Conversely, if f is a univalent function on C∗ then either f = λJ + μ or f = aλ,μ for some λ, μ∈ C with λ = 0. In either case, f is a conformal transformation of C∗ onto C\{μ}. If f is a conformal automorphism of C∗ then μ = 0.
Proof The first statement is obvious. Suppose conversely that f is a univalent function on C∗. Then f has an isolated singularity at 0.
Suppose first that f has a removable singularity at 0. Then there exists ν such that f (z) → ν as z → 0. We shall show that ν ∈ f(C∗). Suppose not, and suppose that f (z0) = ν for some z0 ∈ C∗. Let = |z0|/2. By the open mapping theorem, f (N (z0)) is an open subset of C containing ν. But f (z)→ ν as z → 0, and so there exists w ∈ N ∗(0) with f (w) ∈ f(N (z0)).
Since N ∗(0)∩ N (z0) =∅, this contradicts the univalence of f on C∗. Thus if we set f (0) = ν then f is an entire univalent function on C. By Theorem 25.2.1, f = aλ,μ, for some λ, μ∈ C with λ = 0.
Secondly we show that f cannot have an isolated essential singularity at 0. For if it did, by Weierstrass’ theorem there would be w∈ N1/2∗ (0) with f (w) in the open set f (N1/2(1)), contradicting the univalence of f .
Finally we consider the case where f has a pole at 0. By Corollary 23.6.5, this must be a simple pole. Thus f has a Laurent series expansion
∞
n=−1anzn. By Corollary 23.3.4, there are only finitely many non-zero coefficients an. Thus f (z) = p(z)/z, where p is a polynomial with non-zero constant term a−1. Since C∗ is not conformally equivalent to C, there exists μ ∈ C which is not in f(C∗). Let q(z) = p(z) − μz. Then q(0) = p(0) = a−1 = 0, and q(z) = (f(z) − μ)z = 0 for z = 0. Thus the polynomial q has no zeros, and so must be the non-zero constant func-tion a−1. Thus f (z) = a−1/z + μ. The final two statements now follow
immediately. 2
Exercises
25.3.1 Suppose that S is a discrete closed subset of C and that f is a univa-lent function on C\ S which has a non-removable singularity at each point of S. Show that S has at most one element, and that if f has a singularity then it must be a simple pole.
25.4 The M¨obius group
We next consider univalent meromorphic functions on the unit sphere C∞. Recall that a function f : C∞→ C∞is meromorphic if the restrictions of f and f◦ J to C are meromorphic functions on C.
Theorem 25.4.1 If f is a univalent meromorphic function on C∞ then either f (z) = aλ,μ(z) = λz + μ or f (z) = λ/(z−z0) + μ for some λ, μ and z0 in C, with λ= 0. In either case, f is a homeomorphism of C∞ onto itself.
Proof If the restriction of f to C is holomorphic, then it is a univalent function on C, and so f = aλ,μ, by Theorem 25.2.1. Otherwise, it has one simple pole, at z0 say. Let g(z) = f (z + z0), for z ∈ C∗. Then g is a univalent function on C∗ with a pole at 0, so that g = λJ + μ, by Theorem25.3.1, and f (z) = λ/(z− z0) + μ. In either case, f is a homeomorphism of C∞ onto
itself. 2
A univalent meromorphism of C∞onto itself is called a M¨obius transfor-mation of C∞. If f is a M¨obius transformation, we can write
f (z) = az + b cz + d : in the former case
a = λ, b = μ, c = 0 and d = 1, and in the latter case
a = μ, b = λ− μz0, c = 1 and d =−z0. Note that in either case ad− bc = λ = 0.
Conversely, suppose that ad − bc = 0, and consider the meromorphic function f (z) = (az + b)/(cz + d). If c = 0 then d= 0 and f(z) = λz + μ, where λ = a/d = 0 and μ = b/d. If c = 0, let z0 = −d/c. Then f has a simple pole at z0, and
f (z) = λ z− z0
+ μ, where λ =−ad− bc
c2 = 0 and μ = a c. Thus f is a M¨obius transformation.
Theorem 25.4.2 The setM of M¨obius transformations is a group under composition. If
A =
a b c d
∈ GL2(C), the group of invertible two-by-two matrices, let m(A)(z) = (az + b)/cz + d). Then m is a homomorphism of GL2(C) onto M, with kernel {aI : a = 0}. (m(A))−1= m(B), where
B =
d −b
−c a
.
25.4 The M¨obius group 753 Proof Since A∈ GL2(C) is invertible if and only if det A = ad− bc = 0.
m maps GL2(C) onto M, and m(I)(z) = z. Suppose that A1 =
a1 b1
c1 d1
and A2 =
a2 b2
c2 d2
are in GL2(C). Then
m(A1)m(A2)(z) = m(A1)
a2z+b2 c2z+d2
= a1
a2z+b2 c2z+d2
+ b1 c1
a2z+b2 c2z+d2
+ d1
= a1(a2z+b2) + b1(c2z+d2) c1(a2z+b2) + d1(c2z+d2)
= (a1a2+ b1c2)z + a1b2+ b1d2
(c1a2+ d1c2)z + c1b2+ d1d2
= m(A1A2)(z).
