§5.1 Biholomorphy

5.1 Biholomorphy

In Section 2.4 $Section 2.4 Section 2.4$ , it was asserted that for a holomorphic function $𝑓 (𝑧)$ , the map $𝑤 = 𝑓 (𝑧)$ is conformal when $𝑓^{'} (𝑧) \neq 0$ .

We have the following immediate assertion:

Theorem 5.1.1 (Open Mapping Theorem).

Suppose

𝑈 \subseteq ℂ

is a region (open, nonempty, and connected). Then the image of any holomorphic and non-constant function

𝑓 : 𝑈 \to ℂ

𝑓 (𝑈)

, is a region.

Proof.

The nonemptiness of $𝑓 (𝑈)$ is an immediate conclusion from the fact that $𝑈$ is nonempty and $𝑓$ is defined on all of $𝑈$ .

Let $𝑤_{0}$ be an arbitrary point in $𝑓 (𝑈)$ . Then $\exists 𝑧_{0} \in 𝑈$ such that $𝑓 (𝑧_{0}) = 𝑤_{0}$ . Since $𝑓$ is non-constant, the function $𝑓 - 𝑤_{0}$ has an isolated zero at $𝑧_{0}$ . Thus for sufficiently small $𝜌 > 0$ , the only zero of $𝑓 - 𝑤_{0}$ in $\overline{𝐷 (𝑧_{0}, 𝜌)}$ is at $𝑧_{0}$ .

By Theorem 3.3.7 $Theorem 3.3.7 Theorem 3.3.7$ , then there exists $𝛿 > 0$ such that $\forall 𝜀 \in 𝐷 (0, 𝛿)$ , $𝑓 (𝑧) - 𝑤_{0} - 𝜀$ has exactly one zero in $\overline{𝐷 (𝑧_{0}, 𝜌)}$ . In other words, $\forall 𝑤_{0} \in 𝑓 (𝑈)$ , $\exists 𝛿 > 0$ such that $\forall 𝑤 \in 𝐷 (𝑤_{0}, 𝛿)$ , $\exists! 𝑧 \in \overline{𝐷 (𝑧_{0}, 𝜌)}$ such that $𝑓 (𝑧) = 𝑤$ . Thus, $𝐷 (𝑤_{0}, 𝛿) \subseteq 𝑓 (𝑈)$ . Thus, $𝑓 (𝑈)$ is an open set since each contained point has a fully contained open neighborhood.

Let $𝑤_{1}, 𝑤_{2} \in 𝑓 (𝑈)$ be arbitrary and distinct. Then there exist $𝑧_{1}, 𝑧_{2} \in 𝑈$ such that $𝑓 (𝑧_{1}) = 𝑤_{1}$ and $𝑓 (𝑧_{2}) = 𝑤_{2}$ . By the connectivity of $𝑈$ , there exists a path $𝛾 \subset 𝑈$ that connects $𝑧_{1}$ and $𝑧_{2}$ . Then $𝑓 (𝛾) \subset 𝑓 (𝑈)$ is a curve that joins $𝑤_{1}$ and $𝑤_{2}$ . Thus, $𝑓 (𝑈)$ is connected.

⁠ $□$

Holomorphic injectivity, or univalence, satisfies the proceeding assertion:

Lemma 5.1.2.

Let

𝑈 \subseteq ℂ

be a region and suppose

𝑓 : 𝑈 \to ℂ

is univalent. Then

𝑓^{'}

is non-vanishing on

𝑈

Proof.

Suppose, for the sake of contradiction, that $𝑓$ is univalent on $𝑈$ such that $\exists 𝑧_{0} \in 𝑈$ such that $𝑓^{'} (𝑧_{0}) = 0$ . Let $𝑤_{0} = 𝑓 (𝑧_{0})$ . The previous statement is equivalent to: $𝑓 (𝑧) - 𝑤_{0}$ has a zero at $𝑧_{0}$ with multiplicity $𝑚 \geq 2$ .

