Let $f$ be holomorphic and nonzero on some open set $\Omega\subseteq \mathbb{C}$. Then is $\ln {f}$ well-defined? Usually, in order for $\ln {z}$, we need to choose a branch, e.g., $(-\infty,0]^c$, so that the function is well-defined, but how do we know that the image of $f$ is in some branch?

My thought process is that if the image of $f$ goes all around $0$, then we can choose some path $\gamma$ such that $\Gamma = f\circ \gamma$ circles around 0 and thus

$$ \int_\gamma \frac{f'}{f} dz =\int_{\Gamma} \frac{dw}{w} \ne 0\\ $$

However, since $f$ is holomorphic and nonzero, the most-left integral should be zero.

The problem is that I can't quite show that $\gamma$ is $C^1$ or at least piecewise $C^1$.

  • 2
    $\begingroup$ If $\Omega$ is simply connected you can always define a holomorphic logarithm if $f$ doesn't vanish; if $\Omega$ is not simply connected it depends (see $f(z)$=$z$ on the punctured unit disc for a counterexample) but if the image of $f$ is nice (e.g it lies in a simply connected domain that doesn't contain zero), you can also define a holomorphic logarithm $\endgroup$ – Conrad Feb 17 at 3:47

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