I am trying to show the following statement:
Let $\Omega \subset \mathbb C^*$ be an open subset. We define a branch of logarithm as a continuous function $g:\Omega \to \mathbb C$ such that $e^{g(z)}=z$ for all $z \in \Omega$. Show that if $g$ is a branch of logarithm on $\Omega$, then $\Omega$ can't contain $S^1$.
So here's what I did:
Suppose $S^1 \subset \Omega$, then there is $0<\epsilon<1$ such that $B(1,\epsilon) \subset \Omega$. The functions $$f_1(z)=ln|z|+i\arg_1(z), \space \arg_1(z) \in (-\pi,\pi)$$ $$f_2(z)=ln|z|+i\arg_2(z), \space \space \arg_2(z) \in (0,2\pi)$$
are branches of logarithm in $S_1=B(1,\epsilon)$ and $S_2=B(1,\epsilon) \cap (Im(z)<0)$ respectively.
Now, since $e^{g(1)}=1$, then $g(1)=i2m\pi$ for some $m$ integer. First suppose $m=0$. We have that $g(1)=f_1(1)$, so we must have $g=f_1$ in the ball $B(1,\epsilon)$. And since $g$ is a branch of logarithm in $S_2$ as well, then $$g=f_2+i2k\pi$$for some integer $k$.
Take a sequence $(z_n) \in S_2$ such that $z_n \to 1$. Then we have $g(z_n)=ln|z_n|+i(\arg_2(z_n)+2k\pi)$. It is clear that $\arg_2(z_n) \to 2\pi$.
Since $g$ is continuous, we have $g(z_n) \to g(1)=0$. So we must have $\arg_2(z_n)+2k\pi \to 0$, it follows that $k=-1$.
I wanted to arrive to an absurd but I couldn't. I would appreciate if someone could help me to show the statement.
Let $\log z$ denote the principal branch. Let $S' = S^1\setminus \{-1\}.$ We have $e^{g(z)} = e^{\log z}$ on $S'.$ Define $h(z) = g(z) - \log z.$ Then $e^{h(z)} \equiv 1$ on $S'.$ Thus $h$ maps $S'$ into $2\pi i \mathbb {Z}.$ But $h$ is continuous on $S',$ a connected set. Thus $h(S')$ is a connected subset of $2\pi i \mathbb {Z},$ which implies $h(S')$ is a single point, i.e., $h$ is constant on $S'.$ So we have $g(z) = \log z + C$ on $S',$ which implies $g$ is not continuous on $S^1,$ contradiction.