THE MÖBIUS TRANSFORMATION
LECTURES NOTES IN MAT2410
Definition 1. Let f (z) = (az + b)/(cz + d) with a, b, c, d 2 C and ad bc 6= 0.
Then f is called a fractional linear transformation, or Möbius transformation.
Note: The condition ad bc 6= 0 ensures that the following conditions hold:
• Neither az + b nor cz + d vanish identically.
• a and c cannot both be zero (in which case f would be constant).
• b and d cannot be both zero (in which case f would be constant).
• In general, the denominator cannot be a constant multiple of the numerator.
Indeed, if k(az + b) = cz + d for some k 2 C, then equation coefficients
yields ka = c, kb = d and ad bc = a(kb) b(ka) = 0.
Thus ad bc 6= 0 ensures that az + b and cz + d do not have a common factor, and therefore f (z) is a well-defined non-constant holomorphic
function P ! P.
Let GL2 (C) denote the group of 2 ⇥ 2 matrices M with complex entries and
non-zero determinants.1
⇢

a b
GL2 (C) = M =
| a, b, c, d 2 C, det(M ) = ad bc 6= 0
c d
As before, M 2 GL2 (C) defines fM : P ! P with
az + b
.
cz + d
Recall that fM (1) = a/c if c 6= 0, fM ( d/c) = 1 if c 6= 0 and fM (1) = 1 if
c = 0.
fM (z) =
Theorem 1. Every f 2 Aut(P) can be written in the form f = fM for some
M 2 GL2 (C). Conversely, fM 2 Aut(P) for all M 2 GL2 (C).
Proof. As before, if M, N 2 GL2 (C) then fM fN = fM ·N . Thus each fM 2 Aut(P)
with inverse fM 1 .
Let f 2 Aut(P). By our previous discussion f must be rational, and non-constant
with numerator and denominator of degree at most 1. This is captured precisely by
saying that
⇥ f ⇤(z) = (az + b)/(cz + d) with a, b, c, d 2 C and ad bc 6= 0. So f = fM
for M = ac db . Thus Aut(P) = {z 7! (az + b)/(cz + d) | a, b, c, d 2 C, ad bc 6= 0}.
⇤
Note: Let M 2 GL2 (C) and
2 C⇤ . Then we have
f (z) = fM (z) =
az + b
az + b
=
=f
cz + d
cz + d
M (z),
and ( a)( d) ( c)( b) = 2 (ad bc) 6= 0, so M 2 GL2 (C) defines the same
linear transformation as M for all 2 C⇤ . We are therefore free to choose so
1This group is called the general linear group.
1
2
THE MÖBIUS TRANSFORMATION LECTURES NOTES IN MAT2410
that det M = 1, that is we can find M 2 SL2 (C) with
⇢

a b
SL2 (C) = M =
| a, b, c, d 2 C, det M = 1
c d
such that f = fM . As before, M, M both in SL2 (C) defines the same fM = f
so we identify them to get P SL2 (C). We have
M,
SL2 (C) ⇠ GL2 (C)
Aut(P) ⇠
.
= P SL2 (C) ⇠
=
=
{±I}
C⇤
The group Aut(P).
1. Properties of Möbius transformations
Lemma 1. Every Möbius transformation is the composition of four elementary
maps (each a Möbius transformation):
a)
b)
c)
d)
Translations, z 7! z + z0 , z0 2 C,
Dilations, z 7! z, > 0, 2 R,
Rotations, z 7! ei✓ z, ✓ 2 R.
Inversions, z 7! 1/z.
Proof. Let f 2 Aut(P), f (z) = (az + b)/(cz + d). If c = 0, then f (z) = (a/d)z + b/d
which is a composition of a scaling and a rotation, followed by a translation. If
c 6= 0 write
bc ad
1
a
f (z) =
·
+
c2
z + d/c
c
which is a translation, inversion, scaling, rotation and translation (in that order).
⇤
Fact:
• Let S ⇢ C be a circle. Using the correspondence C $ P \ {1} given by
stereographic projection, S maps to a circle S 0 ⇢ P \ {1}. Conversely, any
circle S 0 in P not through 1 maps to a circle S in C.
• Let L ⇢ C be a straight line. Using C $ P \ {1}, L maps to L0 ⇢ P, a
circle through 1 with the point at infinity removed. Conversely, if L0 ⇢ P
is a circle through 1 with 1 removed, then stereographic projection of L0
gives a line L in C.
Summary:
• Circles in C $ circles in P not through 1.
• Lines in C $ circles in P through 1.
Proposition 1. Möbius transformations preserve circles in P. That is, a Möbius
transformation maps circles in P to circles in P.
Proof. Let S be a circle in C. Clearly, translations, dilations and rotation all map
S to another circle in C. Similarly, if L is a line in C then translations, dilations,
rotations all map L to another line L0 and also map 1 to 1. So these three types
of elementary maps preserve circles in P.
THE MÖBIUS TRANSFORMATION
Let |z
LECTURES NOTES IN MAT2410
3
z0 | = r be a circle in C and let w = 1/z. We get2
|z
z0 |2 = r2 =)(z
=)|z|2
=)
1
|w|2
z0 )(z̄
z¯0 )
r2 = 0
2Re(z̄z0 ) + |z0 |2
2
r2 = 0
Re(wz0 )
+ |z0 |2
|w|2
r2 = 0.
If |z0 | = r, we get 2Re(wz0 ) = 1, letting w = u + iv, z0 = x0 + iy0 , |z0 | = r
implies that 2(ux0 vy0 ) = 1, a line in C. (The circle passes through the origin).
