Riemann Mapping Theorem's Consequence It is well known that the Riemann mapping theorem asserts that for any open simply connected $G\subset \mathbb{C}$ and $z_{0}\in G$, there exists a unique bijective analytic function 
$f:G\to \mathbb{D}$, such that $f(z_{0})=0$ and $f^{\prime}(z_{0})>0$ (here $\mathbb{D}$ denotes the open unit disk).
I would greatly appreciate if you could help me to get an answer to the following.
Question. Supposing, in addition, that $\overline{\mathbb{C}}\setminus \overline{G}$ is simply connected and contains
the point $\infty$, how can be used the Riemann mapping theorem in order to constructed the unique conformal bijection mapping $\Phi:\overline{\mathbb{C}}\setminus \overline{G}\to \overline{\mathbb{C}}\setminus \overline{\mathbb{D}}$ which satisfies $\Phi(\infty)=\infty$ and $\Phi^{\prime}(\infty)>0$  ?
The existence of the above function $\Phi$  appears stated in many books on complex analysis but without to give any explanation to its construction (only mentioned that it  follows from the Riemann mapping theorem).
I might presume that the answer is simple, but unfortunately I cannot figure out it.
Thus, I could take $\Phi(z)=1/f(z)$, where $f:\overline{\mathbb{C}}\setminus \overline{G}\to\mathbb{D}$ is the mapping from the Riemann's theorem (since $\overline{\mathbb{C}}\setminus \overline{G}$ is open and simply connected). But, the problem is that choosing in the Riemann's theorem $z_{0}=\infty$ and $f$ satisfying $f(\infty)=0$, $f^{\prime}(\infty)>0$, we get $\Phi(\infty)=1/0=+\infty$,  but $\Phi^{\prime}(\infty)=-\frac{f^{\prime}(\infty)}{f^{2}(\infty)}=-\infty$. We were expected to get here $\Phi^{\prime}(\infty)>0$, and the questuon is how could be fixed this thing ?
Thank you in advance.
George
 A: You cannot apply the theorem as stated to $z_0=\infty$. Indeed, without too much difficulty we can observe that the conclusion that $f(\infty)=0$ and $f'(\infty)>0$ is impossible. If $f$ is holomorphic at $\infty$ (in particular, $f(\infty)$ is finite), then $f'(\infty)$ must be $0$.
Why? Well, $f$ being "holomorphic at $\infty$" really means that $z \mapsto f(1/z)$ is holomorphic at $0$. (Of course, when I say holomorphic at a point, I really mean a neighborhood of the point.) So $f(1/z) = a_0 + a_1z + a_2z^2 + \cdots$ in a neighborhood of $0$. Then in a neighborhood of infinity, $f(z) = \cdots + a_2\frac{1}{z^2} + a_1\frac{1}{z} + a_0$ . So $f'$ "at infinity" is $0$.
There is no contradiction with the Riemann mapping theorem. Uniqueness of a Riemann map $\phi$ up to $\phi(z_0) = 0$ and $\phi'(z_0)>0$ only works for $z_0 \in \mathbb{C}$, not for $z_0 = \infty$.
[Side note: This is just one reason why I believe the formulation of the Riemann mapping theorem in terms of "uniqueness up to $\phi(z_0) = 0$ and $\phi'(z_0)>0$" is a big red herring. The existence of the conformal bijection $\phi$ is the real content of the theorem. As for the uniqueness, $\phi$ is unique up to composition with automorphisms of $\mathbb{D}$. The analytic automorphisms of $\mathbb{D}$ are the maps of the form
$$
    M_{a,\theta}(z) = e^{i\theta} \frac{z-a}{\overline{a}z-1}
    \qquad (\theta \in \mathbb{R}, a \in \mathbb{D}).
$$
These form a "three parameter family" (one real parameter $\theta$, one complex parameter $a$). Specifying $\phi(z_0)=0$ and $\arg(\phi'(z_0))=0$ just happens to be one way of using up the three degrees of freedom to get a unique $\phi$ (valid for $z_0 \neq \infty$). There's nothing special about it though.]

Your problem can be solved as follows:


*

*Let $U = \overline{\mathbb{C}} \setminus \overline{G}$. Let $g(z)=\frac{1}{z-c}$, where $c$ is some point not in $U$. Let $V$ be the image of $U$ under $g$. Then $g:U \to V$ is a conformal bijection.

*Apply the version of the Riemann mapping theorem as stated at the top of your post: Let $\phi:V \to \mathbb{D}$ be the Riemann mapping with $\phi(0)=0$, $\phi'(0)>0$.

*Map $\mathbb{D}$ to $\overline{\mathbb{C}} \setminus \overline{\mathbb{D}}$ by $h(z)=1/z$.


Then $\Phi = h \circ \phi \circ g$ is a conformal bijection from $U$ to $\overline{\mathbb{C}} \setminus \overline{\mathbb{D}}$, and $\Phi(\infty)=\infty$ and $\Phi'(\infty)>0$ (computations omitted).
