About holomorphic functions

Question

it's true that if I have $\Omega\subset\mathbb{C}$ a connected open non-empty set, then there exists a holomorphic function $f:\Omega\to\mathbb{C}$ such that $[f(z)]^2=z$, for all $z\in\Omega$ and $f(\Omega)$ is in a half plane of $\mathbb{C}$?

I think that it's necesary to "redefine" a branch of the square root function.

Can anyone give me some approach of the answer?

Thanks in advance.

Answer 1

Let $Ω = \{ z : z \in \mathbb{C} \land 0<|z|<2 \}$, which is basically an open annulus surrounding $0$, and let $g(t) = \exp(it)$ for any $t \in [0,2π]$, so that $g$ is basically a circle within $Ω$ with centre $0$. Now let $h = \arg \circ f \circ g$, and so $h$ is continuous and $\frac{1}{2π}(2h(t)-t) \in \mathbb{Z}$ for any $t \in [0,2π]$. (Note that every integer-valued continuous function is constant.) Thus $\frac{1}{2π}(2h(t)-t)$ is constant over all $t \in [0,2π]$, which gives a contradiction at $t = 0$ and $t = 2π$ (which are basically the endpoints of $g$).

Closely related question

$\def\less{\smallsetminus}$If $Ω$ is a simply-connected subset of $\mathbb{C} \less \{0\}$, then there is actually a branch of the square-root on $Ω$, though this branch may not lie within a half-plane. Below is a sketch of the proof, which reveals exactly when such holomorphic branches are possible. (The above example has $Ω$ that cannot be expanded to any simply-connected subset of $\mathbb{C} \less \{0\}$, because of that curve in $Ω$ with non-zero winding number around $0$.)

First prove that you can define a branch of $\ln$ that is holomorphic on your domain. Let $c$ be a point in your domain, and $a$ be such that $e^a = c$. Then let $g(z) = a + \frac{1}{2πi} \int_γ \frac{1}{z}$ where $γ$ is some path from $c$ to $z$. Use simply-connectedness to prove that $g$ is independent of the path and hence prove that it is holomorphic, and also that it satisfies $\exp(g(z)) = z$ for any $z$ in the domain. Then define $f(z) = \exp(\frac{1}{2}g(z))$. Prove that $f$ is a branch of $\sqrt{\ }$ as required.