Suppose we have a Jordan curve $\gamma$ in the plane. Let $\Omega_0$ be the bounded component of $\mathbb{C}\backslash\gamma$ and let $\Omega_{\infty}$ denote the unbounded component. We assume $0\in \Omega_0$. Let $f(z): \mathbb{D}\rightarrow \Omega_0$ be the conformal map taking the unit disk to $\Omega_0$ with $f(0)=0$ and $f'(0)>0$. We write $f(z)=A(z+a_1z^2+a_2z^3+\cdots)$.
Similarly, we can take a uniformization $g: \mathbb{D}^{\ast}\rightarrow \Omega_{\infty}$ of the disk $\mathbb{D}^{\ast}=\{|\frac{1}{z}|<1\}$ with the property $g(\infty)=\infty$ and $g'(\infty)>0$. We write $g(z)=B(z+b_1+b_2z^{-1}+\cdots)$.
In principle, $f(z)$ characterizes $\Omega_{0}$ and therefore $\mathbb{C}\backslash\Omega_0$. My question involves relations between the coefficient $\{A,a_k\}$ and $\{B, b_k\}$. For example, the bounded quantity $\frac{A}{B}$ can be written in terms of $\{a_k, \bar{a}_k, b_k, \bar{b}_k\}$. My question is the following: Given $n\geq 1$, is it possible to relate $b_n$ to $\{A, a_k\}$ ? Or perhaps some type of homogeneous polynomials in ${b_1, \ldots, b_n}$ to $\{A, a_k\}$ ? Really, ANY relation between the coefficients would be helpful.
Notational remark: it's a bad idea to write series like $z+a_1z^2+\dots$ Having normalized the $z$-coefficient, you can still write $z+a_2z^2+\dots$ without being accused of wasting a scarce resource (taxpayer-funded indices). Chances are that it will be convenient to use $a_1=1$ in the computations.
I take it that you know the area formula, but for the sake of completeness I put it here too: the area of the image of conformal map $f(z)=\sum_{n=0}^\infty c_n z^n$, defined on the unit disk, is
$$ \pi\,\sum_{n=1}^\infty n|c_n|^2 $$ while the area of the complement of image of conformal map $f(z)=\sum_{n=-\infty}^{1} d_n z^n$, defined on the complement of the unit disk, is $$ \pi\,\sum_{n=-\infty}^{1} n|d_n|^2 $$ In our situation both areas are the same, so we have a relation between coefficients.
And that's about it. The relation between individual Taylor coefficients of a conformal map and the geometry of its image remains inscrutable. To show you why there is no homogeneous polynomial or any reasonable function that gives $b_n$, I offer the following:
Claim. $b_n$ is not a continuous function of $a_1,\dots,a_m$ for any $m$.
Proof. Let $\Omega$ be the open $\epsilon$-neighborhood of the set $\{|z|\le 1\}\cup [1,2]$. As $\epsilon\to 0$, the normalized conformal maps onto $\Omega$ converge to the identity map, by the Carathéodory kernel theorem. In particular, the coefficients converge. By the same theorem, conformal maps onto the complement of $\Omega$ converge to a conformal map onto the complement of $\{|z|\le 1\}\cup [1,2]$. Which is of course rather different from the map onto the complement of the unit disk.