I have already known the fact: Let $\gamma$ be a path in $\mathbb C \setminus \{c\}$. Then there exists a parametrisation $\gamma: [a,b]\to \mathbb C \setminus \{0\}$ of $\gamma$ for which $t \mapsto \arg (\gamma (t)-c)$ is a continuous function. How to prove that continuous choice of argument exists?
My question: suppose $c\in \mathbb{C}$, $G$ is a region(an open connected set) and $c\notin G$. When can we choose a continuous $\arg$ on $G$?
My thoughts: if there exists a straight line $l$ joining $c$ and the infinity $\infty$, then we can define a continuous arg on $\mathbb{C}\setminus l$.
My Guess: If $G$ is a simply connected region, then we can choose a continuous branch of $\arg$. I'm not sure if this is correct. Since $G$ is simply connected, we can choose a simple curve $\gamma$ in the extended complex plane from $c$ to $\infty$. I think we can choose a continuous $\arg$ on $C\setminus \gamma$, but I don't know how to prove it when $\gamma$ is not a straight line.
