However, as we shall see below, it is conformal at all other points of the complex plane

Conformal Equivalence

Definition Suppose two smooth curves C₁ and C₂ intersect at z₀. Then the angle from C₁ to C₂ at z₀, ∠(C1, C₂), is defined as the angle measured counterclockwise from the tangent line of C₁ at z₀ to the tangent line of C₂ at z₀.

C₂

C₁

θ z0

Definition Suppose f is defined in a neighborhood of z0. f is said to be conformal at z0 if f preserves angles there. That is, for each pair of smooth curves C₁ and C₂ intersecting at z₀,

∠(C1, C₂) = ∠(f (C1), f (C₂)).Similarly, we say f is conformal in a region D if f is conformal at all points z ∈ D.

Note that f (z) = z² is not conformal at z = 0. For example, the positive real axis and the positive imaginary axis are mapped onto the positive real axis and negative real axis, respectively.

However, as we shall see below, it is conformal at all other points of the complex plane.

Definitions

• f is locally 1-1 at z₀ if for some δ > 0 and any distinct z₁, z₂ ∈ D(z₀; δ), f (z₁) 6= f (z₂).

• f is locally 1-1 throughout a region D if f is locally 1-1 at every z ∈ D.

• f is a 1-1 function in a region D, if for every distinct z₁, z₂ ∈ D, f (z₁) 6= f (z₂).

Again, note that f (z) = z² is not locally 1-1 at z = 0 since f (z) = f (−z) for all z. However, f is locally 1 − 1 at all points z = 0.

TheoremSuppose f is analytic at z₀ and f⁰(z₀) 6= 0. Then f is conformal and locally 1-1 at z₀. Proof

Let C : z(t) = x(t) + iy(t) be a smooth curve with z(t₀) = z₀. Then the tangent line to C at z₀ has the direction of z⁰(t₀) = x⁰(t₀) + iy⁰(t₀) so that its angle of inclination with the positive real axis is Arg z⁰(t₀).

If we set Γ = f (C), then Γ is given by Γ : w(t) = f (z(t)) and the angle of inclination of its tangent line at f (z₀) is equal to

Arg w⁰(t₀) = Argf⁰(z₀)z⁰(t₀) = Arg f⁰(z₀) + Arg z⁰(t₀).

Hence the function f maps all curves at z₀ in such a way that the angles of inclination are increased by the constant Arg f⁰(z₀).

Thus, if C1 and C2 meet at z0 and Γ1, Γ2 are their respectively images under f, it follows that

∠(Γ1, Γ₂) = ∠(C1, C₂) and f is conformal at z₀.

To show that f is 1 − 1 in a neighborhood of z₀, let f (z₀) = α and δ₁ > 0 so that f (z) − α 6= 0 ∀ z ∈ ¯D(z0; δ1) \ {z0}.

Such a δ₁ can always be found for otherwise we would have f⁰(z₀) = 0 by the Uniqueness Theorem.

If we let C = C(z0; δ1) and Γ = f (C), it follows by the Argument Principle that

1 = 1

2πi Z

C

f⁰(z)

f (z) − αdz number of zeros (counting multiplicities) of f (z) − α inside C

set w=f (z)

= 1

2πi Z

Γ

dw

w − α = n(Γ; α) winding number of Γ around α

and since the winding number n(Γ; α) is locally constant, there exists an ε > 0 such that 1 = n(Γ; α) = 1

2πi Z

Γ

dw

w − α = 1 2πi

Z

Γ

dw

w − β = n(Γ; β) ∀ β ∈ D(α; ε).

Choose 0 < δ ≤ δ1 so that D(z0; δ) ⊂ f⁻¹(D(α; ε)). Then f is 1 − 1 on D(z0; δ) since for any z₁, z₂ ∈ D(z₀; δ), we have f (z₁), f (z₂) ∈ D(α; ε) and

1 = n(Γ; f (z₁)) = n(Γ; f (z₂))

= 1

2πi Z

Γ

dw

w − f (z₁) = 1 2πi

Z

Γ

dw w − f (z₂)

= 1

2πi Z

C

f⁰(z)

f (z) − f (z₁)dz = 1 2πi

Z

C

f⁰(z) f (z) − f (z₂)dz

= Z(f (z) − f (z1)) = Z(f (z) − f (z2)) ∀ z₁, z₂ ∈ D(z₀; δ),

i.e. the number of zeros counting multiplicities of f (z) − f (z1) and f (z) − f (z2) are both 1 inside C so that f (z₁) 6= f (z₂) if z₁ 6= z₂.

