# Analytic functions of absolute value 1 on the boundary of the unit disc

Is there a characterization of analytic functions $f$ on the unit disc such that $|f(z)|=1$ for $|z|=1$? If $f$ only has a zero $a\in D(0,1)$ of order $n$, then $f(z)=\phi_a(cz^n)$ for some constant $|c|=1$ where $\phi_a$ is the Möbius transformation at $a$. One can iterate this process in the case when $f$ has distinct zeroes, but is there a cleaner formula than something that looks like $\phi_{a_1}(c_1z^{n_1}\phi_{a_2}(c_2 z^{n_2}\cdots \phi_{a_N}( c_N z^{n_N})\cdots)$? Thanks!

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possible duplicate of Mapping that takes unit circle to unit circle – Jonas Meyer Jul 29 '11 at 2:49
All these functions are finite Blaschke products, since they extend from the interior into a continuous function on the boundary , sending the circle/boundary to itself, but I don't know if this characterizes all your functions. – gary Jul 29 '11 at 2:57
Maybe you'd like to know also that the group of conformal automorphisms of the unit disc is $PSL(2, \mathbb{R})$, since the unit disc is conformally equivalent to the upper half-plane. – Bruno Joyal Jul 29 '11 at 7:17

I assume your function is analytic in the open unit disc and continuous on the closed unit disk. It has a finite number of zeros $z_j, j=1 \ldots n$ (counted by mutliplicity). We can extend the definition to the complement of the disc in the Riemann sphere by $f(z) = 1/\overline{f(1/\overline{z})}$ ($\infty$ if $f(1/\overline{z}) = 0$). Thus extended, you have an analytic function from the Riemann sphere to itself, and this must be a rational function (with poles at $1/\overline{z_j}$). Now $f(z) \prod_{j=1}^n \frac{1 - \overline{z_j} z}{z_j - z}$ has no poles or zeros and must be constant. We conclude that $f$ is a finite Blaschke product.
An interesting generalization is the inner function planetmath.org/encyclopedia/OuterFunction.html ... Boundary value of absolute value $1$ is required only almost everywhere on the unit circle. – GEdgar Jul 29 '11 at 13:29
Note that you can also apply the maximum modulus principle to $g(z) = f(z) \prod_{j=1}^n{1 - \bar{z}_j z \over z_j - z}$ and its reciprocal, leading to that $|g(z)| = 1$ on the disc and therefore $g(z)$ is constant by the open mapping theorem. – Zarrax Jul 29 '11 at 14:14