Use that every element of \(\mathop{\mathrm{Aut}}({\mathbb{D}})\) is of the form \(f(z) = \lambda\psi_a(z)\), and the region \(G\) is conformally equivalent to \({\mathbb{D}}\). Thus every element of \(\mathop{\mathrm{Aut}}(G)\) can be conjugated to an element of \(\mathop{\mathrm{Aut}}({\mathbb{D}})\) by a conformal map \(F: G\to {\mathbb{D}}\). One such map is gotten by rotating \(z\mapsto iz\) to get \({\mathbb{H}}\cap{\left\lvert {z} \right\rvert}>1\), then applying a Joukowski map \(z\mapsto z+z^{-1}\) to get \({\mathbb{H}}\). So \begin{align*} f\in \mathop{\mathrm{Aut}}(G) \implies f = F^{-1}\circ \psi_{a} \circ F \text{ for some } \psi_a \in \mathop{\mathrm{Aut}}({\mathbb{D}}) .\end{align*}

Part 1: this is a bigon with vertices \(0, \infty\), so send \(0\to\infty\) with \(1/z\). Orient \(i{\mathbf{R}}\) and the circle \(S\) positively, note that both will be mapped to generalized circles. To find the resulting region, use handedness – it’s on the right of \(i{\mathbf{R}}\) and the right of \(S\). The map preserves \(i{\mathbf{R}}\) and as \(t\) traces out \((-\infty, 0^-, 0^+, \infty)\), \(f(it)\) traces out \((0^+, \infty, -\infty, 0^-)\), so this preserves the orientation of \(i{\mathbf{R}}\). For \(S\), let \(z_0\coloneqq{1\over 2}(1+i)\in S\), then \(f(z_0) = 1-i\). So the arc \((1,z_0, 0)\) maps to \((1, 1-i,\infty)\), so this is a vertical line through \(\Re(z) = 1\) oriented downward. The region is to the right of \(S\), so we have

The rest is standard:

• Dilate and rotate to \(0<\Im(z) < \pi\) using \(z\mapsto i\pi z\).
• Exponentiate using \(z\mapsto e^z\)

  \to 

  get \({\mathbb{H}}\).
• Apply the Cayley map \(z\mapsto {z-i\over z+i}\) to get \({\mathbb{D}}\).

Part 2: a bigon with vertex \(1\), i.e. a lune. Send \(1\to \infty\) with \(f(z) \coloneqq{1\over z-1}\), and check that

• \(1\mapsto \infty\)
• \({1\over 2}(1+i) \mapsto -(1+i)\)
• \(0\mapsto -1\)
• \(i\mapsto -{1\over 2}(1+i)\)
• \(-1\mapsto -{1\over 2}\)

By tracking tangent/normal vectors, this results in the region \(-1<\Re(z) < -{1\over 2}\):

The rest is standard:

• Translate to the right by \(z\mapsto z+{1\over 2}\) to get \(-{1\over 2}<\Re(z) < 0\).
• Rotate and dilate by \(z\mapsto -2i\pi z\) to get \(0<\Im(z) < \pi\)
• Exponentiate by \(z\mapsto e^z\) to get \({\mathbb{H}}\),
• Cayley map \(z\mapsto {z-i\over z+i}\) to get \({\mathbb{D}}\).

Part 3: a bigon in \({\mathbb{H}}\) with vertices \(\pm 1\), with an arc passing through \(z_3 \coloneqq i(\sqrt{2} - 1)\). Take \(z\mapsto {z+1\over z-1}\) to obtain

• \(-1\mapsto 0\)
• \(1\mapsto \infty\)
• \(0\mapsto -1\)

Orienting the bigon positively, we have \((-1, 0, 1)\mapsto (0, -1, \infty)\), i.e. the real axis oriented from \(+\infty\to-\infty\). Similarly \((1, z_3, -1)\mapsto (\infty, \omega_4^3, 0)\), which is a line passing through \(\omega_4^3\), oriented from \(Q_3\to Q_1\). Since the original region was on the left of both curves, we get

Now

• Flip this to \(Q_1\) with \(z\mapsto -z\) to get \(0<\operatorname{Arg}(z) < \pi/4\).
• Rotate clockwise with \(z\mapsto e^{-i\pi\over 8}\) to get \(-\pi/8<\operatorname{Arg}(z) < \pi/8\).
• Dilate the argument to a half-plane with \(z\mapsto z^{\pi\over 2\theta_0}\) where \(\theta_0 = \pi/8\) to get \(-\pi/2<\operatorname{Arg}(z) < \pi/2\).
• Rotate with \(z\mapsto iz\) to get \({\mathbb{H}}\).
• Cayley map, \(z\mapsto {z-i\over z+i}\).

