Let $A \subset \mathbb{C} $ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.
Does it follow that $f(z) = a z^{n} \frac{cz^{m}-b}{1-cz^{m}\bar{b}} $ for some $a,b,c \in \mathbb{C}$ s.t. $|c|=1$ $|a|=1$, $|b|<1$ and some $n,m\ge 0 $?
Quoting part of my answer to this other question is my answer to yours (and Robin basically already said it in the comments):
If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus 1 on the circle, then $f$ is a finite Blaschke product. You can prove this by taking all of the zeros inside the disk counted according to multiplicity, dividing by corresponding holomorphic automorphisms of the disk that have those zeros, and showing that the result is constant. (This quotient and its reciprocal are analytic and bounded by 1 on the disk...)
Of course I'm assuming $A$ is connected.
The answer is No. Recall the question:
Let $A\subset\mathbb{C}$ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.
Does it follow that $$f(z)=az^{n}\,\frac{cz^{m}-b}{1-\bar{b}cz^{m}}$$ for some $a,b,c\in\mathbb{C}$ s.t. $|c|=1$, $|a|=1$, $|b|\le1$ and some $n,m\ge0$?
Let $B$ be the set of analytic functions from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$, and let $C$ the set of rational fractions of the indicated form.
If $f$ is an element of $C$ with $0<|b|<1$ and $m\ge1$, then $z\mapsto f(z)^2$ is in $B$, but, having a multiple zero $z_0\not=0$, is not in $C$.