Consequently m(A1)m(A2)∈ M. Since
m(A−1)m(A) = m(A)m(A−1) = m(I)
and m(I) is the identity mapping on C∞, it follows thatM is a group under composition, and that m is a homomorphism of GL2(C) onto M. Clearly m(A)(z) = z for all z ∈ C∞ if and only if a = d = 0 and b = c = 0. Since AB = BA = (ad− bc)I, m(B) is the inverse of m(A). This last statement can also be verified directly; if w = (az + b)/(cz + d), we can consider this as an equation in z, and solve it to find that z = (dw− b)/(−cw + a). 2 A M¨obius transformation is determined by its action on three distinct points.
Proposition 25.4.3 Suppose that{z1, z2, z3} and {w1, w2, w3} are sets of distinct points of C∞. Then there exists a unique M¨obius transformation m for which m(z1) = w1, m(z2) = w2 and m(z3) = w3.
Proof First we show that if w1, w2 and w3 are distinct points of C∞ then there exists a unique M¨obius transformation m = mw1,w2,w3 for which m(w1) = 0, m(w2) = 1 and m(w3) =∞. If w1, w2, w3 ∈ C we can take
m(z) =
w2− w3
w2− w1
z− w1
z− w3
and otherwise we can take
m(z) = w2− w3
z− w3
if w1 =∞,
= z− w1
z− w3
if w2=∞,
= z− w1
w2− w1
if w3 =∞.
Suppose that n∈ M is another M¨obius transformation for which n(w1) = 0, n(w2) = 1 and n(w3) =∞. Let k = m ◦ n−1. Then k(0) = 0, k(1) = 1 and k(∞) = ∞. Suppose that k(z) = (az + b)/(cz + d). Since k(0) = 0, b = 0.
If c = 0 then k(−d/c) = ∞, giving a contradiction. Thus c = 0. Finally k(1) = a/d = 1, so that a = d and k(z) = z. k is the identity mapping, and so n = m; m is unique.
The M¨obius transformation m−1z1,z2,z3 ◦ mw1,w2,w3 then has the required properties. It is unique, for if n is another M¨obius transformation for which m(z1) = w1, m(z2) = w2 and m(z3) = w3, then n◦ mz1,z2,z3 = mw1,w2,w3, so that n = mw1,w2,w3◦ m−1z1,z2,z3. 2 The following M¨obius transformations are called elementary M¨obius transformations:
• Tb(z) = z + b (translation);
• Dr(z) = rz for r real and positive (dilation);
• Rθ(z) = eiθz for θ ∈ R (rotation);
• J (z) = 1/z (inversion) If λ = reiθ with r > 0 then
λz + μ = Tμ◦ Rθ◦ Dr and λ z− z0
+ μ = (Tμ◦ Rθ◦ Dr◦ J ◦ T−z0)(z), so that these elementary transformations generate M. This is very useful in establishing properties of general M¨obius transformations, as Theorem 25.4.5 will show. Our next aim is to show that a M¨obius transformation
‘maps circles and straight lines into circles or straight lines’.
First we must describe straight lines and circles in C and C∞in terms of the complex structure. A straight line L in C can be written as
L ={z = x + iy ∈ C : ax + by + c = 0},
where a, b and c are real, and a and b are not both zero. Substituting x = (z + ¯z)/2 and y = (z− ¯z)/2i, we find that
L ={z ∈ C : ¯λz + λ¯z + μ = 0},
25.4 The M¨obius group 755 where λ = a + ib= 0 and μ = 2c is real. L is a closed unbounded subset of C. In C∞, we define a straight line L∞to be L∪ {∞}, where L is a straight line in C. Thus L∞ is the closure of L in C∞, and so is closed in C∞.
A circle C in C can be written as
C ={z ∈ C : |z − c| = r} = {z ∈ C : (z − c)(z − c) = r2},
where c∈ C and r > 0. Then C = {z ∈ C : z¯z − ¯cz − c¯z = r2− c¯c}, where c∈ C and r > 0. C is a bounded closed subset of C, and so is closed in C∞. It is clear that translation, dilation and rotation map straight lines to straight lines and circles to circles. What about inversion?
Proposition 25.4.4 Suppose that L∞ is a straight line in C∞ and that C is a circle in C∞.
(i) If 0∈ L∞ then J (L∞) is a straight line in C∞ and 0∈ J(L∞).
(ii) If 0∈ L∞ then J (L∞) is a circle in C∞ and 0∈ J(L∞).
