Let \(\Omega_{1}, \Omega_{2}\) and \(\Omega\) be open subsets of \(\mathbb C\).
A one-to-one holomorphic mapping \(f\) from \(\Omega_{1}\) onto \(\Omega_{2}\) such that \(f^{-1} \in H\left(\Omega_{2}\right)\) is called an analytic homeomorphism of \(\Omega_{1}\) onto \(\Omega_{2}\). If \(\Omega=\Omega_{1}=\Omega_{2}\), then \(f\) is called an analytic automorphism of \(\Omega\).
A one-to-one holomorphic mapping from \(\Omega_{1}\) onto \(\Omega_{2}\) is called a conformal mapping of \(\Omega_{1}\) onto \(\Omega_{2}\). If \(\Omega=\Omega_{1}=\Omega_{2}\), then \(f\) is called a conformal mapping of \(\Omega\).
Recall the inverse mapping theorem:
Suppose that \(\Omega\subseteq \mathbb C\) is open, \(f \in H(\Omega)\), and \(f\) is one-to-one in \(\Omega\). Then \(f^{\prime}(z) \neq 0\) for every \(z \in \Omega\), and the inverse of \(f\) is holomorphic.
Remarks.
By the inverse mapping theorem, we see that a conformal mapping of \(\Omega_{1}\) onto \(\Omega_{2}\) is an analytic homeomorphism of \(\Omega_{1}\) onto \(\Omega_{2}\), and a conformal mapping of \(\Omega\) is an automorphism of \(\Omega\).
We say that \(\Omega_{1}\) and \(\Omega_{2}\) are conformally equivalent whenever there is a conformal mapping \(f\) from \(\Omega_{1}\) onto \(\Omega_{2}\) (or between \(\Omega_{1}\) and \(\Omega_{2}\) ) and we write \(\Omega_{1} \sim \Omega_{2}\), where \(\sim\) is an equivalence relation.
If \(\Omega_{1} \sim \Omega_{2}\), then the corresponding conformal mapping \(f:\Omega_{1} \to \Omega_{2}\) satisfies \(f^{\prime}(z) \neq 0\) for \(z \in \Omega_{1}\) by the inverse mapping theorem.
Some authors take the condition \(f^{\prime}(z) \neq 0\) for \(z \in \Omega_{1}\) as the definition of a conformal mapping.
The latter condition is less restrictive. For example, \(f(z)=z^{2}\) is not one-to-one but \(f^{\prime}\) vanishes nowhere in \(\mathbb{C} \backslash\{0\}\).
Remarks.
There is a geometric consequence of the condition \(f^{\prime}(z) \neq 0\) and it is at the root of this discrepency of terminology in the definitions. A holomorphic map that satisfies this condition preserves angles.
Loosely speaking, if two curves \(\gamma\) and \(\eta\) intersect at \(z_{0}\), and \(\alpha\) is the oriented angle between the tangent vectors to these curves, then the image curves \(f \circ \gamma\) and \(f \circ \eta\) intersect at \(f\left(z_{0}\right)\), and their tangent vectors form the same angle \(\alpha\).
We will always mean, as is the general practice is complex analysis that conformal mapping is one-to-one onto holomorphic function.
The set of all automorphisms of \(\Omega\) is a group under composition of mappings and we denote it by \(\operatorname{Aut}(\Omega)\).
For \(\lambda \in \mathbb{C}\) with \(|\lambda|=1\), we observe that \(z \mapsto \lambda z\) is an automorphism of \(D=D(0,1)\) and this automorphism is called a rotation.
Let \(\Omega_{1}=\mathbb{H}=\{z\in\mathbb C: \operatorname{Im}z>0\}\) and \(\Omega_{2}=D=\{z\in\mathbb C: |z|<1\}\). Then the map \(F:\Omega_1\to \Omega_2\), given by \[F(z)=\frac{i-z}{i+z} \quad \text { for } \quad z\in\Omega_1,\] is a conformal map with the inverse \(G:\Omega_2\to \Omega_1\) given by \[G(w)=i \frac{1-w}{1+w}\quad \text { for } \quad w\in\Omega_2.\] In other words, \(\Omega_{1}\) and \(\Omega_{2}\) are conformally equivalent.
