Show that a certain holomorphic function can be written as the quotient of another function and its derivative

68 Views Asked by At

Currently tasked with the following problem:

Let $D \subset \mathbb{C}$ be a simply connected domain and $S \subset D$ finite.

Furthermore let $f: D \setminus S \rightarrow \mathbb{C}$ be holomorphic and $\forall p \in S: Res_f(p) \in \mathbb{Z}$.

Show that there exists a holomorphic function $h: D \setminus S \rightarrow \mathbb{C} \setminus {0}$, such that:

$$\forall z \in D \setminus S: f(z)=\frac{h'(z)}{h(z)}$$

Well my first thought process was to trying to use that D is simply connected. So for all $z \in D$ we'd have a antiderivative F for f if S was empty and then choosing $h(z) = \exp(F(z))$ would probably work out in some way. But obviously we can't just assume that S is empty. The next thing I tried that by using D simply connected $\implies$ D star-shaped (Edit: realized that this is not true at all. Still would somehow like to change my problem into something that works with star-shapes) I'd know the antiderivative in a point p would be just the curve integral from the star centre to p over f. But then again, I can't guarantee that $D \setminus S$ is also star-shaped!

Is there anything else I can try?

1

There are 1 best solutions below

0
On BEST ANSWER

This is a similar question as Which meromorphic functions are logarithmic derivatives of other meromorphic functions?, only that here it is not required that the function $h$ (of which $f$ is a logarithmic derivative) has removable singularities (or poles) at the points in $S$. But one can proceed similarly.

Fix a point $z_0 \in D \setminus S$ and define $h$ in $D \setminus S$ by $$ h(z) = \exp \left( \int_\gamma f(t) \, dt \right ) $$ where $\gamma$ is any path in $D \setminus S$ connecting $z_0$ to $z$.

First use the residue theorem to show that $h$ is well-defined, i.e. that the definition is independent of the choice of $\gamma$ connecting $z_0$ to $z$. This is where the condition about the residues of $f$ being integers is used.

Then show that $h'/h = f$.