Theorem Suppose f is a 1-1 analytic function in a region D. Then

• f⁻¹ : f (D) → D exists and is analytic in f (D),

• f and f⁻¹ are conformal in D and f (D), respectively.

Remark f is called a biholomorphic map from D to f (D) if f : D → f (D) is holomorphic and has a holomorphic inverse f⁻¹ : f (D) → D.

Proof For all z 6= z0, since f is 1-1 and f⁰ 6= 0, we have

z→zlim0

f⁻¹(z) − f⁻¹(z₀)

z − z₀ = lim

z→z0

1

f (f⁻¹(z)) − f (f⁻¹(z₀)) f⁻¹(z) − f⁻¹(z₀)

= 1

f⁰(f⁻¹(z₀)).

Hence f⁻¹ is analytic with f⁻¹
0

= 1

f⁰ and f⁻¹ also has a nonzero derivative. Thus f and f⁻¹ are both conformal.

Definitions

• A 1-1 analytic mapping is called a conformalmapping.

• Two regions D₁ and D₂ are said to be conformally equivalent if there exists a conformal mapping of D1 onto D2.

Definition A conformal mapping of a region onto itself is called anautomorphismof that region.

Lemma Suppose f : D₁ → D₂ is a conformal mapping of D₁ onto D₂. Then

• any other conformal mapping h : D1 → D2 is of the form g ◦ f, for some automorphism g of D₂;

• any automorphism h of D₁ is of the form f⁻¹◦ g ◦ f, where g is an automorphism of D₂. Proof

If f and h are both conformal mappings of D₁ onto D₂, then g = h ◦ f⁻¹ is an automorphism of D₂ and h = g ◦ f.

If h is an automorphism of D1, then g = f ◦ h ◦ f⁻¹ is an automorphism of D2 and h = f⁻¹◦ g ◦ f.

Remark Let Aut (D1) and Aut (D2) denote the automorphism groups of regions D1 and D2, respectively. If f : D₁ → D₂ is a conformal mapping of D₁ onto D₂, i.e. D₁ and D₂ are conformally equivalent, then, by the Lemma, there is an isomorphisim ϕ : Aut (D₁) → Aut (D₂) defined by

ϕ(h) = f ◦ h ◦ f⁻¹ ∀ h ∈ Aut (D₁) with ϕ⁻¹(g) = f⁻¹◦ g ◦ f ∀ g ∈ Aut (D₂).

Theorem Let D be the unit disk. If f : D → D is an automorphism of the disk and f (0) = 0, then f (z) = e^iθz for some θ ∈ R.

Proof If f : D → D is an automorphism of the disk and f (0) = 0, then by the Schwarz Lemma (∗) |f (z)| ≤ |z| ∀ |z| < 1.

Moreover, since f : D → D is an automorphism, f⁻¹ : D → D is holomorphic and f⁻¹(0) = 0, we have

(†) |f⁻¹(z)| ≤ |z| ∀ |z| < 1.

By combining (∗) and (†), we have

|z|= |f⁻¹(f (z))|≤ |f (z)| ≤ |z| ∀ |z| < 1 =⇒ |f (z)| = |z| ∀ z ∈ D.

So, by the Schwarz Lemma again, there exists a θ ∈ R such that f (z) = e^iθz for all z ∈ D.

Corollary Let

H = {f : D → D | f is holomorphic (extending continuously to ∂D) with f (0) = α}.

Then there exists a g ∈ H such that |g⁰(0)| = max

f ∈H |f⁰(0)|, and g is bijective from D to itself, i.e.

the max

f ∈H |f⁰(0)| is achieved by an automorphism of D.

Proof For each f ∈ H, since B_α(f (z)) : D → D is holomorphic and Bα(f (0)) = 0 and by the Schwarz Lemma, we have

|B_α⁰(α)| |f⁰(0)| ≤ 1 =⇒ |f⁰(0)| ≤ 1

|B_α⁰(α)| = |B_−α⁰ (0)| = |g⁰(0)| =⇒ |B_α⁰(α)| |g⁰(0)| = 1.

