# Proving mapping must be a rational function (complex analysis)

• Grothard

## Homework Statement

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.

## Homework Equations

Riemann mapping theorem

## The Attempt at a Solution

This one really puzzles me, because I find it hard to demonstrate that an arbitrary analytic function cannot be anything other than a rational one simply due to these givens. For k = 1 by the Riemann mapping theorem there only exists a special class of mappings that f(z) is forced to belong to, so that could work. I thought about splitting the domain into k pieces such that each one maps to one copy of the open unit disk and then applying the Riemann mapping theorem, but this doesn't seem possible because one can't subdivide the unit disk into k nonoverlapping open sets (that would imply the unit disk is not connected, which is clearly false).

Additionally, I'm not sure how to approach the degree of the denominator. Any help is much appreciated.