Why is the following variant of the Riemann mapping true?
Let $\infty \in U$ be a simply connected open set in the extended complex plane. Suppose moreover that at least two points in $\mathbb{C}$ is not in $U$. Then, there exists a unique pair $(V,f)$ such that $V=\mathbb{C} \cup \{\infty\} - D$ is the exterior of a closed disk $D \subseteq \mathbb{C}$ with radius $r>0$ and $f:U \rightarrow V$ is a biholomorphic map with $f(z) = z + a_1/z + a_2/z^2 + \cdots$ in a deleted neighborhood of $\infty$.
It seems to be clear that we can "invert" ($z$ to $1/z$) and translate $U$ so that we can indeed use the Riemann mapping theorem for the existence part. But how can we guarantee the uniqueness part and the asserted Laurent expansion?
Consider the simply connected $0 \in W$ given as the image of $U$ under $z \to \frac{1}{z}$ minus possibly infinity (if $0 \in U$); by our assumptions (implying that $W$ is not the full plane) and the RMT there is a unique conformal mapping from $W$ onto the unit disc satisfying $g(0)=0, g'(0)>0$.
Scaling we get a unique disc $D_1$ containing zero and a unique Riemann map from $W$ onto $D_1$ for which $g_1(0)=0, g_1'(0)=1, g_1=z+b_1z^2+...$, so $g_1(\frac{1}{z})=\frac{1}{z}(1+b_1\frac{1}{z}+...)$.
Taking $f_1(z)=\frac{1}{g_1({\frac{1}{z}})}=z(1+b_1\frac{1}{z}+...)^{-1}$, and expanding the Taylor series in $\frac{1}{z}$ of the fraction above, which is obviosuly $1-b_1\frac{1}{z}+a_1\frac{1}{z^2}+..$, for some $a_1, a_2,...$ depending uniquely on the original $b_1, b_2,...$, we almost get the required map above with $D=D_1$; however we may have a free non-zero term $a_0=-b_1$, so $f_1$ may look like $z-b_1+a_1/z + a_2/z^2 + \cdots$.
In order to get rid of that, we just translate, so $f=f_1+b_1$ is the required map and similarly, $D$ is obtained from $D_1$ by translation while the radius is unchanged and the unicity is preserved since $b_1$ is unique due to the unicity of the original Riemann map.