Since $𝑓$ is univalent, neither $𝑓 - 𝑤_{0}$ nor $𝑓^{'}$ may have accumulation points in $𝑈$ . Thus, $\exists 𝜌 > 0$ such that $𝑧_{0}$ is the only zero of either $𝑓 - 𝑤_{0}$ and $𝑓^{'}$ contained in $\overline{𝐷 (𝑧_{0}, 𝜌)} \subset 𝑈$ . By Theorem 3.3.7 $Theorem 3.3.7 Theorem 3.3.7$ , $\exists 𝛿 > 0$ such that $\forall 𝑤 \in 𝐷^{*} (𝑤_{0}, 𝛿) = 𝐷 (𝑤_{0}, 𝛿) ∖ {𝑤_{0}}$ , the equation $𝑓 (𝑧) = 𝑤$ has $𝑚$ solutions in $\overline{𝐷 (𝑧_{0}, 𝜌)}$ , which cannot lie all at a single point (unless that point is $𝑧_{0}$ itself, which cannot be the case as $𝑧_{0}$ already maps to $𝑤_{0} \neq 𝑤$ ), as otherwise $𝑧_{0}$ would not be the only zero of $𝑓^{'}$ in $\overline{𝐷 (𝑧_{0}, 𝜌)}$ . This contradicts the univalence of $𝑓$ . ⁠ $□$

Conversely, we have the following statement on local univalence and invertibility.

Theorem 5.1.3.

Let

𝑈 \subseteq ℂ

be a region and suppose

𝑓 : 𝑈 \to ℂ

is holomorphic. If

𝑓^{'} (𝑧_{0}) \neq 0

for some

𝑧_{0} \in 𝑈

, then there exists an open neighborhood of

𝑧_{0}

on which

𝑓

is univalent.

Proof.

Let

𝑤_{0} = 𝑓 (𝑧_{0})

. Since

{lim}_{𝑧 \to 𝑧_{0}} 𝑓 (𝑧) - 𝑤_{0} = 0

and

{lim}_{𝑧 \to 𝑧_{0}} \frac{𝑓 (𝑧) - 𝑤_{0}}{𝑧 - 𝑧_{0}} \neq 0

, it follows that

𝑧_{0}

is a simple zero of

𝑓 (𝑧) - 𝑤_{0}

. Let

𝑉

be an open neighborhood (relatively compact in

𝑈

) of

𝑧_{0}

whose closure does not contain other zeros of

𝑓 - 𝑤_{0}

. By Theorem 3.3.7

Theorem 3.3.7 Theorem 3.3.7

\exists 𝛿 > 0

such that

\forall 𝑤 \in 𝐷 (𝑤_{0}, 𝛿)

𝑓 (𝑧) = 𝑤

has only one solution for

𝑧

satisfying

𝑧 \in 𝑉

. Therefore, we can choose a relatively compact open subset

𝑊

𝑉

such that

𝑓 (𝑊) \subseteq 𝐷 (𝑤_{0}, 𝛿)

, on which

𝑓

is univalent. ⁠

□

Moreover, if $𝑤 = 𝑓 (𝑧)$ is univalent and surjective, mapping $𝑈$ to $𝐺$ , then its inverse $𝑧 = 𝑓^{- 1} (𝑤)$ is univalent on $𝐺$ . Such bijective holomorphic functions are known as biholomorphisms or biholomorphic functions.

We will now study holomorphic functions from a more geometric perspective.

Theorem 5.1.4.

Let

Ω \subseteq ℂ

be a region, and let

𝛾 \subset Ω

be a rectifiable simple closed counterclockwise-oriented curve that is null-homotopic in

Ω

. Denote

int (𝛾)

𝑈

. If

𝑓 : Ω \to ℂ

is holomorphic and maps

𝛾

injectively to a simple closed curve

Γ

, then

𝑤 = 𝑓 (𝑧)

is univalent in

𝑈

𝑓 (𝑈) = int (Γ)

, and

Γ

is traversed counterclockwise.

Proof.

