f holomorphic in an annulus can be written as $f(z)=z^{n}\,e^{g(z)}$

Question

Here is the problem: for $0<r<1$, let $A_r:= D_1(0)-\overline{D_r(0)}$ and $f$ holomorphic in $A_r$, which does not vanish. Show that there exists $n \in \mathbb{Z}$ and $g$ holomorphic in $A_r$ such that:

$$f(z)=z^{n}\,e^{g(z)} \,\,\,\,\, \forall z\in A_r$$

I really don't know how to come up with the exponential since the domain is not simply connected. Any ideas? Thanks!

Answer 1

First, note that such an $f$ can be written in the form $f(z)=e^{g(z)}$ iff $f'(z)/f(z)$ has a global antiderivative on $A_r$, which happens iff $\int_C f'(z)/f(z) dz=0$ for $C$ a circle around $0$ contained in $A_r$. In general, of course, this may not be true, but $\int_C f'(z)/f(z) dz$ will always be an integer multiple of $2\pi i$. So if $\int_C f'(z)/f(z) dz=2\pi i n$, let $h(z)=f(z)z^{-n}$ and note that $\int_C h'(z)/h(z) dz=0$. Thus $h(z)$ has a logarithm $g(z)$, and then $f(z)=z^ne^{g(z)}$.

Answer 2

Hint:

Try to show that residue of $f'/f$ is an integer. Prove that $\frac{1}{2 \pi i} \int f'/f$ is an integer. Let that integer be $n$.

Now consider $h(z) =z^{-n}f$. Check that $h'(z)/ h(z)$ has residue zero. Therefore, using the Laurent series or otherwise, conclude that there is a holomorphic function $g(z)$ such that $h'(z)/ h(z) = g'(z)$. Which means $ (h\cdot e^{-g})' = 0$. Conclude that $f = c \cdot z^n e^{g(z)}$. You can easily choose another $g$ for which $f = z^n e^g$.