Characterizing the unimodular functions from the closed disk $\overline{\mathbb{D}}$ to $\mathbb{C}$ with constraints

Question

Let $\mathbb{D}$ be the open disc. It is well known that if $f:\mathbb{D}\to\mathbb{C}$ is analytic, continuous on the boundary, and is unimodular (say with a finite number of zeros) then $f$ is a finite Blaschke product up to some unimodular complex constant. I am trying to understand the following set: $$S=\Big\{f:\mathbb{D}\to\mathbb{C}:f\,\text{is a finite Blaschke product and}\,f(0)=f\big(\frac{1}{\sqrt{2}}\big)\Big\}$$

It is clear that every finite Blaschke product that contains the factor $z\frac{z-\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}z}$ is in the set, and each $f$ in the set has to have at least two zeros (otherwise it is one to one on the disk and cannot satisfy the last condition). However, given a fixed point $w_1\in\mathbb{D}$, can I always find a function $f\in S$ and a point in the disk $w_2$ such that $f$ vanishes only at $w_1,w_2$ (both are not $0$ or $\frac{1}{\sqrt{2}}$)? Moreover, is this $w_2$ unique in some sense?

This is obviously equivalent to finding $w_2$ such that $w_1w_2=(\frac{1-\sqrt{2}w_1}{\sqrt{2}-\overline{w_1}})(\frac{1-\sqrt{2}w_2}{\sqrt{2}-\overline{w_2}})$ (the RHS is plugging $\frac{1}{\sqrt{2}}$ into the Blaschke product corresponding to $w_1,w_2$ and LHS is plugging $0$ into it). I wasn't able to solve this equation algebraically, nor show such $w_2$ exists. Any hint/help/reference would be helpful and appreciated.

Answer 1

A description of your class $S$ can be obtained as follows. Let $z_0=1/\sqrt{2}$. Let $H$ be the class of all functions of the form $$f(z)=ze^{it}\frac{z-z_0}{1-z_0z}\prod_{n=1}^N\frac{z-z_n}{1-\overline{z_n}z},$$ where $z_n$ are arbitrary points in the unit disk, and $N\geq 0$ is arbitrary. Now the general form of the function of your class $S$ is $\phi\circ f,$ where $\phi$ is an automorphism of the unit disk and $f\in H$.

Proof. For $f\in H$, evidently $\phi\circ f$ is a finite Blaschke product (times a constant of modulus $1$), and $$\phi\circ f(0)=\phi(0)=\phi\circ f(z_0).$$ In the opposite direction, let $g\in S$. Then there is $a$ in the unit disk such that $g(0)=g(z_0)=a$. Let $\phi$ be an automorphism of the disk which sends $0$ to $a$. Then $\phi^{-1}\circ g=f$, and $g=\phi\circ f$, where $f\in H$.