# Calculate logarithmic index of holomorphic function on open set

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?

• How do you know that $U$ is not a domain? Aug 17, 2015 at 16:35
• @AdamSaltz: $U$ is given as open set. I am not sure if the fact about the image being a subset of an upper semicircle implies connectedness. Aug 17, 2015 at 16:38
• sorry, misread. Aug 17, 2015 at 16:40

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 $D$.
• Thanks. If I got it right the holomorphic function $f'/f$ does not have to be defined on a simply connected domain to have the primitive $F$? I think I confused the concept of "complex logarithm" and "branch of $log f$", i.e. a holomorphic function $F$ s.t. $e^{F} = f$. While the latter would have to be defined on $U$ the one you use is well-defined on H. Aug 17, 2015 at 17:06
• @el_tenedor: A holomorphic function $g$ in $U$ has an antiderivative $G$ in $U$ if and only if $\int_\gamma g(z)dz = 0$ for all closed curves in $U$. This is e.g. always the case if $U$ is simply connected. (Compare en.wikipedia.org/wiki/Antiderivative_(complex_analysis), en.wikipedia.org/wiki/Cauchy%27s_integral_theorem.) – Here we used that $D$ is simply connected, so $1/z$ has an antiderivative in $D$. It follows that $f'/f$ has an antiderivative in $U$, without using that $U$ is simply-connected. Aug 17, 2015 at 17:34