Let \(f\) be meromorphic on \(\DD\) with no poles on \(\bd\DD\). Show that there exists a meromorphic \(g\) with no poles in \(\DD\) such that \(\abs{f(z)} = \abs{g(z)}\) when \(\abs{z} = 1\).
Solution
Write \(\ts{a_1,\cdots, a_n}\) for all of the poles of \(f\), indexed with multiplicity, and define \[ g(z) \da \prod_{1\leq k\leq n} \psi_{a_k}(z) f(z) \da \qty{ \prod_{1\leq k \leq n}{z-a_k\over 1 -\bar{a_k} z}} f(z) .\] Then \(g\) has no poles, and since \(\abs{ \psi_{a_k} } = 1\) on \(\bd \DD\), this works.