Then Bα(g(z)) : D → D is holomorphic satisfying B^α(g(0)) = 0 and    

d

dzBα(g(z))    z=0

= 1, so, by the Schwarz Lemma, there exists a θ ∈ R such that

B_α(g(z)) = e^iθz =⇒ g(z) = B−α(e^iθz) = e^iθz + α

1 + ¯αe^iθz = e^iθ z + α e^−iθ 1 + ¯αe^iθz ∈ H and g(z) is bijective from D to itself.

Corollary The automorphisms of D are of the form g(z) = e^iθBα(z) = e^iθ z − α

1 − ¯αz



, |α| < 1.

Remark : In summary,

• the only automorphisms of the unit disk satisfying f (0) = 0 are given by f (z) = e^iθz for some θ ∈ R,

• the only automorphisms of the unit disk satisfying f (α) = 0 are given by f (z) = e^iθB_α(z) for some θ ∈ R.

Theorem The conformal mappings h of H onto D of the form h(z) = e^iθ z − α

z − ¯α



, Im α > 0.

Proof Let α ∈ H and let f (z) = z − α z − ¯α.

Since |z − α| = |z − ¯α| for all z ∈ R, f maps R onto the unit circle. Also, since f (α) = 0 ∈ D and α ∈ H, it follows that f maps H onto D.

Suppose that h : H → D is a conformal mapping satisfying h(α) = 0. By a preceding Lemma, there exists an automorphism g of D such that h = g ◦ f.

However, since 0 = h(α) = g(0), it follows that g(z) = e^iθz and hence h(z) = g ◦ f (z) = e^iθf (z) = e^iθ z − α

z − ¯α



, Im α > 0.

M¨obius Transformations

Recall that if f : ˆC → C is holomorphic (meromorphic on C), then f is a rational function ofˆ the form P (z)

Q(z). Furthermore, if f (z) is bijective, then α = P (z)

Q(z)

has exactly one solution for each α ∈ C, and hence P (z) − αQ(z) = 0 has exactly one solution for each α ∈ C. This implies that

deg (P − αQ) = 1 =⇒ max{deg P , deg Q} = deg (P − αQ) = 1, and

f (z) = az + b

cz + d, ad − bc 6= 0.

Maps of the form above are calledM¨obius transformations. It is obvious that the inverse f⁻¹(z) is given by

f⁻¹(z) = dz − b

−cz + a, with ad − bc 6= 0.

Proposition If

f (z) = az + b

cz + d, with ad − bc 6= 0 and g(z) = αz + β

γz + δ, with αδ − βγ 6= 0, then letting

A B

C D



=a b c d



·α β γ δ

 , we have

f (g(z)) = Az + B Cz + D.

Recall that we defined a M¨obius transformation to be a function f (z) of the following form f (z) = az + b

cz + d, ad − bc 6= 0.

These are the automorphisms of ˆC, i.e. the holomorphic mapsC →ˆ C with holomorphic inverse.ˆ We showed that composition translates into matrix multiplication

f_M1(f_M2(z)) = f_M1M2(z), where M ’s are the matrix forms of the transformation.

These 2 × 2 invertible matrices with complex coefficients are known as the general linear group of order 2 over C:

GL₂C =a b c d



| a, b, c, d ∈ C, ad − bc 6= 0

 .

GL₂C has Lie group structure (it is an algebraic group endowed with manifold structure, e.g. the group operations are C^∞). LetM = {m¨obius transformations}. Then, modding out by complex scalars 6= 0, we have in fact that

M = GL₂C

C^× = PGL₂C,

where C^× = C \ {0}, and PGL2C is the projective linear group of order 2 over C, which is also a Lie group. We mod out by scalar multiples of C^× because

fa b c d



= fλa λb λc λd



∀ λ ∈ C^×,

has no other coincidences; e.g. multiplication by a complex scalar gives the same transformation.

We now claim that the generators for M are as follows:

Generators for M :

(1) Dilations/rotations: z 7→ az.

(2) Translations: z 7→ z + b.

(3) Inversion: z 7→ 1 z. Proof Note that for c 6= 0,

az + b

cz + d = − ad − bc c

  1 cz + d

 +a

c,

is a composition of dialation, translation, inversion, dilation and translation.

For c = 0,

az + b

d = a

dz + b d, is a composition of dialation and translation.

