Branch points of $z^\frac{1}{2} (z-1)^\frac{1}{2}$ and $z^\frac{1}{3} (z-1)^\frac{1}{3}$

Question

I am reading about branch points from here: https://www-thphys.physics.ox.ac.uk/people/FrancescoHautmann/ComplexVariable/s1_12_sl4.pdf

On Page 7, it mentions that branch points of $z^\frac{1}{2} (z-1)^\frac{1}{2}$ are at $z=0$ and $z=1$, but branch points of $z^\frac{1}{3} (z-1)^\frac{1}{3}$ are at $z=0$, $z=1$ and $z=\infty$

I don't understand why the second function has a branch point at $z=\infty$, but the first one does not. Can someone please explain?

Answer 1

You can build a Laurent series for an appropriate branch of $(z(z-1))^{1/2}$, thus:

$(z(z-1))^{1/2}=z(1-1/z)^{1/2}=z-1/2-(1/8z)- ...$

which converges for $|z|>1$. This identifies $\infty$ as a simple pole instead of a branch point. But when the exponent is $1/3$ instead of $1/2$ the analogous series would begin with $z^{2/3}$, which isn't proper for a Laurent series thus indicating a nonisolated singularity.