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For the function $f(z)=\sqrt{z}+\sqrt{z-1}$, on the one-hand I feel like since we have two branch points at $z=0$ and $z=1$, we would be able to define an admissible branch cut simply going from $z=0$ to $z=1$, as follows:

enter image description here

However on the other hand, this is the sum of two separately multi-valued functions. Therefore, I believe that I should define a branch cut and branch for $\sqrt{z}$ and $\sqrt{z-1}$ separately. This makes sense to me, because also $z=\infty$ is a singular point for both of these functions, so I should define two branch cuts: one connecting $z=0$ and $z=\infty$, another connecting $z=1$ and $z=\infty$. These branch cuts are for all intents and purposes separate. For example, we could choose the following branch cuts:

enter image description here

As a sidenote though, if our function was instead $f(z)=\sqrt{z}\sqrt{z-1}$, then a choice of overlapping branch cuts would lead to an overall cancellation of phases, giving us an "effective" single branch cut:

enter image description here

[Question.] Based on my rant above, is it logical for us to choose Picture #1 as a branch cut for the function $f(z)=\sqrt{z}+\sqrt{z-1}$? Or does such a "branch cut" actually not make sense (by the logic in paragraph #2)? If yes, is Picture #2 an admissible choice of branch cuts for the function $f(z)=\sqrt{z}+\sqrt{z+1}$?

I am asking this question because a friend of mine has said that we can simply choose the branch cut in Picture #1, but I am arguing that it's more complicated (as per my rant above), and that that is not actually an admissible choice of branch cuts.

Arturo don Juan
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  • Infinity is a branch point too. Cuts are between branch points. – Brevan Ellefsen Oct 19 '19 at 10:04
  • @BrevanEllefsen Ok, in this case though it appears as though we infinity as two separate branch points - from the function $\sqrt{z}$ and $\sqrt{z11}$. So we have four branch points, $z={0,1,\infty,\infty}$. I am arguing that we can't just connect branch points with branch cuts arbitrarily, i.e. that we can't connect ${0,1}$ and ${\infty,\infty}$. We must connect ${0,\infty}$ and ${1,\infty}$. Do you agree? – Arturo don Juan Oct 19 '19 at 18:33
  • Arturo, you are mostly correct but you must be a tad careful. $\infty$ is not a "double branch point" just because you are taking a sum; for example, you wouldn't say $\sqrt{z}$ has one branch point at infinity but $\frac{1}{2} \sqrt{z} + \frac{1}{2}\sqrt{z}$ has two branch points at infinity (they are the same function). The most proper way to do this is to consider the associated Riemann Surface, but a quick argument to answer your question can be found here (your post is perhaps technically a duplicate, but your's has details worth keeping) – Brevan Ellefsen Oct 19 '19 at 21:01

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