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.
Best Answer
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 $D$.