Proof: Recall that \(F(z)=\frac{i-z}{i+z}\) for \(z\in\Omega_1\).
\(F\) is one-to-one. Indeed, if \(F(u)=F(v)\), then \[\frac{i-u}{i+u}=\frac{i-w}{i+w}\quad \iff \quad 2iu=2iw.\]
Further we observe that \(F\) is analytic in \(\Omega_{1}\) since \(-i \notin \Omega_{1}\).
Let \(z \in \Omega_{1}\) and write \(z=x+i y\) with \(y>0\). Then \[F(z)=\frac{-x-i(y-1)}{x+i(y+1)}.\]
Thus \[|F(z)|^{2}=F(z) \overline{F(z)}=\frac{x^{2}+(y-1)^{2}}{x^{2}+(y+1)^{2}}<1,\] and hence \(F(z) \in \Omega_{2}\) since \(y>0\).
It remains to show that \(F\) is onto. Indeed, let \(w \in \Omega_{2}\) and we write \(w=u+i v\) with \(u^{2}+v^{2}<1\).
We find \(z \in \Omega_{1}\) such that \(F(z)=w\). Thus \[\frac{i-z}{i+z}=w,\] and then \[z=i\left(\frac{1-w}{1+w}\right)=i\left(\frac{1-u-i v}{1+u+i v}\right)=i \frac{(1-u-i v)(1+u-i v)}{(1+u+i v)(1+u-i v)}.\]
Therefore \[\operatorname{Im}(z)=\frac{1-u^{2}-v^{2}}{(1+u)^{2}+v^{2}}>0,\] and hence \(z \in \Omega_{1}\). $$\tag*{$\blacksquare$}$$
For \(n\in\mathbb N\), let \[\Omega_{1}=\left\{z \in \mathbb{C} : 0<\arg z<\frac{\pi}{n}\right\}\] and \(\Omega_{2}=\mathbb{H}\). We show that \(\Omega_{1}\) and \(\Omega_{2}\) are conformally equivalent.
Proof: For this, consider holomorphic function \[f(z)=z^{n} \quad\text { for } \quad z \in \Omega_{1}.\]
We write \(z=r e^{i \theta}\) with \(0<\theta<\frac{\pi}{n}\), then \(f(z)=r^{n} e^{i n \theta}\).
Thus \(f\) is a function from \(\Omega_{1}\) into \(\Omega_{2}\), since \[\operatorname{Im}(f(z))=r^{n} \sin n \theta >0 \quad \text{ for } \quad 0<\theta<\frac{\pi}{n}.\]
\(f\) is one-to-one. Indeed, let \(z_{1}, z_{2} \in \Omega_{1}\) with \(f\left(z_{1}\right)=f\left(z_{2}\right)\) and \[z_{1}=r_{1} e^{i \theta_{1}}, z_{2}=r_{2} e^{i \theta_{2}} \quad \text { with } \quad 0<\theta_{1} \leq \theta_{2}<\frac{\pi}{n}.\]
Then \[r_{1}^{n} e^{i n \theta_{1}}=r_{2}^{n} e^{i n \theta_{2}},\] implying \[\left(\frac{r_{1}}{r_{2}}\right)^{n}=e^{i n\left(\theta_{2}-\theta_{1}\right)}=\cos \left(n\left(\theta_{2}-\theta_{1}\right)\right)+i \sin \left(n\left(\theta_{2}-\theta_{1}\right)\right).\]
By comparing the imaginary parts on both sides, we obtain that \(\sin \left(n\left(\theta_{2}-\theta_{1}\right)\right)=0\) imply\(\operatorname{ing} \theta_{1}=\theta_{2}\) since \(0 \leq n\left(\theta_{2}-\theta_{1}\right)<\pi\).
