Consider a rational function $f(z)\in\mathbb{R}(z)$ with no poles/zeros on the unit circle $\mathbb{T}=\{z\in\mathbb{C}\,:\, |z|=1\}$. Does there always exist a Möbius transformation $$ \rho\colon \overline{\mathbb{C}}\to \overline{\mathbb{C}}, \ \ z\mapsto \frac{az+b}{cz+d}, \quad a,b,c,d\in\mathbb{C}, ad-bc\neq 0, $$ such that the pole/zero at infinity of $f(z)$ (if present) is mapped to a finite point outside the unit circle and all the other finite poles/zeros of $f(z)$ inside (outside) the unit circle remain finite and inside (outside) the unit circle after the transformation?
Thanks a lot.
Sure; you can even additionally require that $\rho$ map the entire unit disk to itself. Just note that for any point $p$ in the open unit disk, there is a Mobius tranformation $\sigma$ such that $\sigma(p)=0$ and $\sigma$ maps the unit disk to itself. Now pick any $p$ such that $0<|p|<1$ and $1/p$ is not a root of $f$ and let $$\rho(z)=\frac{1}{\sigma(1/z)},$$ where $\sigma$ is as above. Then $\rho$ maps the unit disk to itself since $\sigma$ does, so it preserves whether roots of $f$ are inside or outside the unit circle. Also, $\rho(1/p)=\infty$, so neither $\infty$ nor any of the roots of $f$ map to $\infty$ under $f$.