currently I am trying the solve the following exercise about conformal mappings. Consider the following conformal mapping classes: $$ \mathcal{C}_1 := \{f \text{ conformal on }\mathbb D \mid f(z) = z + a_2 z^2 + a_3z^3+\dots \; \in \mathbb D\} \\ \mathcal{C}_2 := \{g \text{ conformal on } \hat{\mathbb C} - \mathbb D \mid g(z) = z+b_0+\frac{b_1}{z}+\frac{b_2}{z^2}+\dots \; \in \hat{\mathbb C} - \mathbb D\} $$ Consider $f \in \mathcal{C}_1$ then prove that $\frac{1}{f(1/z)} \in \mathcal{C}_2$ and $0 \notin g(\hat{\mathbb C} - \mathbb D$) and that one has $$ g(z) = z -a_2+\frac{a^2_2-a_3}{z}+\dots. $$
What I was able to prove so far: Consider any $f \in \mathcal{C}_1$ with $f(0)=0$ and denote $g(z):=\frac{1}{f(1/z)}$. Since $f(1/z)$ has a zero at infinity, we know that $g$ has a (simple) pole at infinity. But then $g(1/z)$ has a simple pole at $0$, hence $g(z)$ has a simple pole at infinity and since $g$ is analytic in $\hat{\mathbb{C}} - \mathbb{D} \cup \{\infty\}$ one has that the Laurent series around infinity must be given by $$ g(z) = z + b_0 + \frac{b_1}{z}+\frac{b_2}{z^2}+\dots, $$ in other words $g \in \mathcal{C}_2$. Next suppose that $0 \in g(\hat{\mathbb C} - \mathbb D)$ then there exsits some $w \in \hat{\mathbb C} - \mathbb D$ such that $g(w)=0$ but then $\frac{1}{w} \in \mathbb D$ and since $f(z) = \frac{1}{g(1/z)}$ this implies that $\lim_{z \to 1/w} |f(z)| = +\infty$ hence $1/w$ would be a pole for $f$ in $\mathbb D$ but this directly contradicts the given Laurent expansion of $f$ in $\mathbb D$ (since $a_{-1} = 0$) that is given by $\mathcal{C}_1$. But now I am stuck for the last part.
My idea: Since $|g(z) - b_0 - z| \to 0$ as $z \to \infty$ we know that $g(z)-b_0-z$ has a zero at $\infty$ but then $g(1/z) - b_0 - \frac{1}{z}$ has a zero at $0$ exactly similiar to $f$ but I his does not help, so the question is how I can fully relate the coefficients of the Laurent series to one-another although the expansions are given around different points/in different domains.
Thanks in advance!