Hence \(r_{1}=r_{2}\) and then \(z_{1}=z_{2}\), proving that \(f\) is one-to-one.
Now, we show that \(f\) is onto. Let \(w=R e^{i \phi} \in \Omega_{2}\). Then \(0<\phi<\pi\).
Setting \(w^{1 / n}\) by taking the principal value of logarithm, we see that \[w^{1 / n}=e^{\frac{1}{n}\log w}=e^{\frac{1}{n}\log R+i\frac{\phi}{n}}\in \Omega_{1},\] and \(f\left(w^{1 / n}\right)=w\).
Hence \(\Omega_{1}\) and \(\Omega_{2}\) are conformally equivalent.$$\tag*{$\blacksquare$}$$
Let \(\Omega_{1}=\left\{x+i y : y>0, x^{2}+y^{2}<1\right\},\) be the upper half disc and \(\Omega_{2}=\{u+i v : u>0, v>0\}\) be the positive quadrant of \(\mathbb C\). Then the mapping \[f(z)=\frac{1+z}{1-z}\] is a conformal mapping of \(\Omega_{1}\) onto \(\Omega_{2}\).
Proof: Let \(z=x+i y \in \Omega_{1}\). Then \(|x|<1, y>0, x^{2}+y^{2}<1\) and \[\begin{aligned} f(z)=\frac{1+x+i y}{1-x-i y}&=\frac{(1+x+i y)(1-x+i y)}{(1-x-i y)(1-x+i y)}\\ &=\frac{1-\left(x^{2}+y^{2}\right)}{(1-x)^{2}+y^{2}}+i \frac{2 y}{(1-x)^{2}+y^{2}}. \end{aligned}\] Since \(\operatorname{Re}(f(z))>0\) and \(\operatorname{Im}(f(z))>0\), we see that \(f(z) \in \Omega_{2}\).
It is clear that \(f\) is analytic in \(\Omega_{1}\) since \(1 \notin \Omega_{1}\).
\(f\) is one-to-one, since \(f(z)=\frac{1+z}{1-z}=\frac{1+w}{1-w}=f(w)\) implies \(2z=2w\).
It remains to show that \(f\) is onto. Let \(w=u+i v \in \Omega_{2}\). Then \(u>0, v>0\) and we shall find \(z \in \Omega_{1}\) such that \(f(z)=w\).
Thus \(\frac{1+z}{1-z}=w\) and then \[\begin{aligned} z=\frac{w-1}{w+1}&=\frac{(u-1)+i v}{(u+1)+i v}=\frac{((u-1)+i v)((u+1)-i v)}{((u+1)+i v)((u+1)-i v)}\\ &=\frac{u^{2}+v^{2}-1}{(u+1)^{2}+v^{2}}+i \frac{2 v}{(u+1)^{2}+v^{2}}. \end{aligned}\]
Thus \(\operatorname{Im}(z)>0\) since \(v>0\). Further \(|z|<1\) since \(|w-1|<|w+1|\) whenever \(w\) lies in the first quadrant and hence \(z \in \Omega_{1}\).$$\tag*{$\blacksquare$}$$
Let \(\Omega_{1}=\mathbb{H}\) be the upper half-plane and \(\Omega_{2}=\{u+i v : 0<v<\pi\}\) be an open strip. Then \(\Omega_{1}\) and \(\Omega_{2}\) are conformally equivalent.
Proof: Let us consider \[f(z)=\log z,\quad \text{ for } \quad z \in \Omega_{1},\] where the branch of logarithm is principal.
Then \(f \in H\left(\Omega_{1}\right)\) and one-to-one. We show that \(f\) is onto.
Let \(w=u+i v \in \Omega_{2}\). Then \(0<v<\pi\) and we take \[z=e^{w}=e^{u} e^{i v}=e^{u}(\cos v+i \sin v) .\]
Thus \[\operatorname{Im}(z)=e^{u} \sin v>0\] since \(0<v<\pi\). Therefore \(z \in \Omega_{1}\) and \(f(z)=w\).