Let $𝑤_{0} \in ℂ$ . Let $𝑘 = 𝑘 (𝑤_{0})$ be the number of zeros of $𝑓 - 𝑤_{0}$ in $𝑈$ . By the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ), for $𝑤_{0} \notin Γ$ ,

𝑘 = \frac{1}{2 π i} ∳_{𝛾} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧) - 𝑤_{0}} d 𝑧 = \frac{1}{2 π i} ∳_{Γ} \frac{d 𝑤}{𝑤 - 𝑤_{0}} = {Ind}_{Γ} (𝑤_{0}) .

If $𝑤_{0} \in ext (Γ)$ , the expression vanishes since ${Ind}_{Γ} (𝑤_{0}) = 0$ . Then $𝑓 (𝑧) = 𝑤_{0}$ has no solution in $𝑈$ (i.e. $𝑘 = 0$ ).
If $𝑤_{0} \in int (Γ)$ , then $Γ$ winds around $𝑤_{0}$ exactly once, and hence, in other words, $\forall 𝑤_{0} \in int (Γ)$ , $𝑓 (𝑧) = 𝑤_{0}$ has a unique solution in $𝑈$ (i.e. $𝑘 = 1$ ). Moreover, $Γ$ must be oriented counterclockwise for $𝑘 > 0$ to be true.
If $𝑤_{0}$ lies on $Γ$ , then it can be shown that $𝑓 - 𝑤_{0}$ has no zeros in $𝑈$ .
Indeed, for the sake of contradiction, assume that $\exists 𝑧_{0} \in 𝑈$ such that $𝑓 (𝑧_{0}) = 𝑤_{0}$ . By the Open Mapping Theorem (Theorem 5.1.1 $Theorem 5.1.1 Theorem 5.1.1$ ), $\exists 𝛿 > 0$ such that $𝐷 (𝑤_{0}, 𝛿) \subseteq 𝑓 (𝑈)$ , or equivalently, $\forall 𝑤 \in 𝐷 (𝑤_{0}, 𝛿)$ , $𝑓 - 𝑤$ has zeros in $𝑈$ . Since $𝑤_{0}$ lies on $Γ$ , a subset of $𝐷 (𝑤_{0}, 𝛿)$ lies in the exterior of $Γ$ . It was previously established that $𝑓 - 𝑤$ has no zeros if $𝑤 \in 𝐷 (𝑤_{0}, 𝛿) \cap ext (Γ)$ . Thus, we have a contradiction, and no such $𝑧_{0}$ exists, implying $𝑘 = 0$ .

We then have

𝑘 = {\begin{cases} 0 & if 𝑤_{0} \in \overline{ext (Γ)} \\ 1 & if 𝑤_{0} \in int (Γ) \end{cases} .

Hence, $𝑓$ is univalent in $𝑈$ (since for each $𝑤_{0}$ , $𝑘$ is at most one).

Moreover, any point $𝑧_{0} \in 𝑈$ must map to either $int (Γ)$ or $\overline{ext (Γ)}$ . The latter is an impossibility since otherwise $𝑘 \neq 0$ . This $𝑓 (𝑈) = int (Γ)$ . ⁠ $□$

The condition that $𝑓$ be holomorphic on a neighborhood of the boundary can be weakened especially when the domain is a disk.

Theorem 5.1.5.

Let

𝑓 \in 𝐶 (\overline{𝔻})

be holomorphic on

𝔻

, and suppose that the restriction

{𝑓 |}_{𝜕 𝔻}

is injective. If

Γ = 𝑓 (𝜕 𝔻)

, then

𝑓

is univalent on

𝔻

𝑓 (𝔻) = int (Γ)

, and

{𝑓 |}_{𝜕 𝔻}

traverses

Γ

in the same direction as

𝜕 𝔻

Proof.

