Analytic mapping from disk to disk must be rational

[Grothard]

Let f(x) be a function which is defined in the open unit disk (|z| < 1) and is analytic there. f(z) maps the unit disk onto itself k times, meaning |f(z)| < 1 for all |z| < 1 and every point in the unit disk has k preimages under f(z). Prove that f(z) must be a rational function. Furthermore, show that the degree of its denominator cannot exceed k.

If this was limited to k=1 I think we could use the Riemann mapping theorem, but for larger k I am quite lost. How does one go about proving that an arbitrary analytic function with these givens must be rational? I've seen a form of this question in several places, but I still can't grasp how one would tackle such a problem.

 

[mathwonk]

I think under these hypotheses, probably k must be 1. and then we are concerned with automorphisms of the unit disc, all fractional linear transformations.

 

[micromass]

I think under these hypotheses, probably k must be 1. and then we are concerned with automorphisms of the unit disc, all fractional linear transformations.

What about f(z)=z^2. This seems to satisfy the conditions? Except of course that 0 only has 1 preimage. It does have 2 preimage counting multiplicity, so I wonder whether to count the preimages with multiplicity or not.

 

[Grothard]

What about f(z)=z^2. This seems to satisfy the conditions? Except of course that 0 only has 1 preimage. It does have 2 preimage counting multiplicity, so I wonder whether to count the preimages with multiplicity or not.

I do believe we should consider the multiplicity; my language was imprecise as I was paraphrasing the problem I've encountered in several sources before. I think f(z)=z^n are precisely the types of functions we are looking for. But I have no clue how to show that any arbitrary analytic function that covers the unit disk in this way must be rational. I feel like there's a theorem I'm missing.

 

[mathwonk]

i was using the fact that he did not use multiplicities for counting preimages.