Assume that $D_n\subset D$, where $D$ is the unit disk is an increasing sequence of Jordan domains with smooth boundaries such that $\cup_{n=1}^\infty D_n=D$ and let $f_n: D \to D_n$ be conformal mapping such that $f(0)=0$ and $\arg(f_n'(0))=0$. Is $f_n$ uniformly convergent to $Id$.

1$\begingroup$ You mean $f_n(0)=0$ and $\mathrm{arg}f_n'(0)=0$ don't you? $\endgroup$ – Pietro Majer Nov 7 '13 at 21:51

1$\begingroup$ @PietroMajer : Surely this is what is meant. I assumed it in my answer. $\endgroup$ – Malik Younsi Nov 7 '13 at 21:51
The answer is no. Consider the following sequence $D_n$. Take the disc $z<11/n=r_n$, and remove from it the arc $\{ z:z=r_{n1},\arg z<\pi1/n\}$ and the interval $(r_{n1},r_n)$. The result is a simply connected region $D_n$. This $D_n$ contains $z<r_{n1}$, so the union of $D_n$ is the whole unit disc, and they are nested. Small deformation will make these regions Jordan, smooth, etc.
Evidently the convergence of the conformal maps $f:D\to D_n$ is uniform on compact subsets of $D$ (this follows from Caratheodory thm), but not uniform in $D$, because near the point $z=1$ the function $f_n(z)$ takes values close to every point of the unit circle. (A small neighborhood of $1$ blows up under $f_n$ to almost the whole ring $r_{n1}<z<r_n$.)

$\begingroup$ Yes you right, I understand the point. Thanks. $\endgroup$ – user36162 Nov 8 '13 at 7:40

$\begingroup$ You're right of course, I was confused and my answer was wrong. Thank you for pointing it out. Great answer, +1 ! $\endgroup$ – Malik Younsi Nov 8 '13 at 12:33