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I have seen a lot of arguments recently where, for instance: $$ \sqrt{z^2-1} = \sqrt{z-1} \cdot \sqrt{z+1} \hspace{5mm} (*) $$ without ever specifying the chosen branch or the chosen branch cut (these expressions were later used in integrals so I am fairly sure that we need to make them single-valued and we are thus not referring to a simple equality between sets of values).

I would like to know if I am overlooking something fundamental when I am saying that $(*)$ is not correct (or at least not complete without specifying the branch and the branch cut) if we want to integrate say $\sqrt{z^2-1}$ itself.

Moreover, I would like to know how can I evaluate $\sqrt{z^2-1}$, having chosen the branch cut $z \notin [-1,+1]$ and the positive real part branch, as $z$ approaches the cut from below and from above (I can easily do this if I only had $w = \sqrt{z}$ but now it seems to me that there is "another plane" to consider between the z-plane and the w-plane before I can evaluate the expression).

Matteo Menghini
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  • I think it is hard to give a general answer, I would be interested in some examples of arguments that use the identity (*). – Maximilian Janisch Apr 06 '23 at 16:46
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    in any domain where $\sqrt{z^2-1}, \sqrt{z-1} \cdot \sqrt{z+1}$ are analytic functions, we can have only two relations between them, $()$ or $-()$ since if we call LHS, RHS $u(z), v(z)$ we have that $u(z)^2=v(z)^2$ so being analytic functions in a domain this implies either $u(z)=v(z)$ everywhere hence $()$ or $u(z)=-v(z)$ everywhere so $-()$. In particular, if $()$ holds at a point, it holds everywhere so it's possible that the branches were chosen in such a way that $()$ holds at some given point in the domain implicitly, and then we are allowed to use it everywhere – Conrad Apr 06 '23 at 16:55
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    @MaximilianJanisch For instance I have seen people integrate $\int_{\gamma} \frac{1}{\sqrt{z^2-1}}dz$ where $\gamma$ is a circle without specifying the branch they are using. – Matteo Menghini Apr 06 '23 at 16:56
  • @Conrad What you say makes absolutely sense to me. Is it also correct to say that if a value of $\sqrt{z^2-1}$ is given then we can choose the branches of $\sqrt(z-1)$ and $\sqrt(z+1)$ so that $(*)$ is true? Does this also determine the branch cut? – Matteo Menghini Apr 06 '23 at 17:00
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    for the first, it is true with the caveat that $\sqrt {z^2-1}$ may generally have different branches than $\sqrt {z-1}$ (taken separately and similarly with $\sqrt {z+1}$ taken separately) since their branch singularities differ (one has $1,-1$ with $\infty$ as a simple pole, the other $1, \infty$); so I wouldn't really try and make a general statement, but look at the particular situation and see what happens; for example if we are in the half plane $\Re z>1$ where all three functions above have unambiguous branches ($\pm$) then $(*)$ entails free choice of either branch for two of the three – Conrad Apr 06 '23 at 17:56
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    and for the case above $\Re z>1$ you can specify the branches by the value at any point - eg the value at $2$ which is either a positive or a negative real for each of the three – Conrad Apr 06 '23 at 18:04

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