For $0 < 𝑟 < 1$ , define $𝑓_{𝑟} (𝑧) = 𝑓 (𝑟 𝑧)$ . Each $𝑓_{𝑟}$ is holomorphic on the disk $𝐷 (0, \frac{1}{𝑟})$ , which is a neighborhood of $\overline{𝔻}$ . Since $𝑓$ is uniformly continuous on $\overline{𝔻}$ (by Heine–Cantor or Theorem 1.2.26 $Theorem 1.2.26 Theorem 1.2.26$ ),

max_{𝑧 \in 𝜕 𝔻} | 𝑓_{𝑟} (𝑧) - 𝑓 (𝑧) | \to 0

as $𝑟 \to 1^{-}$ . Fix $𝑤_{0} \notin Γ$ and set $𝑑 = dist (𝑤_{0}, Γ) > 0$ . For all $𝑟$ sufficiently close to $1$ , the preceding maximum is less than $𝑑$ . Consequently, (for sufficiently large $𝑟 < 1$ )

{Ind}_{𝑓_{𝑟} (𝜕 𝔻)} (𝑤_{0}) = {Ind}_{Γ} (𝑤_{0})

(heuristically, the condition ${max}_{𝑧 \in 𝜕 𝔻} | 𝑓_{𝑟} (𝑧) - 𝑓 (𝑧) | < 𝑑$ means that $𝑤_{0}$ can not lie in between the two curves; a smooth homotopy or deformation exists between the two curves, leaving the winding indices invariant).

Let $𝑘_{𝑟} = 𝑘_{𝑟} (𝑤_{0})$ be the number of zeros of $𝑓_{𝑟} - 𝑤_{0}$ in $𝔻$ , counting multiplicities. Assume $𝜕 𝔻$ is positively oriented. Then by the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ),

𝑘_{𝑟} = {Ind}_{𝑓_{𝑟} (𝜕 𝔻)} (𝑤_{0}) = {Ind}_{Γ} (𝑤_{0}) .

The zeros of $𝑓_{𝑟} - 𝑤_{0}$ in $𝔻$ correspond, with the same multiplicities, to the zeros of $𝑓 - 𝑤_{0}$ in $𝐷 (0, 𝑟)$ . Moreover, since $𝑓 - 𝑤_{0}$ is nonzero on $𝜕 𝔻$ and is continuous on $\overline{𝔻}$ , it is nonzero throughout some annular neighborhood of $𝜕 𝔻$ in $\overline{𝔻}$ . Thus, for $𝑟$ sufficiently close to $1$ , $𝑘_{𝑟}$ is the number $𝑘 = 𝑘 (𝑤_{0})$ of all zeros of $𝑓 - 𝑤_{0}$ in $𝔻$ . We have therefore shown that (for sufficiently large $𝑟 < 1$ )

𝑘 (𝑤_{0}) = 𝑘_{𝑟} (𝑤_{0}) = {Ind}_{Γ} (𝑤_{0}) .

Since $Γ$ is a Jordan curve, its index vanishes for $𝑤_{0} \in ext (Γ)$ and is $\pm 1$ for $𝑤_{0} \in int (Γ)$ . The latter index cannot be negative because it is equal to the number $𝑘 (𝑤_{0})$ of zeros. It is therefore $1$ , which then shows that ${𝑓 |}_{𝜕 𝔻}$ traverses $Γ$ counterclockwise. Hence $𝑓 - 𝑤_{0}$ has exactly one zero in $𝔻$ when $𝑤_{0} \in int (Γ)$ , and no zeros there when $𝑤_{0} \in ext (Γ)$ .

Finally, suppose $𝑤_{0} \in Γ$ and $𝑓 (𝑧_{0}) = 𝑤_{0}$ for some $𝑧_{0} \in 𝔻$ . By the Open Mapping Theorem (Theorem 5.1.1 $Theorem 5.1.1 Theorem 5.1.1$ ), $𝑓 (𝔻)$ contains a disk centered at $𝑤_{0}$ . Such a disk contains a point in $ext (Γ)$ , contradicting our earlier assertion. Thus no point of $Γ$ lies in $𝑓 (𝔻)$ . It follows that $𝑓 (𝔻) = int (Γ)$ , and as $𝑘 \leq 1$ in all cases, $𝑓$ is univalent therein. ⁠ $□$

We will now give examples of biholomorphisms.

Example 5.1.6.

