A Conformal Riemann mapping problem

Question

Suppose $\{f_n\}$ is a sequence of conformal, one-to-one maps from $\mathbb D$ onto the right half plane $\Bbb A:=\{z \in \mathbb C :\mathfrak R z>0\}$. If $\{f_n\}$ converges to $f$ uniformly on compact subsets of $\mathbb D$ and $f$ is not one-to-one, find $\mathfrak Rf(0)$.

Answer 1

HINT: Call $\Bbb H:=\{z\in\Bbb C\;:\;\Im(z)>0\}$.

Consider the conformal map $g_1:\Bbb D\to\Bbb H$ defined by $$ z\mapsto\frac{z-i}{z+i}\;; $$ then $g_2:\Bbb H\to\Bbb A$ defined by $z\mapsto-iz$ is conformal too; thus composing them, we get that $h:=g_2\circ g_1:\Bbb D\to\Bbb A$ defined by $$ z\mapsto-\frac{iz+1}{z+i} $$ is conformal too.

Then a known result says that all the conformal maps $\Bbb D\to\Bbb A$ are of the form $h\circ f$ when $f\in\operatorname{Aut}(\Bbb D)$; now we know that every $f\in\operatorname{Aut}(\Bbb D)$ is of the form $$ f(z)=\lambda\frac{z-a}{\bar az-1} $$ where $|\lambda|=1$ and $a\in\Bbb D$.

Now \begin{equation} (h\circ f)(0)=-\frac{i(a\lambda)+1}{a\lambda+i}.\;\;\;\;(*) \end{equation}

Thus if $\{f_n\}_n$ is a sequence on conformal maps $\Bbb D\to\Bbb A$, whose limit (wrt compact subsets topology) is $f:\Bbb D\to\Bbb A$, then in particular you have that $$ f(0)=\lim_nf_n(0);. $$ Now, by $(*)$ you know what form $f_n(0)$ have, so try, starting from this, to understand how things work.