Existence of analytic square root of $z^2-1$

Question

Take the complement of a compact connected set containing $\pm 1$ (which we call $\Omega$). Show that there exists an analytic square root of $z^2-1$ for all $z \in \Omega$.

The case in which $\Omega^c \supset [-1,1]$ is clear to me (by investigating where $z^2-1\geq 0$ or $\leq 0$ and choosing a branch of square root). However, I don't understand this more general case. One thought I had was to reduce to the simply connected case, e.g. via some conformal map, but I am still coming short.

Alternatively I thought of maybe starting with an analytic square root in the complement of some big compact disk centered at $0$ and containing $\Omega^c$ and somehow extending to the remainder of the domain (e.g. Monodromy theorem?) but I doubt that level of complexity is needed here.

Would appreciate any thoughts.

Answer 1

Let $\gamma$ be any rectifiable, simple closed curve in $\Omega$. By the Jordan curve theorem, $\mathbb{C}\setminus\operatorname{im}(\gamma)$ consists of two open connected sets, namely the interior and exterior of $\gamma$. Since $\Omega^c$ is disjoint from $\operatorname{im}(\gamma)$, this then implies that $\Omega^c$ lies either in the interior of in the exterior of $\gamma$. In particular, we always have

$$ \int_{\gamma} \frac{1}{2}\left(\frac{1}{z+1} + \frac{1}{z-1} \right) \, \mathrm{d}z \in 2\pi i \mathbb{Z}. \tag{1} $$

From this, we can define the function $f : \Omega \to \mathbb{C}$ by

$$ f(z) = \sqrt{\smash[b]{z_0^2} - 1} \exp\left( \int_{z_0}^{z} \frac{1}{2}\left(\frac{1}{\xi+1} + \frac{1}{\xi-1} \right) \, \mathrm{d}\xi \right) $$

on each connected component $C$ of $\Omega$, where $z_0$ is any chosen point in $C$, $\sqrt{\smash[b]{z_0^2} - 1}$ is any square root of $z_0^2 - 1$, and the integral is taken over any path in $C$ joining from $z_0$ to $z$.

Since $\Omega^c$ is closed, $C$ is open and connected, and hence there is always a rectifiable path from $z_0$ to $z$. This and the condition $\text{(1)}$ altogether show that $f(z)$ is a well-defined, analytic function.

Moreover, it is easy to verify that $\frac{f(z)^2}{z^2-1}$ has zero derivative, and so, we have $f(z)^2 = z^2 - 1$ on all of $\Omega$. Therefore $f(z)$ is an analytic square root of $z^2-1$ on $\Omega$.