Calculate logarithmic index of holomorphic function on open set

Question

Let $U \subseteq \mathbb{C}$ be an open set and $f \,:\, U \to \mathbb{C}$ a holomorphic function, such that $$f(U) \subseteq\{r e^{it} \,:\, r > 0, -\pi < t < \pi\}.$$ I want to show that

$$\int_\gamma \frac{f'(z)}{f(z)} \mathrm{d}z = 0$$

for all closed smooth paths $\gamma$ in U.

Apparently, $f$ does not have any zeros in $U$ and the logarithmic derivative of $f$ should be holomorphic on $U$. However, I cannot use Cauchy's integral theorem since $U$ is no domain and $\gamma$ is not given as null-homotopic. Neither the residual theorem can be applied, which would yield a proof similar to this one.

Is there maybe a variant of the two mentioned theorems that I am missing? Do you have other suggestions?

Answer 1

The logarithm defined by $$ \log z = \log |z| + i \arg z \, , \quad -\pi < \arg z < \pi $$ is holomorphic in $D = \{r e^{it} \,:\, r > 0, -\pi < t < \pi\}$.

Therefore $F := \log \circ f$ is holomophic in $U$ and satisfies $F' = f'/f$. It follows that $$ \int_\gamma \frac{f'(z)}{f(z)} \mathrm{d}z = \int_0^1 \frac{f'(\gamma(t))}{f(\gamma(t))} \gamma'(t) \mathrm{d}z = F(\gamma(1)) - F(\gamma(0)) $$ for every path $\gamma : [0, 1] \to U$, and in particular $$ \int_\gamma \frac{f'(z)}{f(z)} \mathrm{d}z = 0 $$ for every closed path $\gamma$ in $U$.

The same conclusion holds if $f(U) \subset D$ for any simply connected domain $D$ with $0 \notin D$, because then you can define a holomorphic branch of the logarithm in $D$.