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.