Hence \(f\) is a conformal mapping between \(\Omega_{1}\) and \(\Omega_{2}\).$$\tag*{$\blacksquare$}$$
Let \(f\) be holomorphic in the open unit disc \(D\).
If \(f\) satisfies \(|f(z)| \leq 1\), and \(f(0)=0\), then \(|f(z)| \leq|z|\) for \(|z|<1\) and \(\left|f^{\prime}(0)\right| \leq 1\).
If \(|f(z)|=|z|\) for some \(z\in D\) or \(\left|f^{\prime}(0)\right|=1\), then \(f(z)=c z\) in \(D\) for some constant \(c\) whose absolute value is \(1\).
Proof: Let \[g(z)= \begin{cases} \frac{f(z)}{z}, & \text { if } z \neq 0, \\ f^{\prime}(0), & \text { if } z=0. \end{cases}\]
Then \(g\) is analytic in \(|z|<1\) since \(f(0)=0\).
For \(0 \leq r<1\), we derive from the maximum modulus theorem that \[\max _{|z| \leq r}|g(z)|=\max _{|z|=r}|g(z)|=\frac{1}{r}\left(\max _{|z|=r}|f(z)|\right) \leq \frac{1}{r}.\]
Letting \(r\) tend to \(1\), we get \[|g(z)| \leq 1 \quad \text { in } \quad |z|<1 .\]
Hence, \[|f(z)| \leq|z| \quad \text { in } \quad |z|<1.\]
Assume that either \(\left|f\left(z_{0}\right)\right|=\left|z_{0}\right|\) for some \(\left|z_{0}\right|<1\) or \(\left|f^{\prime}(0)\right|=1\).
Then \(g(z)=1\) for some \(|z|<1\). Therefore \(g\) is constant in \(D\) of absolute value \(1\) by the maximum principle.
Then \(f(z)=c z\) in \(|z|<1\), where \(c\) is a constant with \(|c|=1\).$$\tag*{$\blacksquare$}$$
For \(a \in D\), we consider a function \(\phi_{a}:\mathbb{C}_{\infty}\to \mathbb{C}_{\infty}\) given by \[\phi_{a}(z)=\frac{z-a}{1-\overline{a} z}. \tag{*}\]
We observe that \[\begin{aligned} \phi_{a}\left(\frac{1}{\overline{a}}\right)&=\infty, \quad\: \phi_{-a}\left(-\frac{1}{\overline{a}}\right)=\infty,\\ \phi_{a}(\infty)&=-\frac{1}{\overline{a}},\quad \quad \phi_{-a}(\infty)=\frac{1}{\overline{a}}. \end{aligned}\]
Let \(a\in D\) and \(\mathbb T=\{z\in\mathbb C: |z| =1\}\) be the torus.
\(\phi_{a}\) is one-to-one, onto and the inverse of \(\phi_{a}\) is \(\phi_{-a}\) and \(\phi_{a}(a)=0\).
\(\phi_{a}\) is analytic in \(D\left(0, \frac{1}{\overline{a}}\right)\) containing \(\overline{D}(0,1)\).
\(\phi_{a}(\mathbb T)=\mathbb T\).
\(\phi_{a}(D)=D\).
\(\phi_{a}^{\prime}(0)=1-|a|^{2}\), and \(\phi_{a}^{\prime}(a)=\frac{1}{1-|a|^{2}}\).
Proof (i): We show \[\phi_{-a} \circ \phi_{a}(z)=z,\quad \text{ and } \quad \phi_{a} \circ \phi_{-a}(z)=z\quad \text { for }\quad z \in \mathbb{C}.\] We prove the first and the proof for the latter is similar.