Part 4: See part 5. The critical step is a Blaschke map \(\psi_a\) which sends \(a\to 0\). For \(a\in {\mathbf{R}}\), \(\psi_a({\mathbf{R}}) = {\mathbf{R}}\) and this will map the partial slit from \(a\) to the boundary to a usual slit from \(0\) to the boundary.

Part 5: Dealing with the slit:

• Use a Blaschke factor to send \(a\coloneqq-1/2\to 0\), so \(z\mapsto {a-z\over 1-\overline{a} z}\). Checking that \((-1/2, 0, 1)\to (0, -1/2, -1)\), the image is \({\mathbb{D}}\setminus(-1, 0]\).
• Rotate with \(z\mapsto e^{-i\pi}z\) to get \({\mathbb{D}}\setminus[0, 1)\).
• Unfold with \(z\mapsto z^{1\over 2}\) to get \({\mathbb{D}}\cap{\mathbb{H}}\), noting that the slit becomes \([-1, 1]\) and is erased here.
• Use \(z\mapsto -1/z\) to get \({\mathbb{D}}^c \cap{\mathbb{H}}\).
• Use the Joukowski map \(z\mapsto z+z^{-1}\) to map to \(Q_{34}\)
• Use \(z\mapsto -z\) to get \({\mathbb{H}}\).

It’s well known that \(z\mapsto \sin(z)\) sends \(-\pi/2<\Re(z) < \pi/2\) with \(\Im(z) > 0\) to \({\mathbb{H}}\):

So take \(z\mapsto \arcsin(z)\).

This is a lune-type region:

The usual strategy is to blow up the tangency, so send \(1\to\infty\) with \begin{align*} f(z) \coloneqq{1\over z-1} .\end{align*}

From here, it’s a standard exercise. In steps:

• Map \(R\) to the vertical strip \(-1< \Re(z) < -{1\over 2}\) using \(z\mapsto {1\over z-1}\).
• Shift using \(z\mapsto z+{1\over 2}\) to send this to \(-{1\over 2}< \Re(z) < 0\).
• Rotate using \(z\mapsto -iz\) to get \(0<\Im(z) < {i\over 2}\), a horizontal strip.
• Dilate using \(z\mapsto 2\pi z\), which sends \({i\over 2}\to \pi i\), so the resulting region is \(0 < \Im(z) < \pi\).
• Apply \(z\mapsto e^z\) to map the horizontal strip to \({\mathbb{H}}\).
• Apply the Cayley map \(z\mapsto {z-i\over z+i}\) to map \({\mathbb{H}}\to{\mathbb{D}}\).

The region:

Note that you can find that \(i, -i\) are the intersection points by noting that \(i{\mathbf{R}}\) is the perpendicular bisector through the line segment connecting the centers of the circles, then expanding \({\left\lvert {z-1} \right\rvert}^2 = (x-1)^2 + y^2 = 2\) and setting \(x=0\) to get \(y=\pm i\).

First rotate this by \(\pi/2\):

Call the upper circle \(C_1\) and the lower \(C_2\). Send \(-1\to 0, 1\to \infty\) by taking \begin{align*} f(z) \coloneqq{z+1\over z-1} .\end{align*} This sends the circles to lines through zero, and the lune now spans a triangular sector. Finding the angles of the lines: write \(c\coloneqq 1 + \sqrt{2}\). Note that \(f\) fixes \({\mathbf{R}}\), so the image regions are symmetric about \({\mathbf{R}}\), and it suffices to find the angle of the line \(f(C_1)\). Note that \(C_1 \cap i{\mathbf{R}}= ic\), so we compute \(\operatorname{Arg}(f(ic))\): \begin{align*} f(ic) &= {ic+1\over ic-1} \\ &= {(1+ic)^2 \over c^2 + 1} \\ &= {-1 + c^2 - 2ic \over c^2 + 1} \\ &= {c^2-1 \over c^2+1} -i {2c\over c^2+1} ,\end{align*} so \begin{align*} \operatorname{Arg}(f(ic)) &= \arctan\qty{ 2c\over -1 + c^2} \\ &= \arctan\qty{2(1+\sqrt 2) \over 1 - (3 + \sqrt 2)} \\ &= \arctan(-1) \\ &= {\pi\over 4} \text{ or } {3\pi \over 4} .\end{align*}

Thus \(f(C_1) = \left\{{te^{-i\pi\over 4} {~\mathrel{\Big\vert}~}t\in (-\infty, \infty)}\right\}\). Note that \(f(ic)\in Q_4\), since \(c^2-1, c^2+1 > 0\) and \({-2c\over c^2+1} < 0\). For the orientation of \(f(C_1)\), note that \((1, ic, -1) \mapsto (\infty, f(ic), 0)\), so the line is oriented from \(Q_4\) to \(Q_2\).

