Convergence of the sequence of inverses of automorphisms Let$\lbrace f_n \rbrace_{n\in \mathbb{N}}$ be a sequence of biholomorphisms(bijective holomorphic maps with a holomorphic inverse) on a bounded region(open connected set) $\Omega$. If this sequence converges to $f$ uniformly on every compact subset of $\Omega$, does it also follow that $\lbrace f_n^{-1} \rbrace_{n\in \mathbb{N}}$ converges uniformly? (to $f^{-1}$, if defined)
Thank you in advance.
 A: Basic Idea: You can use the integral representation of $f_n^{-1}$ to decide this question. I.e. use
$$f_n^{-1}(\omega) = \frac1{2\pi i}\int_{|\zeta-f_n^{-1}(\omega)|=r} \frac{\zeta f_n'(\zeta)}{f_n(\zeta) - \omega} \ d\zeta \label{\ast}$$
which holds for appropriate $r>0$. 

Derivation of this formula: This follows from the residue theorem: Let $\omega, n$ be fixed and let $z_0 = f_n^{-1}(\omega)$. Then $f_n(z) -\omega = (z-z_0)g(z)$ for some meromorphic function $g:\Omega \to \Omega$, which satisfies $g(z) \ne 0$ on all of $\Omega$. Thus 
$$\frac{zf_n'(z)}{f_n(z) - \omega} = z\frac{g(z) + (z-z_0)g'(z)}{(z-z_0)g(z)} = \frac{z}{z-z_0} + z\frac{g'(z)}{g(z)}$$
so we have
$$\mathrm{Res}\left(\frac{zf_n'(z)}{f_n(z) - \omega}; z_0\right) = z_0 = f_n^{-1}(\omega)$$
In particular the above holds as long as $D_r(z_0)$ is contained in $\Omega$ (we can of course take any other path in $\Omega$ going around $z_0$ exactly once!).

Answer to you question: Now let $\omega$ be arbitrary and choose $z_0$ such that $f(z_0) = \omega$. (I'm assuming that $f$ is surjective... I'll think about a proof)
Note that for all $\delta > 0$ and for sufficiently large $n$, we have that 
$$|f_n(\zeta) - f(\zeta)|< \delta$$
for all $\zeta$ on the circle $|\zeta - z_0| = r$. So we can actually make the path of integration independent of $n$, i.e. we have
$$f_n^{-1}(\omega) = \frac{1}{2\pi i}\int_{|\zeta-z_0|=r} \frac{\zeta f_n'(\zeta)}{f_n(\zeta) - \omega}  d\zeta$$
for all sufficiently large $n$.
Using this, we see that $f_n^{-1}$ converges locally uniformly to some holomorphic function $h:\Omega \to \Omega$ (consider a small enough disc around $\omega$). 
We have $z = f_n^{-1}(f_n(z))$ for all $z, \ n$. This together with locally uniform convergence implies that $$f_n^{-1}(f(z)) \to z$$ as $n \to \infty$.
But since $$f_n^{-1}(f(z)) \to h(f(z))$$ we must have $h(f(z)) = z$ ! So $h$ is a holomorphic left inverse for $f$. 
But now, let's summarize what we have shown above:
If $f_n$ is a sequence of biholomorphic functions converging to $f$ locally uniformly, then $f$ has a left inverse $h$. In particular, $h$ is surjective.
So arguing in the same way, we see that since $f_n^{-1}$ is a sequence of biholomorphic function converging to $h$ locally uniformly, $h$ has a left inverse $g$. In particular $h$ is injective.
But then $h$ is bijective and therefore $f$ must be its inverse. This shows that $f$ is itself biholomorphic.
