Computing the domain of analyticity of $f(z)=\sqrt{z^2-1}$

Question

In this question, it is said that the domain of analyticity of the function $f(z)=\sqrt{z^2-1}$ over the branch $(0,2\pi)$ is $\mathbb{C} \setminus ((-\infty,-1) \cup (1,\infty))$.

My question: I would like to know how to show this affirmation.

My attempt: We have $f(z)=\sqrt{z^2-1}=e^{\frac{1}{2}\ln(z^2-1)}$, where $ln(z)$ is analytic in $\mathbb{C} \setminus [0,\infty)$. Let $z=x+iy$. So, $f(z)$ it is not analytic if $z^2-1=a$ for $a \in \mathbb{R}$ and $a\geq 0$ and we have the following system:

$$ \begin{split} x^2-y^2-1=a \hbox{ and } xy=0. \end{split} $$ But I don't know how to complete the proof.

Answer 1

Given that the function is a root of a polynomial, it does not have any singularities, but it does have branch cuts. Since it is specified that $\arg(z^2-1)\in(0,2\pi)$, this means that points lying on $\mathbf R^+$ are discontinuous, that is, those points satisfying $\mathfrak{Re}\{z^2-1\}>0$ and $\mathfrak{Im}\{z^2-1\}=0$.

We know $z^2-1=(x+iy)^2-1=x^2-y^2-1+2ixy$, therefore: $$ \left\{ \begin{aligned} &x^2-y^2-1>0\\ &2xy=0 \end{aligned} \right. $$ If $x=0$, we clearly have a contradiction: $y^2+1<0$. Thus, the correct branch would be $y=0\wedge|x|>1$.

Finally, we can conclude $f(z)$ is analytic at $\mathbb C \setminus \{(-\infty,-1)\cup(1,\infty)\} $.

Answer 2

There is a common misunderstanding that there is a unique "domain" for a potentially multi-valued (but locally holomorphic, away from a few bad points) function on $\mathbb C$. True, there are "traditional choices" in many of the "traditional examples", but they are by no means the only ones.

Already this example shows that there is typically more than one choice of "domain" on which there is a (single-valued) version of "the function". Namely, among other choices, removing ("slitting") the plane along the interval $[-1,1]$ succeeds, as does slitting along $(-\infty,-1]\cup [1,+\infty)$. Also, many more...

The way to really/robustly understand this is the Monodromy Theorem: the "family" of "local" instances of the function, by meromorphic continuation along paths, gives a representation of the first homotopy group of the plane punctured at the bad points. (This is a free group on a number of generators equal to the number of bad points.) The point of "slitting" the plane is to kill off enough of the homotopy group so that that representation is trivial on what's left.

In general, there is more than one way to do this... :)