Let $D$ be a domain in $ \mathbb{C} \setminus \{0\}$ such that the annulus $\{z\in \mathbb{C} : 1<|z|<2\}$ is contained in $D$. Prove that there is no branch of the logarithm defined in $D$.

My Attempt: Select a point $Z'\in \{z\in \mathbb{C} :1<|z|<2 \}$ such that $Z'=-x$ where $1 < x< 2$. Then $\log(Z')= \ln x +i \pi$.

Choose a sequence $\{ Z_n\}$ such that $Z_{n}= x(\cos(-\pi+ \frac{1}{n})+ \sin(-\pi+\frac{1}{n}))$. Then $Z_{n}\neq Z'$ for all $n$ and $\lim Z_{n}=Z'$ and $\lim{\log(Z_{n})}=\ln x+i(-\pi)\neq \log(Z')$.

Therefore there exist no continuous function from $D$. Hence there is no branch of logarithm defined in $D$.

Please tell me whether this approach is right or wrong and if its wrong please be kind enough to state how to solve this.


I don't think your approach is good. Your proof has issues and maybe it can be cured (but I don't find any cure).

The main issue is that you do assumption on a particular branch of the logarithm that wouldn't work. It doesn't rule out the possibility that another candidate would do. Note that a branch doesn't require the cut to be along the negative $x$ axis (or in a straight line at all). Actually the proof will be part of proof that there must exist a cut (being a path from origin to infinity).

What you should use is to consider a branch of $\log z$ and it's variation on a path in $D$, for example $|z|=3/2=r$. You can for example write $z$ on polar form and consider the function $f(\varphi) = \log(re^{i\varphi})$ and by taking derivative:

$$f'(\varphi) = {ire^{i\varphi}\over re^{i\varphi}} = i$$

This means that $f(\varphi) = i\varphi + C$, but since $C = f(0) = \log(r) = \log(re^{2i\pi}) = f(2\pi) = C + i2\pi$ you have a contradiction.

What this technique shows is that for a branch of the logarithm there must for each path around the origin be at least one point for which the branch is not defined. This can then be used to see that there exists a path from origin to infinity that is outside of the domain of the branch.


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