Unique Conformal map satisfying normalization at $\infty$ Let $H$ be the upper half plane in $\mathbb C$. Let $A$ be a compact subset of $\bar H$ with $H\setminus A$ simply connected. I am trying to prove that there is a unique biholomorphism $f : H\setminus A \rightarrow H$ such that $\lim_{z\rightarrow \infty} f(z) - z = 0$. I started by trying to prove uniqueness. If the maps were defined on all of $\mathbb C$, we would have various forms of Liouville's theorem at our disposal, as well as power series expansions about $\infty$, but neither of these are valid on the half plane, and I wasn't sure how to adapt. Does anyone have any suggestions?
 A: Uniqueness: If $f,g:\mathbb{H}\setminus A\to \mathbb{H}$ are both biholomorphic, then $\varphi=g\circ f^{-1}:\mathbb{H}\to \mathbb{H}$ is also biholomorphic, i.e. $\varphi(z)=\frac{az+b}{cz+d}$, where $a,b,c,d\in\mathbb{R}$ with $ad-bc=1$. Then from $\lim_{z\to\infty}(f(z)-z)=\lim_{z\to\infty}(\varphi\circ f(z)-z)=0$ you can conclude that $\varphi$ must be identity map.
Existence: Following the notations in $\S$ 6.5. Reflection principle of Ahlfor's Complex Analysis(third edition), denote $\Omega^+=\mathbb{H}\setminus A$, $\sigma=\partial\Omega^+\cap\partial \mathbb{H}$,  $\Omega^-=\{z\in\mathbb{C}:\bar{z}\in\Omega^+\}$ and $\Omega=\Omega^+\cup\sigma\cup\Omega^-$. 
By Riemann mapping theorem, there exists $f_0:\mathbb{H}\setminus A\to \mathbb{H}$, biholomorphic. Denote the imaginary part of $f_0$ by $v$. Since $\lim_{z\to\sigma}v(z)=0$, $v$ can be extend to a continuous function on $\Omega^+\cup\sigma$ with $v(z)=0$ on $\sigma$. Then according to Theorem 24 in $\S$ 6.5. Reflection principle of Ahlfor's Complex Analysis(third edition), $f_0$ can be extended to a holomorphic function $f_0:\Omega\to\mathbb{C}$, which satisfies that $f_0(z)=\overline{f_0(\bar{z})}$ on $\Omega^-$ and $f_0(\sigma)\subset\partial \mathbb{H}$. 
Since $A$ is compact, $\Omega\supset\{z\in\mathbb{C}:|z|>R\}$ for some $R>0$, so we can consider the Laurent expansion of $f_0$ on  $\{z\in\mathbb{C}:|z|>R\}$. Since $f_0$ is injective on $\Omega\setminus\sigma$ and $f_0(\sigma)\subset\partial \mathbb{H}$, $\infty$ cannot be an essential singularity of $f_0$. Therefore, $f_0(z)=\sum_{k=-\infty}^na_kz^k$ on $\{z\in\mathbb{C}:|z|>R\}$ for some $n\in\mathbb{Z}$ with $a_n\ne 0$. From $\lim_{z\to\infty}\frac{f_0(z)}{z^n}=a_n$ and $f_0(\Omega^+)\subset \mathbb{H}$, we know that $-1\le n\le 1$. Moreover, and if $n=1$, then $a_1>0$. In addition, if $n=0$ or $1$, from $f_0(\sigma)\subset\partial \mathbb{H}$ we know that $a_0\in\mathbb{R}$. 
Therefore, if $n=1$, we can choose $f=\frac{f_0-a_0}{a_1}$; if $n=-1$, by letting $f_1=-\frac{1}{f_0}$, it is reduced to the case $n=1$; if  $n=0$, by letting $f_1=f_0-a_0$, it is reduced to the case $n=-1$.