A similar computation shows \begin{align*} f(-ic) = {c^2-1\over c^2+1} + i{2\over c^2+1} \in Q_1 ,\end{align*} and \((-1, -ic, 1)\mapsto (0, f(-ic), \infty)\), so \(f(C_2)\) is oriented from \(Q_3\) to \(Q_1\).

Since the origin region was to the left of the curves, it remains to the left, so the resulting region is \(\left\{{z{~\mathrel{\Big\vert}~}3\pi/4 < \operatorname{Arg}(z) < 5\pi/4}\right\}\):

From here, it’s a standard exercise, so to sum up:

• Rotate \(R\to \tilde R\) by \(z\mapsto iz\) to get a horizontal lune with intersection points \(\pm 1\).
• Send \(-1\to \infty, 1\to 0\) by \(z\mapsto {z+1\over z-1}\) to send \(\tilde R \to 3\pi/4 < \operatorname{Arg}(z) < 5\pi/4\).
• Reflect by \(z\mapsto -z\) to get \(-\pi/4 < \operatorname{Arg}(z) < \pi/4\).
• “Open” by \(z\mapsto z^{\pi \over 2 \theta_0}\) to map to \(-\pi/2<\operatorname{Arg}(z) < \pi/2\), where here \(\theta_0 \coloneqq{\pi \over 4}\).
• Rotate by \(z\mapsto iz\) to get \(0 < \operatorname{Arg}(z) < {\pi \over 2}\), i.e. \({\mathbb{H}}\).
• Use the Cayley map \(z\mapsto {z-i\over z+i}\) to send \({\mathbb{H}}\to {\mathbb{D}}\).

The main step: blow up the tangency.

The individual maps:

• Send \(0\to \infty\) by \(z\mapsto {1\over z}\) to map \(R\) to \(0<\Re(z)<1\)
• Rotate with \(z\mapsto iz\) to map this to \(0< \Im(z) < 1\)
• Dilate by \(z\mapsto \pi z\) to get \(0 < \Im(z) < \pi\)
• Apply \(z\mapsto e^z\) to map to \({\mathbb{H}}\).

That steps 1 and 2 work requires a bit of analysis. Use that \(f(z) \coloneqq 1/z\) satisfies \(f({\mathbf{R}}) = {\mathbf{R}}\) and \(f(i{\mathbf{R}}) = i {\mathbf{R}}\). To see where the circle \(C_1\) gets mapped to, parameterize it as \begin{align*} \gamma(t) \coloneqq\left\{{{1\over 2}\qty{1+e^{it}} {~\mathrel{\Big\vert}~}t\in [-\pi, \pi]}\right\} .\end{align*}

Now computing its image: \begin{align*} f(\gamma(t)) &= {2 \over 1 + e^{it}} \\ &= {2e^{-it\over 2} \over e^{-it\over 2}(1 + e^{it})} \\ &= {2e^{-it} \over e^{-it\over 2} + e^{it\over 2} } \\ &= { 2e^{-it\over 2} \over 2\cos\qty{t\over 2} } \\ &= \sec\qty{t\over 2}\qty{\cos\qty{-t\over 2} + i\sin\qty{-t\over 2}} \\ &= \sec\qty{t\over 2}\qty{\cos\qty{t\over 2} - i\sin\qty{t\over 2}} \\ &= 1 - i\tan\qty{t\over 2} ,\end{align*} and as \(t\) runs from \(-\pi\) to \(\pi\), \(-\tan\qty{t\over 2}\) runs from \(+\infty\) to \(-\infty\). So this is a line along \(z=1\), oriented top to bottom as in the image.

Similarly, computing \(f(i{\mathbf{R}})\): parameterize \(\gamma(t) = it\) with \(t\in (-\infty, \infty)\), then \begin{align*} f(\gamma(t)) = {1\over it} = -{i\over t} ,\end{align*} and as \(t\) runs from \(0\) to \(\infty\), \(-1/t\) runs from \(-\infty\) to \(0\), and as \(t\) runs from \(-\infty\) to \(0\), \(-1/t\) runs from \(0\) to \(\infty\). So \(f(i{\mathbf{R}})\) is oriented from bottom to top, as in the image.

