A Conformal Riemann mapping problem 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 $\mathfrak Rf(0)$. 
 A: 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.