Otherwise, we get 1 2Re(wz0 ) + (|z0 |2 r2 )|w|2 = 0
=)
|w|2
=)
w
2Re(wz0
(|z0 |2
+
|z0 |2 r2
(|z0 |2
2
z¯0
r2 )
=0
r 2 )2
r2
=0
|z0
(|z0
r 2 )2
which is the equation of a circle in w.
Otherwise, if we have a line L : 2Re(z z¯0 ) = a, a 2 R, then w = 1/z gives
2Re(wz0 ) = a|w|2 . If a = 0 (so that L is through the origin), then 2Re(wz0 ) = 0 is
another line (through the origin).
Otherwise, we get wz0 + w̄z¯0 aww̄ = 0
|2
r2
|2
w̄z¯0 /a + |z0 |2 /a2 |z0 |2 /a2 = 0
⇣ z ⌘2
0
2
=)
|w z¯0 /a| =
,
a
a circle in C. Thus all 4 elementary maps preserve circles in P. Hence every
composition of elementary maps also preserves circles in P, and this gives all Möbius
transformations by the previous lemma.
⇤
=)
ww̄
wz0 /a
2. How many fixed points can a Möbius transformation have?
Let f 2 Aut(P), f (z) = (az + b)/(cz + d). Suppose z is fixed by f ,
az + b
f (z) = z =
) cz 2 + dz = az + b ) cz 2 + (d a)z b = 0.
cz + d
If c 6= 0 this is a quadratic and has either 1 or 2 solutions in C.3 If c = 0 and d 6= a
then this is linear and has 1 solution in C. But then f (z) = (a/d)z + b/d satisfies
f (1) = 1 and 1 is also a fixed point. If c = 0 and a = d, then f (z) = z + b/d.
If b 6= 0 then f (1) = 1 is the only fixed point of f . (Otherwise f (z) = z is the
identity map and fixes every point of P). Thus every f 2 Aut(P), f 6= IdP has
either 1 or 2 fixed points in P.
Lemma 2. A Möbius transformation is completely determined by its action on
three distinct points.
Proof. Let S, T 2 Aut(P) and suppose there exists points a, b, c 2 P such that
S(a) = T (a) = ↵, S(b) = T (b) = and S(c) = T (c) = . But then S T 1 2 Aut(P)
is a Möbius transformation, and S T 1 (↵) = S(a) = ↵. Similarly, S T 1 ( ) =
and S T 1 ( ) = . So S T 1 has 3 distinct fixed points hence S T 1 = Id
which implies that S = T .
⇤
2z̄ = 1/w̄ = w/|w|2 .
3Also note that 1 is not fixed.
4
THE MÖBIUS TRANSFORMATION LECTURES NOTES IN MAT2410
Let z1 , z2 and z3 be distinct points in P. Define S 2 Aut(P) by
8 z z1 z2 z3
z1 , z2 , z3 2 C
>
z z3 · z2 z1
>
>
< z2 z3
z1 = 1
S(z) = zz zz13
>
z2 = 1
>
>
: zz zz31
z3 = 1
z2 z1
Then S(z1 ) = 0, S(z2 ) = 1, S(Z3 ) = 1, and S is the only Möbius transformation
with this property.
Definition 2. Let z0 , z1 , z2 z3 2 P. Their cross-ratio is defined as
z0 z1 z2 z3
[z0 : z1 : z2 : z3 ] =
·
= S(z0 )
z0 z3 z2 z1
where S 2 Aut(P) satisfies S(z1 ) = 0, S(z2 ) = 1 and S(z3 ) = 1.
Examples:
• [z1 : z1 : z2 : z3 ] = 0
• [z2 : z1 : z2 : z3 ] = 1
• [z : 0 : 1 : 1] = z
• [z : 1 : 1 : 0] = z1
Proposition 2. If z1 , z2 , z3 2 P are distinct points and T 2 Aut(P) is a Möbius
transformation, then
[z0 : z1 : z2 : z3 ] = [T (z0 ) : T (z1 ) : T (z2 ) : T (z3 )]
for all z0 2 P. (Möbius transformations preserve cross-ratios).
Proof. Let S(z) = [z : z1 : z2 : z3 ], then S 2 Aut(P). Letting M = S T
M (T (z1 )) = 0, M (T (z2 )) = 1 and M (T (z3 )) = 1, hence
M (w) = S
T
1
1
we have
(w) = [w : T (z1 ) : T (z2 ) : T (z3 )] .
Letting w = T (z0 ), we get
S
T
1
(w) = S
T
1
(T (z0 )) = S(z0 ) = [T (z0 ) : T (z1 ) : T (z2 ) : T (z3 )] .
⇤
Proposition 3. Let z1 , z2 , z3 2 P be distinct, and w1 , w2 , w3 2 P be distinct. There
is a unique S 2 Aut(P) such that
S(zj ) = wj ,
j = 1, 2, 3.
Proof. Let T (z) = [z : z1 : z2 : z3 ] and R(z) = [w : w1 : w2 : w3 ]. Then S = R
T 2 AutP has the desired property. By earlier lemma, S is unique.
1
⇤
Recall: 3 points in C determine a circle or a line. So 3 points in P determine a circle
in P.
Proposition 4. Given two circles S, S 0 in P, there is a Möbius transformation T
taking S to S 0 .
Proof. Let z1 , z2 , z3 2 S and w1 , w2 , w3 2 S 0 and let T 2 Aut(P), T (z1 ) =
w1 , T (z2 ) = w2 and T (z3 ) = w3 . Then T (S) is a circle passing through w1 , w2 , w3
hence T (S) = S 0 .
⇤