(iii) If 0∈ C then J(C) is a straight line in C∞ and 0∈ J(C).
(iv) If 0∈ C then J(C) is a circle in C∞ and 0∈ J(C).
Proof This is a matter of straightforward verification.
(i) If 0∈ L∞, then
L∞={z ∈ C : ¯λz + λ¯z = 0} ∪ {∞}, so that
J (L∞) = J−1(L∞) ={∞} ∪ {z ∈ C \ {0} : λ¯ z +λ
¯
z = 0} ∪ {0}
={z ∈ C : λz + ¯λ¯z = 0} ∪ {∞},
which shows that J (L∞) is a straight line in C∞ and that 0∈ J(L∞).
(ii) If 0∈ L∞, then
L∞={z ∈ C : ¯λz + λ¯z + μ = 0} ∪ {∞}, with μ= 0. Arguing as above,
J (L∞) = J−1(L∞) ={0} ∪ {z ∈ C \ {0} : ¯λ z +λ
¯
z + μ = 0}
={z ∈ C : z¯z + λz μ +
λ¯¯z μ = 0}
={z ∈ C : z¯z − ¯cz − c¯z = r2− c¯c}
where c =−¯λ/μ and r2 = c¯c. This shows that J (L∞) is a circle in C∞ and that 0∈ J(L∞).
(iii) If 0∈ C, then
C ={z ∈ C : z¯z − ¯cz − c¯z = 0}, so that
J (C) = J−1(C) ={z ∈ C \ {0} : 1 z ¯z − ¯c
z − c
¯
z = 0} ∪ {∞}
={z ∈ C : cz + ¯c¯z = 1} ∪ {∞},
which shows that J (C) is a straight line in C∞ and that 0∈ J(C).
(iv) If 0∈ C, then
C ={z ∈ C : z¯z − ¯cz − c¯z = d}, with d = r2− c¯c = 0, and so
J (C) = J−1(C) ={z ∈ C \ {0} : 1 z ¯z − ¯c
z − c
¯ z = d}
={z ∈ C : z¯z +cz d +¯c¯z
d = 1 d},
which shows that J (C) is a circle in C∞ and that 0∈ J(C). 2 Theorem 25.4.5 Suppose that m is a M¨obius transformation, that L∞is a straight line in C∞ and that C is a circle in C∞.
(i) If ∞ ∈ m(L∞) then m(L∞) is a straight line in C∞. (ii) If ∞ ∈ m(L∞) then m(L∞) is a circle in C∞. (iii) If ∞ ∈ m(C) then m(C) is a straight line in C∞.
(iv) If∞ ∈ m(C) then m(C) is a circle in C∞.
Proof Translations, dilations and rotations map straight lines to straight lines and circles to circles. Since m is a product of elementary transforma-tions, it follows from Proposition25.4.4 that m(L∞) is either a straight line or a circle. m(L∞) is a straight line if∞ ∈ m(L∞) and is a circle if not, and m(C) is a straight line if∞ ∈ m(C) and is a circle if not. 2 This result suggests that we may think of a straight line in C∞ as an unbounded circle, or as a circle of infinite radius.
The complement of a straight line in C∞has two connected components, as does the complement of a circle (the inside, and the union of the outside and{∞}). Since a M¨obius transformation m is a homeomorphism of C∞, if S is a circle or straight line and U and V are the connected components of C∞\ S then m(U) and m(V ) are the connected components of C∞\ m(S).
25.4 The M¨obius group 757
0
–1 0 1 0 1
J 1
m1
m2
0 1
Figure 25.4.
As an example, the M¨obius transformation m(z) = (−z +1)/(z +1) maps the extended y-axis Y∞ = {z = iy : y ∈ R} ∪ {∞} onto the unit circle T.
Although this can be verified directly, it is more informative to construct the mapping in several steps. First, the mapping m1(z) = z + 1 maps Y∞ onto the extended line L∞ ={z = 1 + iy : y ∈ R} ∪ {∞}. Secondly, since 0 ∈ L∞, J maps L∞ onto a circle C passing through 0 and 1. J maps the extended x-axis onto itself. Since L∞ is orthogonal to the x-axis, and since M¨obius transformations are conformal, the tangent to J (L∞) at 1 is orthogonal to the x-axis. Thus C is the circle with centre 1/2 and radius 1/2. The mapping m2(z) = 2z− 1 then maps C to a circle passing through 1 and−1, with centre 0, so that m2(C) = T. Then m = m2◦ J ◦ m1. Since m(1) = 0, m maps the right-hand half-plane onto the unit disc D and the left-hand half-plane onto{z : |z| > 1} ∪ {∞}.
Exercises
25.4.1 Show that the mapping m(z) = (z− i)/(z + i) defines a conformal transformation of the upper half-plane H+ onto the open unit disc D, and maps i to 0.