The only biholomorphisms which map $𝔻$ to itself are in the form of

𝑤 = e^{i 𝜃} \frac{𝑧 - 𝑎}{1 - \overline{𝑎} 𝑧}, 𝑎 \in 𝔻, 𝜃 \in ℝ . (5.1 . 1)

This follows directly from Theorem 3.5.3

Theorem 3.5.3 Theorem 3.5.3

Example 5.1.7.

The only biholomorphisms which map $ℍ^{+}$ to $𝔻$ are in the form of

𝑤 = e^{i 𝜃} \frac{𝑧 - 𝑎}{𝑧 - \overline{𝑎}}, 𝑎 \in ℍ^{+}, 𝜃 \in ℝ . (5.1 . 2)

Proof.

First assume $𝑦 = ℑ𝔪 (𝑧) > 0$ . It follows that

| 𝑤 | = | \frac{𝑧 - 𝑎}{𝑧 - \overline{𝑎}} | = \sqrt{\frac{{(𝑥 - ℜ𝔢 (𝑎))}^{2} + {(𝑦 - ℑ𝔪 (𝑎))}^{2}}{{(𝑥 - ℜ𝔢 (𝑎))}^{2} + {(𝑦 + ℑ𝔪 (𝑎))}^{2}}} < 1 .

Therefore, this transformation maps $ℍ^{+}$ to $𝔻$ . The inverse mapping is equal to

𝑧 = \frac{𝑤 \overline{𝑎} - 𝑎 e^{i 𝜃}}{𝑤 - e^{i 𝜃}} . (5.1 . 3)

Assume $𝑤 \in 𝔻$ . We then have

\begin{aligned} ℑ𝔪 (𝑧) & = ℑ𝔪 (\frac{(𝑤 \overline{𝑎} - 𝑎 e^{i 𝜃}) (\overline{𝑤} - e^{- i 𝜃})}{{| 𝑤 - e^{i 𝜃} |}^{2}}) \\ = \frac{{| 𝑤 |}^{2} ℑ𝔪 (\overline{𝑎}) - ℑ𝔪 (𝑎 e^{i 𝜃} \overline{𝑤}) - ℑ𝔪 (𝑤 \overline{𝑎} e^{- i 𝜃}) + ℑ𝔪 (𝑎)}{{| 𝑤 - e^{i 𝜃} |}^{2}} \\ = \frac{(1 - {| 𝑤 |}^{2}) ℑ𝔪 (𝑎)}{{| 𝑤 - e^{i 𝜃} |}^{2}} > 0 . \end{aligned}

Hence, $𝑧$ maps $𝔻$ to $ℍ^{+}$ univalently and surjectively since it is also an element in $Aut (\hat{ℂ})$ .

Let $𝜓 (𝑧)$ be the biholomorphism from $ℍ^{+}$ to $𝔻$ in the form of $𝜓 (𝑧) = \frac{𝑧 - i}{𝑧 + i}$ (for $𝜃 = 0$ and $𝑎 = i$ , known as the Cayley transform). Let $𝑓$ be an arbitrary biholomorphism from $ℍ^{+}$ to $𝔻$ . It follows that $𝜑 = 𝑓 \circ 𝜓^{- 1}$ is a holomorphic automorphism on $𝔻$ . Since $𝜑 \in Aut (𝔻)$ , we have

\begin{aligned} 𝑓 (𝑧) = 𝜑 \circ 𝜓 (𝑧) & = e^{i 𝜃} \frac{𝑧 (1 - 𝑎) - i (𝑎 + 1)}{𝑧 (1 - \overline{𝑎}) + i (\overline{𝑎} + 1)} \\ = e^{i 𝜃} \frac{𝑧 \frac{1 - 𝑎}{1 - \overline{𝑎}} + i \frac{𝑎 + 1}{\overline{𝑎} - 1}}{𝑧 - i \frac{\overline{𝑎} + 1}{\overline{𝑎} - 1}} \\ = e^{i 𝜃} \frac{1 - 𝑎}{1 - \overline{𝑎}} \frac{𝑧 - i \frac{𝑎 + 1}{1 - 𝑎}}{𝑧 - \overline{i \frac{𝑎 + 1}{1 - 𝑎}}} . \end{aligned}

