Does there exist a holomorphic function $f:\mathbb{C} \setminus [-1,1] \rightarrow \mathbb{C}$ such that $f(z)^{2} = z^{2}-1$ holds for all z in the domain?

I figured that the answer should be NO.

If yes, then define $g(z) = \sqrt{z^{2}-1}$, which is holomorphic on $\Omega = $ some branch of $\mathbb{C}$. Then $f = g$ on $\mathbb{C} \setminus (\Omega \cup [-1,1])$ by uniqueness. But then $\Omega \neq \mathbb{C} \setminus [-1,1]$ and no analytic continuation of $g$ is possible. Therefore, $f$ cannot be holomorphic on $\mathbb{C} \setminus (\Omega \cup [-1,1])$, contradition.

It just seems to me that something is missing in this reasoning and I don't find it convincing enough. Any other better approach?



The answer is yes. Note that $w = 1/2 + 1/(z-1)$ maps $\mathbb C \cup \{\infty\} \backslash [-1,1]$ to $\mathbb C \backslash (-\infty, 0]$, and $z^2 - 1 = \dfrac{8w}{(2w-1)^2}$. Let $f(z) = \dfrac{\sqrt{8w}}{2w-1}$ using the principal branch of the square root.


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