# Mapping that takes unit circle to unit circle

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 $n,m\ge 0$?

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What about $z\mapsto(z^n-b)/(1-\overline b z^n)$? –  Robin Chapman Sep 17 '10 at 17:27
Thanks Robin. I changed my question in response –  Digital Gal Sep 17 '10 at 17:30
You define two sets of functions and ask if they coincide. The first set is closed under multiplication and composition. You should check that the second set has the same property. –  Pierre-Yves Gaillard Sep 17 '10 at 17:37
@Moron - f is not supposed to be entire. –  Pierre-Yves Gaillard Sep 17 '10 at 17:41
Not really a duplicate, as there is no reason to suppose the function entire. But the OP should look at Blaschke products. –  Robin Chapman Sep 17 '10 at 17:42

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.

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Community wiki because I'm just copying my other answer, and because the answer was already given in the comments. But I thought this question might as well have an "Answer". –  Jonas Meyer Sep 17 '10 at 22:39

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 $C$.

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