If \(z=\infty\), then \[\phi_{-a} \circ \phi_{a}(\infty)=\phi_{-a}\left(-\frac{1}{\overline{a}}\right)=\infty.\]
If \(z=\frac{1}{\overline{a}}\), then \[\phi_{-a} \circ \phi_{a}\left(\frac{1}{\overline{a}}\right)=\phi_{-a}(\infty)=\frac{1}{\overline{a}}.\]
Therefore we may assume that \(z \neq \frac{1}{\overline{a}}\) and \(z \neq \infty\). Now \[\phi_{-a} \circ \phi_{a}(z)=\phi_{-a}\left(\frac{z-a}{1-\overline{a} z}\right)=\frac{\frac{z-a}{1-\overline{a} z}+a}{1+\overline{a} \frac{z-a}{1-\overline{a} z}}=\frac{z(1-a \overline{a})}{1-a \overline{a}}=z,\] since \(a \overline{a}=|a|^{2}<1\) for \(a \in D\). $$\tag*{$\blacksquare$}$$
Proof (ii): Note that \(\phi_{a}(z)\) is analytic in \(\mathbb{C}\) except at \(z=\frac{1}{\overline{a}}\) with \(\frac{1}{|\overline{a}|}>1\).
Let \(1<r< \frac{1}{|\overline{a}|}\), then \(\phi_{a}(z)\) is analytic in \(D(0, r)\).
Since \(\overline{D} \subseteq D(0, r)\), the assertion follows. $$\tag*{$\blacksquare$}$$
Proof (iii): For \(t \in \mathbb{R}\), we have \[\phi_{a}\left(e^{i t}\right)=\frac{e^{i t}-a}{1-\overline{a} e^{i t}}=\frac{e^{i t}-a}{e^{i t}\left(e^{-i t}-\overline{a}\right)} .\]
Therefore \[\left|\phi_{a}\left(e^{i t}\right)\right|=1 .\]
Thus \(\phi_{a}(\mathbb T) \subseteq \mathbb T\). Similarly \(\phi_{-a}(\mathbb T) \subseteq \mathbb T\), implying \[\mathbb T \subseteq \phi_{a}(\mathbb T).\]
Hence \(\phi_{a}(\mathbb T)=\mathbb T\). $$\tag*{$\blacksquare$}$$
Proof (iv): It follows from (iii) by the maximum modulus principle.$$\tag*{$\blacksquare$}$$
Proof (v): We have \[\begin{aligned} \phi_{a}^{\prime}(z)&=\frac{(1-\overline{a} z)-(z-a)(-\overline{a})}{(1-\overline{a} z)^{2}}\\ &=\frac{1-a \overline{a}}{(1-\overline{a} z)^{2}}=\frac{1-|a|^{2}}{(1-\overline{a} z)^{2}} . \end{aligned}\]
Thus \[\phi_{a}^{\prime}(0)=1-|a|^{2},\] and \[\phi_{a}^{\prime}(a)=\frac{1-|a|^{2}}{\left(1-|a|^{2}\right)^{2}}=\frac{1}{1-|a|^{2}} .\]
This completes the proof the lemma. $$\tag*{$\blacksquare$}$$
Let \(f\) be non-constant and analytic in \(D\), and satisfy \(|f(z)|<1\) for \(z\in D\). Let \(w \in D\) with \(f(w)=a\). Then \[\left|f^{\prime}(w)\right| \leq \frac{1-|a|^{2}}{1-|w|^{2}} .\] Moreover equality occurs only when \[f=\phi_{-a} \circ\left(c \phi_{w}\right) \text { in } D,\] for some constant \(c\) whose absolute value is \(1\).
