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?
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.
(Apologies that this is likely three years too late...)
Let our function be $f(z)$ and recall that a function $f(z)$ has a branch point at $\infty$ if $f(\frac{1}{\zeta})$ has a branch point at $\zeta = 0$. We have $$f\left(\frac{1}{\zeta}\right) = \left(\frac{1}{\zeta} \right)^{\frac{1}{n}} \left(\frac{1-\zeta}{\zeta} \right)^\frac{1}{n}.$$ Let $\zeta = re^{i \theta}, \ 1- \zeta = R e^{i \Theta}$, then $$f\left(\frac{1}{\zeta}\right) = (r^{-2})^\frac{1}{n} R^{\frac{1}{n}} e^{i \cdot \frac{1}{n} (-2 \theta + \Theta)}.$$ Now, consider any small path encircling $\zeta= 0$ and apply the same argument given in the .pdf; it should be clear why $n=2$ doesn't introduce a discontinuity as $\theta$ jumps from $-2\pi$ to $0$, but $n=3$ does.
