what does it mean for $D(z)$ to be rational here
imgur.com/a/Fyxa7Ok ? I can see that for some branch of $\sqrt{}$ , $D(z)$ is holomorphic in an open disk centered at the origin, and so I understand the part about its maclaurin series having integer coefficients, is this result saying that there is some rational function $\frac{P(z)}{Q(z)}$ with $P,Q \in \mathbb{Z}[X]$ such that $D$ is equal to this in that open disk?