1Analytic functions
IB Complex Methods
1.5 M¨obius map
We are now going to consider a special class of maps, namely the M¨obius maps, as
defined in IA Groups. While these maps have many many different applications,
the most important thing we are going to use it for is to define some nice
conformal mappings in the next section.
We know from general theory that the M¨obius map
z 7→ w =
az + b
cz + d
with
ad − bc 6
= 0 is analytic except at
z
=
−
d
c
. It is useful to consider it as a
map from C
∗
→ C
∗
= C ∪ {∞}, with
−
d
c
7→ ∞, ∞ 7→
a
c
.
It is then a bijective map between C
∗
and itself, with the inverse being
w 7→
−dw + b
cw − a
,
another M¨obius map. These are all analytic everywhere when considered as a
map C
∗
→ C
∗
.
Definition (Circline). A circline is either a circle or a line.
The key property of M¨obius maps is the following:
Proposition. M¨obius maps take circlines to circlines.
Note that if we start with a circle, we might get a circle or a line; if we start
with a line, we might get a circle or a line.
Proof. Any circline can be expressed as a circle of Apollonius,
|z − z
1
| = λ|z − z
2
|,
where z
1
, z
2
∈ C and λ ∈ R
+
.
This was proved in the first example sheet of IA Vectors and Matrices. The
case
λ
= 1 corresponds to a line, while
λ 6
= 1 corresponds to a circle. Substituting
z in terms of w, we get
−dw + b
cw − a
− z
1
= λ
−dw + b
cw − a
− z
2
.
Rearranging this gives
|(cz
1
+ d)w − (az
1
+ b)| = λ|(cz
2
+ d)w − (az
2
+ b)|. (∗)
A bit more rearranging gives
w −
az
1
+ b
cz
1
+ d
= λ
cz
2
+ d
cz
1
+ d
w −
az
2
+ b
cz
2
+ d
.
This is another circle of Apollonius.
Note that the proof fails if either
cz
1
+
d
= 0 or
cz
2
+
d
= 0, but then (
∗
)
trivially represents a circle.
Geometrically, it is clear that choosing three distinct points in
C
∗
uniquely
specifies a circline (if one of the points is
∞
, then we have specified the straight
line through the other two points).
Also,
Proposition.
Given six points
α, β, γ, α
0
, β
0
, γ
0
∈ C
∗
, we can find a M¨obius
map which sends α 7→ α
0
, β 7→ β
0
and γ → γ
0
.
Proof. Define the M¨obius map
f
1
(z) =
β − γ
β − α
z − α
z − γ
.
By direct inspection, this sends α → 0, β → 1 and γ → ∞. Again, we let
f
2
(z) =
β
0
− γ
0
β
0
− α
0
z − α
0
z − γ
0
.
This clearly sends
α
0
→
0
, β
0
→
1 and
γ
0
→ ∞
. Then
f
−1
2
◦ f
1
is the required
mapping. It is a M¨obius map since M¨obius maps form a group.
Therefore, we can therefore find a M¨obius map taking any given circline to
any other, which is convenient.