I'm trying to the following problem from my Complex Analysis class:
Let $a,b\in \mathbb{C}$, with $a\neq b$ and put $\Omega = \mathbb{C}\setminus [a,b]$, where $[a,b]$ is the segment joining $a$ and $b$. Show that $\Omega$ is not simply connected. Show that there exists an holomorphic function $f$ defined on $\Omega$ such that $[f(z)]^2=(z-a)(z-b)$ for all $z\in \mathbb{C}$.
I've managed to show that $\Omega$ is simply connected, but I'm struggling with showing that such an $f$ exists. Obviously we'll end up with $f(z)=\sqrt{(z-a)(z-b)}$ with some appropriate branch of the square root function, the issue is finding said branch (as in showing that is is holomorphic in the appropriate domains)
My plan was finding some closed set $F$ such that $\mathbb{C}\setminus F$ is simply connected, so we know that there exists a branch of the logarithm defined on $\mathbb{C}\setminus F$, which lets us define a branch of the square root in said set. The thing is we need $F$ to verify the following:
$(1)$ $\Omega \subset{\left ( \mathbb{C} \setminus F \right )}$
$(2)$ If $z\in \Omega$ then $(z-a),(z-b)\not \in F$.
However it seems that no sets containing $\Omega$ are simply connected, so $(1)$ can't hold, and hence this strategy is doomed to fail. Maybe we could use analytic continuation to show that the relation $[f(z)]^2=(z-a)(z-b)$ holds in some small set, but this seems unlikely.
Any ideas?
Thanks in advance!