I want to construct a conformal map from the closed unit disc onto itself that maps given three points on the unit circle to another given set of three points on the unit circle.

I know that the word "conformal" can mean different things: sometimes it means the derivative is nonzero; sometimes the function should be injective; and sometimes the word is synonymous to biholomorphic. I also know that if the word mean "biholomorphic", such a map has to be a linear fractional transformation of the form $e^{i\theta}(z - a)/(1 - \bar a z)$, in which case the conditions are so restricted that I imagine such a map doesn't exist.

Do maps that satisfied the condition above exist, possibly depending on what you mean by "conformal"?

EDIT: clarification in bold.

  • $\begingroup$ Möbius. $\endgroup$ – k.stm Feb 27 '15 at 7:50
  • $\begingroup$ @k.stm Does applying that usual method to points of modulus one guarantee that closed unit disc is mapped to itself? $\endgroup$ – Pteromys Feb 27 '15 at 7:53
  • $\begingroup$ Conformal mappings map circles to circles and a circle is determined completely by any three points on it. Then use the maximum modulus principle. $\endgroup$ – k.stm Feb 27 '15 at 8:10
  • $\begingroup$ @k.stm Is it possible for you to elaborate? For example, what do you mean by conformal maps? Do you mean that conformal maps between circles are determined by three points on one circle and other three points on the other? $\endgroup$ – Pteromys Feb 27 '15 at 8:27

You are asking yourself whether such a map exists at all. The question is: Can a "circle" $\gamma$ be mapped onto itself such that three arbitrarily given points on $\gamma$ are mapped onto another given triple of points on $\gamma$?

Now a familiar example of such a "circle" is the real axis. Given three different points $a<b<c$ in ${\mathbb R}$, the Moebius map $$S:\quad z\mapsto{z-a\over z-c}$$ maps the real axis (incl. $\infty$) onto itself; furthermore $S(a)=0$, and $S(c)=\infty$. Let $S(b)=:q<0$. Then the map $$T:\quad z\mapsto {1\over q}\>{z-a\over z-c}\tag{1}$$ maps the arbitrary triple $(a,b,c)$ onto the "special" triple $(0,1,\infty)$. When $(a',b',c')$ is another such triple, and $T'$ the corresponding map $(1)$, then $$T_*:=T'^{-1}\circ T$$ will map $(a,b,c)$ onto $(a',b',c')$.

There is one proviso however: When $(a,b,c)$ and $(a',b',c')$ do not "induce the same orientation" on ${\mathbb R}$ the map $T_*$ will map the upper halfplane onto the lower halfplane. The same is true for your two triples on the unit circle $\partial D$: If they don't induce the same orientation of $\partial D$ the resulting Moebius map $T_*$ does not map $D$ onto itself, but interchanges the interior and the exterior of $D$.

  • $\begingroup$ Thank you. What is the easy (calculation-less?) way of showing the fact about orientation? $\endgroup$ – Pteromys Feb 27 '15 at 9:50
  • $\begingroup$ This is a pure coincidence: I have answered the same question yesterday here (in a real setting): math.stackexchange.com/questions/1166279/… $\endgroup$ – Christian Blatter Feb 27 '15 at 9:54

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