Let $f$ be schlicht, (univalent on the unit disc with power series $f(z)= z + \sum_{n=2}^ \infty a_n z^n$ ) and let $g(z)= [f(z^{-1})]^{-1}$ I am trying to find the Laurent series of $g$ for all $z$.
My attempt is as follows: \begin{align*} g(z)&=\left[ z^{-1} + \sum_{n=2}^ \infty a_n z^{-n} \right]^{-1} =\frac{z}{1+ \sum_{n=2}^ \infty a_n z^{1-n}} \end{align*}
And using that for $\omega$ with modulus $|\omega|< 1$ $$\frac{1}{1+\omega}=\sum_{k=0}^\infty (-\omega)^k,$$
we have that if $\left| \sum_{n=2}^ \infty a_n z^{1-n} \right|<1$ then
\begin{align*} g(z)&=z \left( \sum_{k=0}^\infty \left(-\sum_{n=2}^ \infty a_n z^{1-n} \right)^k \right)\\ &=z\left(1-a_2z^{-1}-a_3z^{-2}+a_2^2z^{-2} + \cdots \right)\\ &=z - a_2 + \left( a_2^2-a_3 \right)z^{-1} + \cdots \end{align*}
I know this is the correct power series, as it is given in a book, but I don't see how I can assume that $\left| \sum_{n=2}^ \infty a_n z^{1-n} \right|<1$.
If you need more context, I am trying to prove The Bieberbach conjecture for the coefficient $a_2$.
Answer compiled from the comments by Daniel Fischer.
Since $$h(z) = \sum_{n=2}^\infty a_n z^{1-n}$$ has a zero in $\infty$, you have $\lvert h(z)\rvert < 1$ for all large enough $\lvert z\rvert$.
Note that although the Laurent series of $g$ converges (at least) on $\{ z : \lvert z\rvert > 1\}$, the derivation of the Laurent series by expanding $\frac{1}{1+h(z)}$ into a geometric series is usually valid only in a smaller domain $\{ z : \lvert z\rvert > R\}$, where $R$ may be large.
Concerning the convergence of the Laurent series for $g$:
If a function $f$ is holomorphic on an annulus $K(a;r,R) = \{ z : r < \lvert z-a\rvert < R\}$, where $0 \leqslant r < R \leqslant \infty$, then $f$ has a Laurent series expansion $$f(z) = \sum_{n=-\infty}^\infty c_n (z-a)^n$$ converging (locally uniformly) on the annulus $K(a;r,R)$. The coefficients are given by $$c_n = \frac{1}{2\pi i}\int_{\lvert z-a\rvert = \rho} \frac{f(z)}{(z-a)^{n+1}}\,dz,$$ where $r < \rho < R$ is arbitrary. If $f$ is also holomorphic on an annulus $K(a;r_1,R_1)$ with $r_1 < r,\; R < R_1$, then the Laurent series representing $f$ on that larger annulus has the same coefficients, i.e., it is the same series. Thus this Laurent series converges in the largest annulus centered at $a$ containing the circle $\lvert z-a\rvert = \rho$. Since $g$ is holomorphic on $\{ z : 1 < \lvert z\rvert < \infty\}$, the Laurent series representing $g$ on $R < \lvert z\rvert < \infty$ for any $R < \infty$ converges (at least) on the complement of the closed unit disk.