Conformal map and Jordan curve Here is my question :
Suppose you have a simple (analytic) closed curve $\gamma$ in an open simply connected domain $\Omega \neq \mathbb{C}$. Does there exist a conformal bijection $f : \Omega \rightarrow U \subset \mathbb{C}$, such that $\gamma$ is sent to the unit circle $S^1$ (the unit disc $D$ would then be contained in $U$) ?
The Riemann mapping theorem tells you that you can find a conformal map from the interior of $\gamma$ to $D$, and the conformal geometry of an annulus tells you that you can also find a conformal map from the annulus $\Omega - int(\gamma)$ to an annulus $A(r_1,r_2) = \{z \in \mathbb{C} | r_1 < |z| < r_2 \}$. But this does not answer the question...
Note that in the question I don't ask for the open set $U$ to be bounded by a circle (this would clearly be too restrictive). 
This question is somehow related to a previous question I asked, but I think this one is quite different though. The answer must be well-know, but I can not find it anywhere - neither from myself.
 A: If $G\neq\mathbb{C}$ is a s.c. domain, then a conformal map to a disc extends analytically beyond the boundary if and only if $G$ is bounded by an analytic Jordan curve. In particular, if you were to let your domain $\Omega$ depend on the curve $\gamma$, then the answer would be positive.
However, as stated, the answer is negative. Indeed, let $\Omega$ be a simply-connected domain; we shall construct an analytic curve $\gamma$ such that no conformal isomorphism of the interior of $\gamma$ to a disc extends to all of $\Omega$.
Let $G$ be any bounded simply-connected domain whose closure is contained in $\Omega$, such that $G$ has non-analytic boundary. Let $\phi$ be a conformal isomorphism to the unit disc $\mathbb{D}$ and let $\gamma$ be the preimage of a round circle under $\phi$; say
$$\gamma := \phi^{-1}( \partial B(0,1/2) )$$
(where $B(z,\delta)$ is the ball of radius $\delta$ centred at $z$).
Suppose $f:\Omega \to U\subset\mathbb{C}$ was a conformal isomorphism with $f(\gamma)=\partial\mathbb{D}$. Since a conformal isomorphism from the inside of $\gamma$ to a disc is unique up to post-composition by a Möbius transformation, it follows that there is Möbius $M$ such that $\phi=f\circ M$. But then $\phi$ extends analytically beyond the boundary of $G$, a contradiction.
(Edited as requested to provide further details and adjust notation slightly.)
A: No. Let $\gamma$ be the image of $S^1$ under $z\mapsto e^z$. Any conformal isomorphism $g$ from $D$ to the region bounded by $\gamma$ must have the form $z\mapsto e^{R(z)}$, where $R$ is some degree one rational function (such that $R(D)=D$). Let $\Omega$ be any simply connected region having both $\gamma$ and $0$ in its interior. A conformal map $f$ from $\Omega$ to a region containing $D$ such that $f(\gamma)=S^1$ would have to have some such map $g$ as its inverse; but $0$ is never in the image of $z\mapsto e^{R(z)}$.