That the region outside the disc is mapped to the strip shown: points \(x\in {\mathbf{R}}\) with \({\left\lvert {x} \right\rvert} > 1\) map to \({\left\lvert {x} \right\rvert}<1\), which is in the strip. One can also conclude this by handedness: the original region is on the right with respect to \(C_1\) and also on the right with respect to \(i{\mathbf{R}}\), so the new region should be on the right with respect to both \(f(i{\mathbf{R}})\) and \(f(C_1)\) with their induced orientations.

This is a lune with a single intersection vertex at \(z=i\). Orient the circles positively.

• Take \(f(z) = {z+i\over z-i}\) to send

  • \(i\to \infty\)
  • \(-i\to 0\)
  • \(1\to {1+i\over 1-i} = i\)
  • \(0\to -1\)

  So \({\left\lvert {z} \right\rvert} = 1\) is sent to the imaginary axis \(\left\{{it}\right\}\) for \(t\in (-\infty, \infty)\) oriented positively and \({\left\lvert {z- {i\over 2}} \right\rvert} = {1\over 2}\) is sent to \(\left\{{-1 + it}\right\}\) also oriented positively. The region then maps to \(-1 < \Re(z) < 0\).

• Rotate by \(z\mapsto -i\pi z\) to get \(0 < \Im(z) < \pi\).

• Take \(z\mapsto e^z\) to get \({\mathbb{H}}\).

• Take \(z\mapsto {z-i\over z+i}\) to get \({\mathbb{D}}\).

Note: this seems unusually difficult for a UGA question! This is a bigon with one vertex \(z_2 = 0\), so send it to infinity and keep track of the slit.

In steps:

• Use \(z\mapsto 1/z\) to get a vertical strip \(0<\Re(z) < 1\), where the slit maps to \([1/2, 1]\).

• Shift and dilate with \(z\mapsto \pi(z-1/2)\) to get \(-\pi/2<\Re(z) < \pi/2\), where the slit is now at \([0, \pi/2]\).
• Apply \(z\mapsto \sin(z)\) to get \({\mathbf{C}}\setminus[0, 1]\).
• Move the slit with \(z\mapsto 4(z-1/2)\) to get \({\mathbf{C}}\setminus[-2, 2]\).
• Apply \(z\mapsto z+z^{-1}\) to get \({\mathbf{C}}\setminus{\mathbb{D}}\), where the slit is now eliminated.
• Use \(z\mapsto 1/z\) to get \({\mathbb{D}}\)
• Use the inverse Cayley map \(z\mapsto i{1-z\over 1+z}\) to get \({\mathbb{H}}\).

The region looks like the following:

Following the general strategy for lunar regions, send the intersection points to \(0\) and \(\infty\) to get triangular sector. So choose to send \(0\to 0\) and \(1\to \infty\) by taking \begin{align*} f(z) \coloneqq{z\over z-1} .\end{align*}

Note: mistake here, really we need to compose with \(z\mapsto -z\) to get the picture, so take \(f(z) \coloneqq{z\over 1-z}\) instead!!

From here it is easy to map to the disc:

• \(z\mapsto {z\over z-1}\) sends \(R\) to \({\left\lvert {\operatorname{Arg}(z)} \right\rvert} < \theta_0\)
• \(z\mapsto z^{\pi \over 2\theta_0}\) maps \({\left\lvert {\operatorname{Arg}(z)} \right\rvert}<\theta_0 \to {\left\lvert {\operatorname{Arg}(z)} \right\rvert} < {\pi \over 2}\), the right half-plane.
• \(z\mapsto iz\) rotates the right half-plane into \({\mathbb{H}}\).
• \(z\mapsto {z-i\over z+i}\) maps \({\mathbb{H}}\to {\mathbb{D}}\).

Part a: That \(f: {\mathbf{C}}\setminus\left\{{0}\right\}\to {\mathbb{D}}\setminus\left\{{0}\right\}\) is injective: compute the derivative as \begin{align*} f'(z) = {1\over 2}\qty{1 - {1\over z^2}} ,\end{align*} which only vanishes at \(z=\pm 1\). Away from 0 in \({\mathbb{D}}\), \(f'\) is nonzero and continuous, so by the inverse function theorem \(f\) is a local homeomorphism onto its image, and in particular is injective.

