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My question is the following : suppose you have a doubly-connected open set $\Omega \subset \mathbb{C}$, that is a domain bounded by 2 non-intersecting circles $C_1$ (the interior) and $C_2$ (the exterior).

This is well known (an exercise in Ahlfors's book) that there exists a conformal mapping $\varphi$ from $\Omega$ to an annulus $A(r) = \{ z \in \mathbb{C}, r< |z| < 1 \}$.

My question is the following : is it possible to extend $\varphi$ to a conformal map from the interior of $C_2$ to $D = \{ z \in \mathbb{C} , |z|<1 \}$? Another way to ask the question is the following : is it possible to find a conformal map from the interior of $C_2$ to $D$, which restricts to a conformal map from $\Omega$ to $A(r)$?

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  • $\begingroup$ As you saw in two conflicting answers, the answer depends on what do you mean by a circle (round or topological). $\endgroup$ – Misha Apr 23 '15 at 12:14
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    $\begingroup$ Yes, I realised the question wasn't well posed. I was meaning an arbitrary doubly connected region (that is bounded by 2 non-intersecting Jordan curves). So my question can be modified a bit : suppose I know the modulus of $\Omega$ (that is the real $r$). Take an arbitrary conformal map $\psi$ from the interior of $C_2$ to $D$. In general $C_1$ is not sent to a circle. But can I have a control of this Jordan curve though ? For example, if I know $\Omega$ is not thin (say $r \leq 1-\varepsilon$), is it possible to say that $\psi(C_1)$ is contained in some disc $D(1-\delta)$ ($\delta>0$) ? $\endgroup$ – Clem. Apr 23 '15 at 13:22
  • $\begingroup$ The question more precise should be : is there a conformal map $\psi$ such that a control in the sense I gave is possible (this is not so for arbitrary maps : one can find easy examples sending à circle to a region close to the boundary) ? $\endgroup$ – Clem. Apr 23 '15 at 14:59
  • $\begingroup$ What do you mean by "circles"?? $\endgroup$ – Alexandre Eremenko Apr 23 '15 at 22:36
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As Robert writes, the answer is positive for round circles, which follows from the fact that Möbius transformations map circles to circles (and you can map any circle contained in the unit disc to a circle with its centre at 0).

As Neil mentions, the answer is negative for arbitrary curves. Indeed, this is trivial if the curves in question are not analytic. If the curves are analytic, but not round circles, then you can extend the conformal map across the boundary to a larger annulus by Schwarz reflection, but not to the whole disc - again, because Möbius transformations map circles to circles, and any conformal isomorphism between discs is Möbius!

Finally, to respond to the new version of the question raised in your comment - Suppose that your annulus has modulus M, and let $\psi$ be a conformal map that takes the interior of $C_2$ to the unit disc, normalised in such a way that $\psi^{-1}(0)\notin \Omega$. (This assumption was missing from your question, but without it it is clear that you cannot say anything.)

Consider the annulus $A := 1/\psi(\Omega)$. This is a doubly connected region of modulus M, separating the unit circle from $\infty$. Let $R>1$ be the smallest number such that $A$ omits a point of modulus $R$; wlog this point is $R$ itself. What you are asking for is a lower bound on $R$.

It is well-known that, among all such annuli, the Grötzsch annulus, which is the complement of the closed unit disc and the segment $[R,\infty)$, has the largest modulus, which we denote $M(R)$. (See Ahflors, "Conformal Invariants".) Hence we see that $M(R)\geq M$.

In other words, let $\rho>1$ be such that $M(\rho) = M$. (Such $\rho$ exists because $M$ is clearly monotone and continuous, and tends to $0$ and $R\to 1$ and to $\infty$ as $R\to\infty$.) Then $R\geq \rho$, and hence $$\psi(\Omega) \subset D(0,1/\rho),$$ which answers your question. Clearly this bound is sharp, by construction.

There are explicit estimates for the modulus of the Grötzsch annulus; see e.g. Ahlfors's book cited above.

Another interpreation of your question can be given when your inner boundary $C_1$ is not a round circle, but a quasicircle. (This is true whenever the curve is smooth.) In this case, you can extend the conformal isomorphism to the round annulus to a map that will in general not be conformal, but at least be quasiconformal, with constants depending on the constants in the quasicircle definition. Quasiconformal maps satisfy some nice properties, so again these will give you some control over the image of your curve.

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  • $\begingroup$ Thanks a lot for this answer. I'm still missing something though : there is a mobius transformation (conformal bijection of $D$) which sends 1/2 to $1-\varepsilon$, so the disc $D(0,1/2)$ is sent to a circle arbitrarly "close" to the boundary, which seems impossible by your argument. Where am I wrong ? $\endgroup$ – Clem. Apr 24 '15 at 7:33
  • $\begingroup$ Ok, I see the problem : in my example, 0 won't be in the resulting circle. $\endgroup$ – Clem. Apr 24 '15 at 12:01
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If I understand your question correctly then the answer is negative. Indeed, $C_2$ could be the the unit circle and $C_1$ could be a square. Now $\varphi$ has to be a conformal automorphism of $D$, and any such automorphism has the form $\varphi(z)=\lambda(z-\alpha)/(1-\overline{\alpha}z)$ for some $\alpha,\lambda$ with $|\alpha|<1$ and $|\lambda|=1$. In particular, $\varphi$ is a Mobius map and so sends circles to circles and non-circles to non-circles. In particular, it does not send $C_1$ to a circle, and so does not send the original region $\Omega$ to an annulus.

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Yes: if there is a conformal map from $\Omega$ to $A(r)$ that takes $C_1$ to $\{|z|=r\}$ and $C_2$ to $\{|z|=1\}$, then there is a Möbius transformation that does so, and the pole of this function is outside $C_2$ so it does extend.

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    $\begingroup$ Thanks a lot for the answer, but I'm not a specialist in conformal geometry ; so could you please be a bit more precise (and give some more details) ? Thanks again ! $\endgroup$ – Clem. Apr 23 '15 at 8:54
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    $\begingroup$ I don't think that this is correct. Why should there be a Mobius function? $\endgroup$ – Neil Strickland Apr 23 '15 at 10:48
  • $\begingroup$ Hi Neil, the answer is correct when the question is taken literally, i.e. "circle" actually means "round circle". $\endgroup$ – Lasse Rempe-Gillen Apr 23 '15 at 14:42
  • $\begingroup$ Yes, I took "circle" in the geometric sense. $\endgroup$ – Robert Israel Apr 23 '15 at 15:33

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