I wish to prove that there is no branch of the complex logarithm holomorphic in the domain $D=\{z \in \mathbb{C}:0 \lt \vert z\vert \lt 1\}$.
Is it correct to say that as $0$ is the branch point of $\log(z)$ all branches of $\log(z)$ will have discontinuity on some line from $0$ in the complex plane and hence will be discontinuous in $D$?
Thank you in advance!
Here is a most simple solution : compute $\int_{\gamma} \frac{dz}{z}$ using this determination of the logarithm, where $\gamma$ is a small loop around zero. You should get zero assuming there is a holomorphic branch of $\log$. On the other hand, explicit computations show that $\int_{\gamma} \frac{dz}{z} = 2\pi i$, hence such branch can't exist.
On the other hand, your argument can be fixed. Any branch of logarithm should be on the form $\log(z) = \log(|z|) + i\arg(z) $, where $\log(x)$ is the usual logarithm for $x \in \mathbb R_{>0}$. In particular, a branch of logarithm gives a branch of $\arg$, but it's impossible to have a continuous determination of the argument, so there is no branch of $\log$ on $D \backslash \{0\}$.
Suppose on the contrary that $f:C^*=C-\{0\}\to C$ is a branch of the logarithm. Consider the principal branch of the logarithm, i.e., $Log: G=C-(-\infty, 0]\to C$ such that $\exp(Log(z))= z$ for all $z\in G$.
$f|_G$ is also a branch of the logarithm defined on $G$ so it differs from $Log$ by $\mod 2\pi$. That is, there exists a $k\in \mathbb Z$ such that $f|_G(z)=Log (z)+i2\pi k$ for all $z\in G$. Since $Log$ is not continuous at $-2$, $f|_G$ is not continuous at $-2$. This contradicts the fact that $f$ is continuous at $-2$.
