Is this an admissible branch cut for $f(z)=\sqrt{z}+\sqrt{z-1}$?

292 Views Asked by At

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.