Proof: We consider \[g(z)=\phi_{a} \circ f \circ \phi_{-w}(z)\quad \text { in } \quad |z|<1.\]
Then by the previous lemma (i), (ii) and (iv), we see that \(g(z)\) is analytic in \(|z|<1\) satisfying \(|g(z)|<1\) for \(|z|<1\) and \[g(0)=\phi_{a} \circ f \circ \phi_{-w}(0)=\phi_{a} \circ f(w)=\phi_{a}(a)=0 .\]
Thus the assumptions of Schwarz lemma are satisfied and hence we conclude that \[\left|g^{\prime}(0)\right| \leq 1.\]
Now we compute \(g^{\prime}(0)\) by using the previous lemma (v): \[\begin{aligned} g^{\prime}(0) & =\left(\phi_{a} \circ f\right)^{\prime}\left(\phi_{-w}(0)\right) \phi_{-w}^{\prime}(0)\\ &=\left(1-|w|^{2}\right)\left(\phi_{a} \circ f\right)^{\prime}(w) \\ & =\left(1-|w|^{2}\right) \phi_{a}^{\prime}(f(w)) f^{\prime}(w)\\ &=\left(1-|w|^{2}\right) \phi_{a}^{\prime}(a) f^{\prime}(w) \\ & =\frac{1-|w|^{2}}{1-|a|^{2}} f^{\prime}(w). \end{aligned}\]
By \(\left|g^{\prime}(0)\right| \leq 1\), we obtain \[\left|f^{\prime}(w)\right| \leq \frac{1-|a|^{2}}{1-|w|^{2}} .\]
Suppose that we have equality above. Then \(\left|g^{\prime}(0)\right|=1\).
Now we derive from Schwarz lemma that there exists a constant \(c\) with \(|c|=1\) such that \[g(z)=c z\quad \text { for } \quad |z|<1.\]
Therefore \[\phi_{a} \circ f \circ \phi_{-w}(z)=\chi_{c}(z) \quad \text { for }\quad |z|<1,\] where \(\chi_{c}(z)=c z\). Thus \[f=\phi_{-a} \circ \chi_{c} \circ \phi_{w}=\phi_{-a} \circ (c \phi_{w}) \text { in } D. \qquad \tag*{$\blacksquare$}\]
Let \(f\) be an automorphism of \(D\) and \(w\in D\) such that \(f(w)=0\). Then \[f=c \phi_{w} \text { in } D\] where \(c\) is a constant of absolute value \(1\).
Remark. We observe that rotation is an automorphism of \(D\). On the other hand, the theorem with \(w=0\) yields that all the automorphisms of \(D\) carrying the centre to centre are given by rotations.
Proof of the theorem: Let \(h\) be the inverse of \(f\). Then \[h(f(z))=z \quad \text { for } \quad z \in D, \quad \text{ and } \quad h(0)=w.\]
By the inverse mapping theorem, \(h \in H(D)\) and \(h^{\prime}(z) \neq 0\) for \(z \in D\).
By differentiating both sides of \(h(f(z))=z\), we obtain \[h^{\prime}(f(z)) f^{\prime}(z)=1 \quad \text { for } \quad z \in D.\]
Setting \(z=w\), we have \(h^{\prime}(f(w)) f^{\prime}(w)=h^{\prime}(0) f^{\prime}(w)=1\).
Now we derive from the previous lemma with \(a=0\) that \[\left|f^{\prime}(w)\right| \leq \frac{1}{1-|w|^{2}},\qquad \left|h^{\prime}(0)\right| \leq 1-|w|^{2},\] and hence \[\left|f^{\prime}(w)\right|=\frac{1}{1-|w|^{2}}\]
By the previous lemma applied again, we conclude that \[f=\phi_{0} \circ c \phi_{w}=c \phi_{w} \text { in } D\] where \(c\) is a constant of absolute value \(1\). $$\tag*{$\blacksquare$}$$
Denote by \(G=\mathrm{SL}_{2}(\mathbb{R})\) the set of all \(2 \times 2\) matrices \[M=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\] with \(a, b, c, d \in \mathbb{R}\) such that its determinant \(a d-b c\) is equal to \(1\).
It is clear that \(G\) is a group under matrix multiplication.
For \(M \in G\), we denote \[f_{M}(z)=\frac{a z+b}{c z+d}.\]
For \(M \in G\) and \(N \in G\), we check by direct computation that
\[f_{M N}=f_{M} \circ f_{N}.\]
Let \(M \in G\). Then \(f_{M} \in \operatorname{Aut}(\mathbb{H})\).