The images of circles: parameterize one as \(\gamma(t) = Re^{it}\) for \(t\in [-\pi, \pi]\). Note that if \(R=1\), \(f(\gamma(t)) = {1\over 2}\qty{e^{it} + e^{-it}} = \cos(t)\), so as \(t\) increases from \(-\pi\to \pi\), the interval \([-1, 1]\) is covered twice. For \(0<R<1\), \begin{align*} f(\gamma(t)) &= {1\over 2}\qty{ Re^{it} + {1\over Re^{it}}} \\ &= {1\over 2}\qty{Re^{it} + R^{-1}e^{-it} } \\ &= {1\over 2}\qty{R\cos(t) + iR\sin(t) + R^{-1}\cos(-t) + iR^{-1}\sin(-t) } \\ &= {1\over 2}\qty{R\cos(t) + iR\sin(t) + R^{-1}\cos(t) - iR^{-1}\sin(-t) } \\ &= {1\over 2}\qty{R+R^{-1}}\cos(t) + i{1\over 2}\qty{R-R^{-1}}\sin(t) \\ &\coloneqq H_R \cos(t) + iV_R\sin(t) ,\end{align*} which is generally the equation of an ellipse of horizontal radius \(H_R\) and vertical radius \(V_R\). As \(R\) varies, these sweep out ellipses of vertical radii from 0 to \(\infty\). One can compute the foci: their distance from \(z=0\) is given by \(c\), where \begin{align*} c^2 = H_R^2 - V_R^2 = {1\over 4}(R+R^{-1})^2 - {1\over 4}(R-R^{-1})^2 = 1 ,\end{align*} so the foci are all at \(\pm 1\in {\mathbf{R}}\). One can check that these are clockwise when \(0<R<1\) and counterclockwise when \(R>1\).

In general, you take the coefficient for the major axis squared minus that of the minor axis squared. The foci are along the major axis.

Part b: The claim is that \(f({\mathbf{C}}\setminus{\mathbb{D}}) = {\mathbf{C}}\setminus[-1, 1]\).

Note that \(f(z) = f(1/z)\), so for \(z\neq 1/z\) there are exactly two preimages. These points are exactly \(z=\pm 1\), so we need to take the domain \(\Omega \coloneqq{\mathbf{C}}\setminus({\mathbb{D}}\cup\left\{{\pm 1}\right\})\) to get injectivity. Otherwise, for every \(z\in \Omega\), exactly one of \(z\) or \(1/z\) is in \({\mathbb{D}}\), so \(f(z)\) takes on unique values in \(\Omega\). By part 1, the images of circles of radius \(R\) are ellipses, and these sweep out the entire plane outside of \([-1, 1]\):

To be explicit, one can just solve for the two preimages. Setting \(w=f(z)\) and solving for \(z\) yields \begin{align*} w &= {1\over 2}\qty{z + {1\over z}} \\ \implies 2wz &= z^2 + 1 \\ \implies z^2 - 2wz + 1 &= 0 \\ \implies z &= {2w \pm \sqrt{4w^2 -4} \over 2} \\ \implies z &= w\pm \sqrt{w^2-1} ,\end{align*} where in order to define \(\sqrt{w^2-1}\), one needs a branch cut for \(\operatorname{Log}\) along \([-1, 1]\), which is precisely what we’re deleting from the image.

Part c: as in a usual conformal map problem, find a map to \({\mathbf{C}}\setminus[-1,1]\to {\mathbb{D}}\).

• Send \(-1\to 0\) and \(1\to \infty\) with \(z\mapsto {z+1\over z-1}\). Checking that \(f(0) = -1\), this yields \({\mathbf{C}}\setminus{\mathbf{R}}_{\leq 0}\).

• Unwrap with \(z\mapsto \sqrt{z}\) to obtain the right half-plane \(-\pi/2<\operatorname{Arg}(z) < \pi/2\).

• Apply the rotated Cayley map \(z\mapsto {z-1\over z+1}\) to map this to \({\mathbb{D}}\).

