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Like most us, I've always consider the domain of holomorphy of $\sqrt{z^2-1}$ as $\mathbb{C}\backslash [-1,1].$

However, I'm used to work with a fixed determination of the argument, that is $\arg(z) \in (-\pi,\pi].$ With this definition, the square root is holomorphic on $\mathbb{C}\backslash (-\infty,0].$ And now if I look for the $z$ such that $z^2-1 \in (-\infty,0]$, I find $[-1,1]\cup i\mathbb{R}.$

By definition, the domain of holomorphy is the biggest open set such that the function is holomorphic, so I know that $z \mapsto \sqrt{z^2-1}$ is holomorphic on $\mathbb{C} \backslash [-1,1]\cup i\mathbb{R}$ but maybe it is not the biggest set. Thus, I tried to proove that the function is holomorphic on $i\mathbb{R}_0$. The problem is the following : if I choose $z_0 \in i(0,+\infty)$, we have $$\lim_{z\to z_0, \Re(z)<0} \sqrt{z^2-1} = -i \sqrt{|z_0^2-1|}, \quad \lim_{z\to z_0, \Re(z)>0} \sqrt{z^2-1} = i \sqrt{|z_0^2-1|}.$$ hence the function is not even continuous (and hence, not holomorphic) on $i\mathbb{R}_0$. So, how can we say that $$(z\mapsto \sqrt{z^2-1}) \in \mathcal{O}(\mathbb{C} \backslash [-1,1]) ?$$

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  • $\begingroup$ Relevant $\endgroup$
    – A.Γ.
    Nov 25 '16 at 13:21
  • $\begingroup$ Well, with the new function defined in the answer, we have the good holomorphic domain. But are we agree that with, $f(z) = \sqrt{z^2-1}$, it doesn't work ? Are people usually making an "abuse of langage" by talking of $\sqrt{z^2-1}$ instead of the function defines by R. Israel ? $\endgroup$
    – C. Dubussy
    Nov 25 '16 at 13:27
  • $\begingroup$ It all depends on what you mean by the square root sign. If you mean “principal square root” then $\sqrt{z^2-1}$ will just be what it is, and have a jump singularity at those points where $z^2-1$ is a negative real number. But “a branch of $f(z)=\sqrt{z^2-1}$” may in many contexts also mean “a function such that for every $z$ the value $f(z)$ is one of the two possible square roots of $z^2-1$, and one has chosen the square root individually for each $z$ in order to make $f$ as nice as possible globally”. $\endgroup$ Nov 25 '16 at 14:25
  • $\begingroup$ By the way, I happen to have made a picture of this particular function for a talk I've given a few times. It's on page 35 of these slides, where $Q(z)$ denotes the principal square root of $z$. The text is in Swedish, but the pictures might be of interest anyway... $\endgroup$ Nov 25 '16 at 14:29
  • $\begingroup$ Very useful image, thank you. In my country, we are not used to see "branch theory", we only use the principal argument and the functions related to it, but this new point of view is pretty helpful. $\endgroup$
    – C. Dubussy
    Nov 25 '16 at 17:16

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