Proof: Fix \(M \in G\), then it is clear that \(f_{M}\) is analytic in \(\mathbb{H}\) since \(-\frac{d}{c} \notin \mathbb{H}\).
Let \(z \in \mathbb{H}\) with \(z=x+i y\). Then \(y>0\) and \[\begin{aligned} \operatorname{Im}\left(f_{M}(z)\right) & =\operatorname{Im}\left(\frac{a z+b}{c z+d}\right)=\operatorname{Im}\left(\frac{a x+b+i a y}{c x+d+i c y}\right) \\ & =\operatorname{Im}\left(\frac{(a x+b+i y a)(c x+d-i y c)}{(c x+d)^{2}+c^{2} y^{2}}\right)\\ &=\frac{(c x+d) y a-(a x+b) y c}{(c x+d)^{2}+c^{2} y^{2}} \\ & =\frac{y}{(c x+d)^{2}+c^{2} y^{2}}>0, \end{aligned}\] since \(a d-b c=1\). Thus \(f_{M}\) is analytic function of \(\mathbb{H}\) into \(\mathbb{H}\).
Let \(f_{M}\left(z_{1}\right)=f_{M}\left(z_{2}\right)\). Then \((a d-b c) z_{1}=(a d-b c) z_{2}\) implying \(z_{1}=z_{2}\), since \(a d-b c =1\). Therefore \(f_{M}\) is one-to-one as desired.
We show that \(f_{M}\) is onto. Let \(N \in G\) be the inverse of \(M\). Then \[N=\left(\begin{array}{cc} d & -b \\ -c & a \end{array}\right),\]
and we check that for \(z \in \mathbb{H}\) we have \[f_{M}\left(\frac{d z-b}{-c z+a}\right)=f_{M}\left(f_{N}(z)\right)=\frac{a \frac{d z-b}{-c z+a}+b}{c \frac{d z-b}{-c z+a}+d}=z\] since \(a d-b c=1\). $$\tag*{$\blacksquare$}$$
Let \(z, w \in \mathbb{H}\). Then there exists \(M \in G\) such that \(f_{M}(z)=w\).
Proof: Let \(z=x+i y \in \mathbb{H}\), then \(y>0\). Let \(b\) and \(c \neq 0\) be real numbers to be specified later.
Set \(M=M_{2} M_{1}\), where \[M_{1}=\left(\begin{array}{cc} 0 & -c^{-1} \\ c & 0 \end{array}\right), \qquad M_{2}=\left(\begin{array}{ll} 1 & b \\ 0 & 1 \end{array}\right).\]
It is clear that \(M_{1}, M_{2} \in G\) and thus \(M \in G\).
Then, we have \[f_{M}(z)=f_{M_{2}} \circ f_{M_{1}}(z)=f_{M_{2}}\left(-\frac{1}{c^{2} z}\right).\]
We observe that \[\operatorname{Im}\left(-\frac{1}{c^{2} z}\right)=\frac{y}{c^{2}|z|^{2}}=1,\] by choosing \(c\) suitably. Further we write \[-\frac{1}{c^{2} z}=\chi+i, \quad \text{ where } \quad \chi \in \mathbb{R}.\]
Then, we also see that \[f_{M}(z)=f_{M_{2}}(\chi+i)=\chi+b+i\]
Choosing \(b=-\chi\) we obtain \[f_{M}(z)=i.\]
Let \(w \in \mathbb{H}\). Similarly, as above, there exists \(M_{3} \in G\) such that \(f_{M_{3}}(w)=i\). Let \(M_{4}=M_{3}^{-1} M\). Then \(M_{4} \in G\) and \[f_{M_{4}}(z)=f_{M_{3}^{-1}}\left(f_{M}(z)\right)=f_{M_{3}^{-1}}(i)=w. \qquad \tag*{$\blacksquare$}\]
For \(\theta\in\mathbb R\), let \[M_{\theta}=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right),\] and \[F(z)=\frac{i-z}{i+z}\quad \text { for }\quad z \in \mathbb{H}.\] Then \[F \circ f_{M_{\theta}} \circ F^{-1}(z)=e^{-2 i \theta} z \quad \text { for } \quad z \in D.\]
Proof: We know that \(F\) is a conformal mapping of \(\mathbb{H}\) onto \(D\).
