Branch cut for $\sqrt{1-z^2}$ - Can I use the branch cut of $\sqrt{z}$?

Solution 1:

When taking a branch of $\sqrt z$, you can choose any ray emanating from the origin. In this case, for $\sqrt{1-z}$ and $\sqrt{1+z}$, we need to choose two rays emanating from $-1$ and $1$, and the author chooses them to be $[-1, \infty)$ and $[1,\infty)$.

This seems to rule out the entire interval $[-1, \infty)$ from being part of the domain. But it can be shown that the "jumps" in the branch cuts of the square root functions cancel on the interval $[1,\infty)$, so we get an analytic function on $\mathbb{C}-[-1,1]$.

It's not too hard, and I invite you to try it as an exercise, that you get an analytic continuation across $[1,\infty)$.

This is how I think about it. Andrew's answer in the comments is fantastic and probably better, though. For a complete theory of making such branch cuts, you need to learn about Riemann surfaces. If I recall correctly, Forster's book on Riemann surfaces has a treatment of such functions, but it requires some background to access.

Solution 2:

If we choose a branch of $\sqrt{z}$ with branch cut along the positive axis $[0,\infty)$, then we won't need to integrate on the Riemann sphere, since $1-z^2\ge 0$ if and only if $1\ge z^2$, which happens if and only if $z\in [-1,1]$.

This is equivalent to Potato's answer, which is a nice geometrical interpretation of the branch cut.