The Möbius transformations are the maps of the form $$ f(z)= \frac{az+b}{cz+d}.$$ Can we characterize the Möbius transformations that map the unit circle $\{z\in \mathbb C: |z| = 1\}$ into the (closed) unit disk $\{z\in \mathbb C: |z| \leq 1\}$?

See the related post, but not similar post: Can we characterize the Möbius transformations that maps the unit disk into itself?

  • $\begingroup$ It seems to me that your question is answered here: math.stackexchange.com/a/209434/42969. $\endgroup$ – Martin R Jul 2 '20 at 14:26
  • $\begingroup$ Well here you only know what happens on the unit \emph{circle}. $\endgroup$ – Frederik Ravn Klausen Jul 2 '20 at 14:29
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    $\begingroup$ If $|z|=1$ is mapped into the unit circle then the same is true for $|z| \le 1$, according to the maximum modulus principle. $\endgroup$ – Martin R Jul 2 '20 at 14:30
  • $\begingroup$ Well Martin R the Möbius transformation is not holomophic in the unit disc - so I guess this argument is not applicable? $\endgroup$ – Frederik Ravn Klausen Jul 8 '20 at 12:45

Notice$$\left|\frac{ae^{it}+b}{ce^{it}+d}\right|^2=\frac{(ae^{it}+b)(a^\ast e^{-it}+b^\ast)}{ce^{it}+d)(c^\ast e^{-it}+d^\ast)}=\frac{aa^\ast+bb^\ast+2\Re(ab^\ast e^{it})}{cc^\ast+dd^\ast+2\Re(cd^\ast e^{it})}$$for $t\in\Bbb R$. We wish to identify those $a,\,b,\,c,\,d\in\Bbb C$ for which the above has modulus $\le1$, i.e. $\min_{t\in\Bbb R}(A+B\cos t+C\sin t)\ge0$ with$$A=cc^\ast+dd^\ast-aa^\ast-bb^\ast,\,B=2\Re(cd^\ast-ab^\ast),\,C=2\Im(ab^\ast-cd^\ast).$$The desired constraint is $A\ge\sqrt{B^2+C^2}=|ab^\ast-cd^\ast|$.


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