Theorem

{M¨obius transformations} : {circles and lines} → {circles and lines}, i.e.

f (z) = az + b

cz + d, ad − bc 6= 0, maps circles and lines onto circles and lines.

Proof It suffices to consider the case when f is a generator for M , and since the conclusion is obvious when f is either a dilation or translation, so we only consider the case when f is an inversion.

Suppose that S = C(α; r) = {z | |z − α| = r} is a circle in the z-plane, and let f (S) =

 w = 1

z | z ∈ S

 . Since

(z − α)(¯z − ¯α) = r² ⇐⇒ z ¯z − α¯z − ¯αz = r²− |α|², we have, in terms of w,

1 w ¯w − α

¯ w − α¯

w = r²− |α|². (1) If r = |α|, i.e. if S passes through the origin, (1) is equivalent to

1 − αw − ¯α ¯w = 0 ⇐⇒ Re αw = 1 2.

By setting α = x₀+ iy₀ and w = u + iv, the equation Re αw = 1

2 becomes x₀u − y₀v = 1

2, i.e., f (S) is a line in the w-plane.

If r 6= |α|, then (1) is equivalent to w ¯w −

 α¯

|α|²− r²



¯ w −

 α

|α|²− r²



w = −1

|α|²− r², and setting β = α¯

|α|²− r², we obtain

w ¯w − β ¯w − ¯βw + |β|² = r²

(|α|²− r²)² ⇐⇒ |w − β|² =

 r

|α|²− r²

2

,

so that f (S) is a circle with center β and radius r

|α|²− r².

If S is a straight line, then there exist a, b, c ∈ R such that if z = x + iy ∈ S, ax + by = c. (2)

Letting α = a − ib, (2) is equivalent to

Re αz = c ⇐⇒ αz + ¯α¯z = c.

It follows then, as above, that f (S) is either a circle or a line.

Definition A point w ∈ C is called afixed pointof the function f if f (w) = w. The point ∞ ∈ ˆC is called a fixed point of the function f if lim

z→∞f (z) = ∞.

Lemma Given f ∈M and f 6= id, f has at most 2 fixed points in {∞, 0, 1}.

Proof If c 6= 0, since

z→∞lim f (z) = lim

z→∞

az + b cz + d = a

c 6= ∞ and f (z) = z ⇐⇒ az + b = cz²+ dz, so ∞ is not a fixed point and f has at most 2 fixed points in C.

If c = 0, since lim

z→∞f (z) = lim

z→∞(az + b) = ∞, ∞ is a fixed point of f, and since az + b = z

has at most 1 solution, so f has at most 2 fixed points in {∞, 0, 1}.

Proposition There exists a unique M¨obius transformation f (z) sending sending z₁, z₂, z₃ into

∞, 0, 1, respectively, and f (z) is given by

f (z) = (z − z2)(z3− z1) (z − z₁)(z₃− z₂).

Proof The existence of f (z) is trivial.

If f, g are M¨obius transformations satisfying the above properties, then g ◦ f⁻¹ is a M¨obius transformation sending

∞ 7→ ∞, 0 7→ 0, 1 7→ 1, i.e g ◦ f⁻¹ has three fixed points. Hence

g ◦ f⁻¹(z) = z =⇒ f (z) = g(z) ∀ z ∈ C.

Definition The cross-ratio of the four complex numbers z1, z2, z3, z4, denoted (z1, z2, z3, z4),is the image of z₄ under the bilinear map which maps z₁, z₂, z₃ into ∞, 0, 1, respectively.

By the preceding lemma

(z₁, z₂, z₃, z₄) = T (z₄) = (z₄− z₂)(z₃− z₁) (z₄− z₁)(z₃− z₂).

Proposition The cross-ratio is preserved by M¨obius transformations, i.e. if f (z) = az + b cz + d with ad − bc 6= 0, then

(f (z1), f (z2), f (z3), f (z4)) = (z1, z2, z3, z4).

Proof Let g be a M¨obius transformation sending

z₁ 7→ ∞, z₂ 7→ 0, z₃ 7→ 1.

We have (z1, z2, z3, z4) = g(z4). Then g ◦ f⁻¹ maps

f (z₁) 7→ ∞, f (z₂) 7→ 0, f (z₃) 7→ 1, and by definition

(f (z₁), f (z₂), f (z₃), f (z₄)) = g ◦ f⁻¹(f (z₄)) = g(z₄) = (z₁, z₂, z₃, z₄).

