# 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.

• Not in general, of course, just because the curve $\gamma$ needs to be analytic if there is such a conformal map $f$. May 21, 2015 at 14:18
• Sorry, I forgot this. I edit my question. May 21, 2015 at 14:21
• Your edit asks for $\gamma$ to be smooth (presumably meaning $C^\infty$), but that is still not restrictive enough: you need $\gamma$ to at least be analytic, as Malik pointed out. May 21, 2015 at 15:14
• Well what if $\Omega$ is the whole plane and $\gamma$ is not a circle? In this case, such an $f$ would be linear, so that $f(\gamma)$ cannot be a circle. May 21, 2015 at 17:54
• Thanks for the comments ; these hypothesis were a bit implicit for me. I re-edit the question. May 21, 2015 at 21:38

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.)

• Your solution is hard to follow because your notations clash with the notation in the problem as stated by the OP. May 22, 2015 at 14:24
• Also, the OP DOES require the curve to be analytic, so does the first sentence of your answer imply that the answer is YES? May 22, 2015 at 15:03
• No. If you look carefully, the curve $\gamma$ itself is analytic. I shall see whether I can clarify the answer. May 22, 2015 at 20:35
• I think it's pretty clear. +1 May 23, 2015 at 15:19

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 $z\mapsto e^{R(z)}$.