Obviously, $e^{i 𝜃} \frac{1 - 𝑎}{1 - \overline{𝑎}}$ attains every value on the unit disk for varying $𝑎$ and $𝜃$ . Similarly, the values attained by $i \frac{𝑎 + 1}{1 - 𝑎}$ cover the upper half-plane for $𝑎 \in 𝔻$ (since it is in the form of (5.1.3) $(5.1.3) (5.1.3)$ ). Thus, all biholomorphisms from $ℍ^{+}$ to $𝔻$ are in the form of (5.1.2) $(5.1.2) (5.1.2)$ . ⁠ $□$

Let us now introduce some important properties of linear fractional transformations. By Proposition 4.4.2.6 $Proposition 4.4.2.6 Proposition 4.4.2.6$ , it follows that the composition of two linear fractional transformations is also a linear fractional transformation.

Theorem 5.1.8.

Let

𝒞︀

be the collection of subsets of

\hat{ℂ}

that are circles or

𝐿 \cup {\infty}

, where

𝐿

is a straight line in

ℂ

(known as generalized circles). Then every linear fractional transformation

𝑓 : \hat{ℂ} \to \hat{ℂ}

maps elements of

𝒞︀

to elements of

𝒞︀

Proof.

Since each linear fractional transformation is a composition of maps in the form of $𝑧 \mapsto 𝑎 𝑧$ , $𝑧 \mapsto 𝑧 + 𝑏$ , and $𝑧 \mapsto \frac{1}{𝑧}$ , it suffices to show that these maps preserve the property of being a circle or a straight line. Consider a circle defined implicitly with

𝛼 (𝑥^{2} + 𝑦^{2}) + 𝛽 𝑥 + 𝛾 𝑦 + 𝛿 = 0, 𝑥, 𝑦 \in ℝ, 𝛼, 𝛽, 𝛾, 𝛿 \in ℝ .

For $𝑧 = 𝑥 + i 𝑦$ , this can be rewritten as

\begin{aligned} 𝛼 𝑧 \overline{𝑧} + 𝛽 \frac{𝑧 + \overline{𝑧}}{2} + 𝛾 \frac{𝑧 - \overline{𝑧}}{2 i} + 𝛿 \\ \begin{matrix}  \end{matrix} = 𝛼 𝑧 \overline{𝑧} + 𝜉 𝑧 + \overline{𝜉} \overline{𝑧} + 𝛿 = 0 for 𝜉 = \frac{𝛽}{2} + \frac{𝛾}{2 i} . (5.1 . 4) \end{aligned}

If $𝛼 = 0$ , the equation represents a straight line. It is easy to see that a complex dilation or a translation of $𝑧$ will preserve the property of being a straight line or a circle. Indeed, by letting $𝑧 = 𝑎 𝜁$ for nonzero $𝑎$ in (5.1.4) $(5.1.4) (5.1.4)$ , we have

𝛼 {| 𝑎 |}^{2} 𝜁 \overline{𝜁} + 𝜉 𝑎 𝜁 + \overline{𝜉} \overline{𝑎} \overline{𝜁} + 𝛿 = 0,

which is trivially in the form of (5.1.4) $(5.1.4) (5.1.4)$ . Similarly, if we substitute $𝑧 = 𝜁 + 𝑏$ , we have

\begin{matrix} 𝛼 (𝜁 + 𝑏) (\overline{𝜁} + \overline{𝑏}) + 𝜉 (𝜁 + 𝑏) + \overline{𝜉} (\overline{𝜁} + \overline{𝑏}) + 𝛿 = 0 \\ 𝛼 𝜁 \overline{𝜁} + (𝜉 + 𝛼 \overline{𝑏}) 𝜁 + (\overline{𝜉} + 𝛼 𝑏) \overline{𝜁} + 𝛼 {| 𝑏 |}^{2} + 2 ℜ𝔢 (𝜉 𝑏) + 𝛿 = 0 . \end{matrix}

If we substitute $𝑧 = \frac{1}{𝜁}$ , we have

𝛿 𝜁 \overline{𝜁} + 𝜉 \overline{𝜁} + \overline{𝜉} 𝜁 + 𝛼 = 0,

which is in the form of (5.1.4) $(5.1.4) (5.1.4)$ . ⁠ $□$

Remark.