Theorem There exists a unique M¨obius transformation w = f (z) mapping z1, z2, z3 into w1, w₂, w₃ respectively, is given by

(w − w₂)(w₃− w₁)

(w − w₁)(w₃− w₂) = (z − z₂)(z₃− z₁) (z − z₁)(z₃− z₂). Proof If T₁, T₂ are M¨obius transformations such that

T₁ : z₁, z₂, z₃ 7→ ∞, 0, 1 T₂ : w₁, w₂, w₃ 7→ ∞, 0, 1,

then f = T₂⁻¹◦ T₁ is a M¨obius transformation mapping z₁, z₂, z₃ into w₁, w₂, w₃, respectively.

Also since

(f (z₁), f (z₂), f (z₃), f (z)) = (z₁, z₂, z₃, z) ∀ z ∈ C,

so, by setting w = f (z) and w_i = f (z_i) for i = 1, 2, 3, we have (w₁, w₂, w₃, w) = (z₁, z₂, z₃, z) ∀ z ∈ C, and by definition

(w − w₂)(w₃− w₁)

(w − w₁)(w₃− w₂) = (z − z₂)(z₃− z₁) (z − z₁)(z₃− z₂). Remark Note that if z_i = ∞, then T_i(z) is defined to be

• T1(z) = z − z₂

z₃− z₂ maps ∞, z2, z3 into ∞, 0, 1, respectively.

• T₂(z) = z₃− z₁

z − z₁ maps z₁, ∞, z₃ into ∞, 0, 1, respectively.

• T₃(z) = z − z2

z − z₁ maps z₁, z₂, ∞ into ∞, 0, 1, respectively.

Examples

(1) Let D = {z ∈ C | |z| < 1}, H = {z ∈ C | Im z > 0} and f (z) = i1 − z

1 + z for z ∈ D. Then f (z) is a conformal map from D onto H. (Note that f maps −1, 1, i ∈ ∂D into ∞, 0, 1 ∈ ∂H, respectively.)

(2) Let H = {z ∈ C | Im z > 0} and g(z) = 1

πlog z, for z ∈ H. Then g(z) is a conformal map from H onto R × [0, 1]. Hence

g(f (z)) = 1 πlog

 i1 − z

1 + z



: D → R × [0, 1]

is a conformal map from D onto R × [0, 1].

Corollary The automorphisms of D are the M¨obius transformation taking D to itself.

Corollary The automorphisms of H are the M¨obius transformation taking H to itself.

Theorem The automorphisms of H are of the form f (z) = az + b

cz + d, a, b, c, d ∈ R and ad − bc > 0, i.e. the M¨obius transformations taking H to itself.

Proof Suppose f is an automorphism of H, since f maps R onto itself, f is the M¨obius trans- formation that maps 3 distinct numbers x₁, x₂, x₃ in R ∪ {∞} to ∞, 0, 1, respectively. So, by the definition of cross-ratio,

f (z) = (z − x₂)(x₃− x₁)

(z − x₁)(x₃− x₂) = az + b cz + d, with a, b, c, d ∈ R and ad − bc = (x3 − x₁)(x₃− x₂)(x₂− x₁) 6= 0.

Furthermore, since f is an automorphism of H, f (i) = ai + b

ci + d = (bd + ac) + i(ad − bc)

c²+ d² ∈ H =⇒ ad − bc > 0.

Hence the automorphism of H are of the form f (z) = az + b

cz + d with a, b, c, d ∈ R and ad − bc > 0, Lemma Let D ⊆ C be a region, a ∈ D and f be 1 − 1 holomorphic on D \ {a}. Then

• either a is a pole of order one

• or it is a removable singularity and the continuation of f to D is 1 − 1.

TheoremEvery injective holomorphic function f : C → C is an automorphism of C and can be written in the form

z 7→ az + b for some a ∈ C^×; b ∈ C.

Corollary C is not conformally equivalent to any proper subset U ( C.

Riemann Mapping Theorem Let U ( C be open and simply connected proper subset of C, and let α ∈ U. Then ∃! conformal mapping (a bijective analytic mapping) f : U → D such that f (α) = 0 and f⁰(α) > 0 is a positive real number, where D denotes the open unit disk.