Can we characterize the Möbius transformations that map the unit circle into the unit disk?

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?

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