As in Example 5.1.7

Example 5.1.7 Example 5.1.7

, we can consider extended straight lines in the form of

𝐿 \cup {\infty}

as generalized circles in the Riemann sphere. In other words, the extended line can be geometrically visualized by a circle with infinite radius. In fact, when a circle on the Riemann sphere is projected stereographically onto the complex plane, the result is always either a circle or a straight line.

Definition 5.1.9 (Cross-Ratio).

Let $𝑧_{1}, 𝑧_{2}, 𝑧_{3}, 𝑧_{4} \in \hat{ℂ}$ be points such that at least three of them are distinct. The cross-ratio of these points is defined as

(𝑧_{1}, 𝑧_{2}; 𝑧_{3}, 𝑧_{4}) = \frac{(𝑧_{1} - 𝑧_{3}) (𝑧_{2} - 𝑧_{4})}{(𝑧_{1} - 𝑧_{4}) (𝑧_{2} - 𝑧_{3})} .

If at least one of the four points is $\infty$ , then the cross-ratio is defined by the limit:

\begin{aligned} (\infty, 𝑧_{2}; 𝑧_{3}, 𝑧_{4}) & = \frac{𝑧_{2} - 𝑧_{4}}{𝑧_{2} - 𝑧_{3}}, & (𝑧_{1}, \infty; 𝑧_{3}, 𝑧_{4}) & = \frac{𝑧_{1} - 𝑧_{3}}{𝑧_{1} - 𝑧_{4}} \\ (𝑧_{1}, 𝑧_{2}; \infty, 𝑧_{4}) & = \frac{𝑧_{2} - 𝑧_{4}}{𝑧_{1} - 𝑧_{4}}, & (𝑧_{1}, 𝑧_{2}; 𝑧_{3}, \infty) & = \frac{𝑧_{1} - 𝑧_{3}}{𝑧_{2} - 𝑧_{3}} \end{aligned}

One important property of the cross-ratio is that it is invariant under linear fractional transformations. In other words, if $𝑓$ is a linear fractional transformation, then

(𝑓 (𝑧_{1}), 𝑓 (𝑧_{2}); 𝑓 (𝑧_{3}), 𝑓 (𝑧_{4})) = (𝑧_{1}, 𝑧_{2}; 𝑧_{3}, 𝑧_{4}) .

The proof is trivial and can be verified by substituting the definition of the linear fractional transformation into the definition of the cross-ratio.

Furthermore, if a function $𝑓 (𝑧_{1}, 𝑧_{2}, 𝑧_{3}, 𝑧_{4})$ is invariant under the group of linear fractional transformations, then it is a function of the cross-ratio. In other words, the cross-ratio is the only invariant under the group of linear fractional transformations $Aut (\hat{ℂ})$ . Indeed, suppose that

𝑓 (𝜑 (𝑧_{1}), 𝜑 (𝑧_{2}), 𝜑 (𝑧_{3}), 𝜑 (𝑧_{4})) = 𝑓 (𝑧_{1}, 𝑧_{2}, 𝑧_{3}, 𝑧_{4}) .

We aim to show that $𝑓$ is a function of a cross-ratio. Let

𝜑 (𝑧) = \frac{(𝑧 - 𝑧_{3}) (𝑧_{2} - 𝑧_{4})}{(𝑧 - 𝑧_{4}) (𝑧_{2} - 𝑧_{3})}

be a linear fractional transformation. Then we have

𝑓 (𝜑 (𝑧_{1}), 𝜑 (𝑧_{2}), 𝜑 (𝑧_{3}), 𝜑 (𝑧_{4})) = 𝑓 ((𝑧_{1}, 𝑧_{2}; 𝑧_{3}, 𝑧_{4}), 1, 0, \infty),

which is a function of the cross-ratio.