For \(\theta \in \mathbb{R}\), let \[\lambda_{\theta}(z)=e^{-2 i \theta} z\quad \text { for } \quad z \in D.\]
It suffices to show that \[F \circ f_{M_{\theta}}=\lambda_{\theta} \circ F \text { in } \mathbb{H}.\]
Indeed, if \(F \circ f_{M_{\theta}}=\lambda_{\theta} \circ F\), then for \(z \in D\), we have \[\begin{aligned} F \circ f_{M_{\theta}} \circ F^{-1}(z)&=F \circ f_{M_{\theta}}\left(F^{-1}(z)\right)\\ &=\lambda_{\theta} \circ F\left(F^{-1}(z)\right)=\lambda_{\theta}(z)=e^{-2 i \theta} z. \end{aligned}\]
Now for \(z \in \mathbb{H}\) we have \[\begin{aligned} F \circ f_{M_{\theta}}(z) & =F\left(f_{M_{\theta}}(z)\right)=F\left(\frac{z \cos \theta-\sin \theta}{z \sin \theta+\cos \theta}\right)\\ &=\frac{i-\frac{z \cos \theta-\sin \theta}{z \sin \theta+\cos \theta}}{i+\frac{z \cos \theta-\sin \theta}{z \sin \theta+\cos \theta}} \\ & =\frac{z(i \sin \theta-\cos \theta)+(i \cos \theta+\sin \theta)}{z(i \sin \theta+\cos \theta)+(i \cos \theta-\sin \theta)} \\ & =\frac{-z e^{-i \theta}+i e^{-i \theta}}{z e^{i \theta}+i e^{i \theta}}=e^{-2 i \theta} \frac{i-z}{i+z}=e^{-2 i \theta} F(z).\qquad \end{aligned}\tag*{$\blacksquare$}\]
We have \[\operatorname{Aut}(\mathbb{H})=\left\{f_{M} : M \in G\right\} .\]
Proof: By Lemma (A) we have \(\{f_M: M\in G\}\subseteq \operatorname{Aut}(\mathbb{H})\) it suffices to show that \(\operatorname{Aut}(\mathbb{H}) \subseteq \{f_M: M\in G\}\).
Let \(f \in \operatorname{Aut}(\mathbb{H})\). Then there exists \(\beta \in \mathbb{H}\) such that \(f(\beta)=i\).
By Lemma (B) there exists \(N \in G\) such that \(f_{N}(i)=\beta\). We set \[g=f \circ f_{N} .\]
Then \(g(i)=f\left(f_{N}(i)\right)=f(\beta)=i\). Now we consider \(F \circ g \circ F^{-1}\), where \(F(z)=\frac{i-z}{i+z}\). We observe that \(F \circ g \circ F^{-1} \in \operatorname{Aut}(D)\) and \[F \circ g \circ F^{-1}(0)=F(g(i))= F(i)=0.\]
Hence we conclude that there exists \(\theta \in \mathbb{R}\) so that \[F \circ g \circ F^{-1}=t_{\theta} \text { in } D,\] where \[t_{\theta}(z)=e^{-2 i \theta} z \quad \text { for }\quad z \in D.\]
Further, we see from Lemma (C) that \[F \circ f_{M_{\theta}} \circ F^{-1}=t_{\theta} \text { in } D.\]
Thus we conclude that \[f \circ f_{N}=g=f_{M_{\theta}}.\]
Hence \[f=f_{M_{\theta} N^{-1}}\] and \(M_{\theta} N^{-1} \in G\). $$\tag*{$\blacksquare$}$$