Recently I've been working on branch cuts of square root functions, and come to problems like this:
Find a single-valued analytic branch $f$ of $\sqrt{z^2+z}$ on the set $\{ z \in \mathbb{C}: |z| >1 \}$ such that $f(2) = -\sqrt{6}$. Evaluate the integral of $f$ around the contour $|z| =2$ (positively oriented) using the Laurent series of the square root.
I can find an $f$: the function $-\sqrt{z}\sqrt{z+1}$ where the square roots are the principal branch of the square roots function.
However, the only example of using Laurent series I've seen (in these notes) for contour integrals like this works by factoring a $z$ out of a square root; i.e. writing $ (z^2+z)^{1/2} = z(1+1/z)^{1/2} $ and then using the Laurent expansion for the (principal branch) square root $\sqrt{z}$ about $z=1$ to expand $(1+1/z)^{1/2} $.
My question is about the signs here. The approach above assumes that $\sqrt{z^2} = z$, but it seems that in my branch of the square root, I have $\sqrt{z^2} = -z$.
And then the square root I'm left with in the factorization is still defined based on the branch of $\sqrt{z^2+z} \;$ I'm using, so would I use the Laurent expansion for $- \sqrt{z}$ instead?
I understand that if both signs need to be negative, there's no net effect on the result; I'm just trying to understand the details of the computation. Any insight would be appreciated.