Branch point at infinity?

Question

I have to find the branch points of $f(z)=\left( z(z+1)\right )^{1/3}$. It is clear that $0$ and $-1$ are branch points, but I am not sure about infinity. Making the substituition $x=\frac{1}{z}$ and examining $f(x)=\left(\frac{1}{x}\left(\frac{1}{x}+1\right)\right )^{1/3}$, I do not really see what happens as x tends to 0, it just seems like the whole thing blows up. (Maybe this just simply means that infinity is not a branch point?)

Answer 1

The answer by @Stacks above appears to contain a mistake. The function $(1+x^{-1})^{1/3}$ has a branch point at zero. Indeed, consider a small circular contour around 0: $z=\frac{1}{\rho}e^{i\theta}$, $|\rho|>1$, $\theta \in [0, 2\pi)$. We can represent $$(z^{-1}+1)^{1/3}=\exp(\ln((z^{-1}+1)^{1/3}))=\exp\left(\frac{1}{3}\ln\left(1+z^{-1}\right)\right)$$ If $z=\frac{1}{\rho}e^{i\theta}$, $z^{-1}=\rho e^{-i\theta}$, and since the contour in the $z$-plane is taken arbitrary small, $\rho$ is arbitrary large and so is the circular contour $\rho e^{-i\theta}$ in the $z^{-1}$ plane. For sufficiently large $\rho$, $1+\rho e^{-i\theta}\approx \rho e^{-i\theta}$, and the contour contains 0 inside it. But 0 is the branch point of the logarithm, which, if we make a cut from 0 to infinity along $\mathbb{R}_+$, changes its value from $\ln|\rho|$ at the upper bank of the cut to $\ln|\rho|-2\pi i$ at the lower bank. The value of the function $\exp\left(\frac{1}{3}\ln(1+z^{-1}) \right)$ thus changes from $\exp(\frac{1}{3}\ln|R|)$ to $\exp(\frac{1}{3}\ln|R|-\frac{2\pi i}{3})$, i.e. obtains a complex phase. Thus, we have a branch point at 0.

Answer 2

When you make the substitution $x=1/z$, you try to see if $x=0$ is a branch point, which would mean $z=\infty$ is a branch point. In this case, we get $$f(x) = x^{-1/3}\left(x^{-1} +1 \right)^{1/3}.$$

The part on the right, $(x^{-1}+1)^{1/3}$, has a branch point when $x=-1$, but not at $x=0$. The leading part, $x^{-1/3}$, does have a branch point at $x=0$. This means $z=\infty$ is a branch point.

If you want to think of it as investigating the behavior of $f(x)$ near zero, we could argue that $$ f(x) = x^{-2/3} (1+x)^{1/3},$$ so that near $x=0$ we have $f(x) \approx x^{-2/3}$. This has a branch point at $x=0$, again showing that $z=\infty$ is a branch point.