z
In fact, there is a really powerful theorem telling us most things are conformally
equivalent to the unit disk.
Theorem
(Riemann mapping theorem)
.
Let
U ⊆ C
be the bounded domain
enclosed by a simple closed curve, or more generally any simply connected domain
not equal to all of
C
. Then
U
is conformally equivalent to
D
=
{z
:
|z| <
1
} ⊆ C
.
This in particular tells us any two simply connected domains are conformally
equivalent.
The terms simple closed curve and simply connected are defined as follows:
Definition
(Simple closed curve)
.
A simple closed curve is the image of an
injective map S
1
→ C.
It should be clear (though not trivial to prove) that a simple closed curve
separates C into a bounded part and an unbounded part.
The more general statement requires the following definition:
Definition
(Simply connected)
.
A domain
U ⊆ C
is simply connected if every
continuous map from the circle
f
:
S
1
→ U
can be extended to a continuous
map from the disk
F
:
D
2
→ U
such that
F |
∂D
2
=
f
. Alternatively, any loop
can be continuously shrunk to a point.
Example.
The unit disk is simply-connected, but the region defined by 1
<
|z| <
2 is not, since the circle
|z|
= 1
.
5 cannot be extended to a map from a disk.
We will not prove this statement, but it is nice to know that this is true.
If we believe that the unit disk is relatively simple, then since all simply
connected regions are conformally equivalent to the disk, all simply connected
domains are boring. This suggests we will later encounter domains with holes to
make the course interesting. This is in fact true, and we will study these holes
in depth later.
Example. The exponential function
e
z
= 1 + z +
z
2
2!
+
z
3
3!
+ ···
defines a function
C → C
∗
. In fact it is a conformal mapping. This sends
the region
{z
:
Re
(
z
)
∈
[
a, b
]
}
to the annulus
{e
a
≤ |w| ≤ e
b
}
. One is simply
connected, but the other is not — this is not a problem since
e
z
is not bijective
on the strip.
a
b
e
z