This composes to \begin{align*} g(z) &= {\sqrt{z+1\over z-1} -1 \over \sqrt{z+1\over z-1} +1} \\ &= { \qty{ \sqrt{z+1} - \sqrt{z-1}}^2 \over (z+1)-(z-1) } \\ &= z - \sqrt{z^2-1} ,\end{align*} and checking that it has the right inverse: \begin{align*} w &= z-\sqrt{z^2-1} \\ \implies (z-w)^2 &= z^2-1 \\ \implies w^2+1 &= 2wz \\ \implies z = {1\over 2}\qty{w + {1\over w}} .\end{align*}

Consider the images of arcs \(\gamma_R(t) \coloneqq Re^{it}\) for \(t\in [0, \pi]\) and \(0<R<1\), which fill out the upper half disc: \begin{align*} f(Re^{it}) = - {1\over 2}\qty{ \qty{R+R^{-1}}\cos(t) + i(R-R^{-1})\sin(t) } \coloneqq H_R\cos(t) + V_R\sin(t) ,\end{align*} where \(H_R \coloneqq-{1\over 2}{R+R^{-1}}\) and \(V_R \coloneqq-{1\over 2}(R-R^{-1})\). Note that \(H_R\in (-\infty, -1]\) and \(V_R \in (0, \infty)\) – in particular, since \(V_R>0\), as \(t\) ranges through \((0, \pi)\), this traces out the top half of an ellipse (noting that \(V_R<0\) would trace out the bottom).

As \(R\) ranges from \((0, 1)\), \(V_R\) ranges from \((0, \infty)\), so this traces out all such elliptic arcs in \({\mathbb{H}}\) passing through \(H_R <0, iV_R \in {\mathbb{H}}, -H_R>0\). So these trace out all of \({\mathbb{H}}\).

In steps:

• Unfold with \(z\mapsto z^2\) to get \({\mathbb{D}}\cap{\mathbb{H}}\).
• Joukowski it with \(z\mapsto -{1\over 2}(z+z^{-1})\) to get \({\mathbb{H}}\).
• Fold with \(z\mapsto z^{1\over 2}\) to get \(Q_1 = A\).

Main idea: both circles can be uniformized in ways that send \(z_2, z_2'\) to zero. Handling \(C\):

• \(f_1\): Uniformize by translating and scaling \(C\) to \(S^1\). Note that the image of \(z_1\) is in \(S^1\)
• \(f_2\): if \(z_2\) was not enclosed by \(C\), apply \(z\mapsto 1/z\) to move \(z_2\) into \({\mathbb{D}}\). Otherwise just take \(f_2 = \operatorname{id}\). In either case, \(z_1\in S^1\) is preserved.
• \(f_3\): apply a Blaschke factor \(\psi_a(z)\) to move \(z_2\to 0\) in \({\mathbb{D}}\). Since \(\psi_a\) preserves \(S^1\), \(z_1\) is moved to another point in \(S^1\).

Define \(F\coloneqq f_3 \circ f_2 \circ f_1\). Similarly define \(g_1,\cdots, g_3\) and \(G\) for \(C'\), so \(z_2'\to 0\) in \({\mathbb{D}}\). Now \(F(z_1), G(z_1')\in S^1\), so there is a rotation \(h: z\mapsto \lambda z\) that sends \(F(z_1)\to G(z_1')\).

Take the final map to be \(f\coloneqq G\circ h \circ F^{-1}\).

We have \(f: \Delta^* \to \left\{{\Re(z) > 0}\right\}\), so let \(T: \left\{{\Re(z) > 0}\right\}\to \Delta\) be the rotated Cayley map \(T(z) = {z-1\over z+1}\). Then \(G\coloneqq T^{-1}\circ f: \Delta^* \to \Delta\), and since \({\left\lvert {T} \right\rvert} < 1\), \(z=0\) is a removable singularity for \(F\) and \(F\) extends holomorphically to \(G:\Delta\to {\mathbf{C}}\), since a priori \(G(0)\) may not be bounded by 1. Supposing \(G\) is not constant, \(G(\Delta) \subseteq \overline{\Delta}\) by continuity and \(G(\Delta)\) is open by the open mapping theorem, and \(\Delta = f(\Delta) \subseteq G(\Delta)\), so \(G:\Delta\to \Delta\) is a map of the disc. Then \(F \coloneqq T\circ G: \Delta\to \left\{{\Re(z) > 0}\right\}\) is an extension of \(F\) which is bounded in neighborhoods of \(z=0\), making zero a removable singularity for \(f\).

Write \(T:{\mathbf{C}}\to {\mathbb{D}}\) for the Cayley map, then \(F\coloneqq f\circ T\) satisfies \(F({\mathbf{C}}) = T(f({\mathbf{C}})) \subseteq T({\mathbb{H}}) = {\mathbb{D}}\), so \(F\) is a bounded entire function and thus constant. So \(c = F(z) = T(f(z)) \implies f(z) = T^{-1}(c)\), making \(f\) constant.