Find the domain in $\mathbb{C}$ where the function $f(z) = \sqrt{z^2 - 1}$ is analytic.
So far I've tried to take the derivative of $f$ and got $f'(z) = \frac{z}{\sqrt{z^2 - 1}}$ which is not defined in $\mathbb{C}$ at $z = \pm1$ However that is not the correct answer. The solutions say it's the whole interval from -1 to 1.
I speculate that this question is a misrepresentation of a possibly better question, or else a bad question in an unfortunate context... namely, as most of us can easily see, except at $z=\pm 1$, locally there are two holomorphic square roots of $z^2-1$.
A possibly operational point is that much traditional curriculum does not even attempt to talk about local holomorphic functions, a.k.a. "germs", or anything else operationally equivalent, despite clear use of such ideas 150+ years ago, in Riemann's work and others'.
So, for example, a benighted "correct answer" to the question would likely involve choosing a global "domain", that is, a maximal open on which a well-defined section (!) exists. But, of course, there is no single such thing. And the question fails to ask for this.
Still, yes, it can be interesting to ask for two different things: first, to ask for the bad points, which have no neighborhood on which there can be a holomorphic square root; second, to ask for an example of a maximal (and not too hard to describe) set on which there is a global section...