Branch cuts for $(z^2+1)^{1/3}$

Question

I'm just learning about branch cuts so I'm hoping to get some clarification on this.

As in the title, I'm looking at $f(z)=(z^2+1)^{1/3}$. The obvious way to write this is $f(z)=\exp(\frac{1}{3}\ln(z^2+1))$. There are two main candidates for my branch cut (of $\ln w$) here: $A:[0,-\infty), B:[0,\infty)$. I know I can choose any ray from the origin for $\ln w$, but I think these are the only 2 interesting ones.

If I choose $A$ as my branch cut then that corresponds to branch cuts of $[i,i\infty)\cup[-i,-i\infty)$ in $f(z)$. I believe this is the natural choice as it uses the principle branch of the logarithm and also a 'nice' branch for $f$.

On the other hand, if I choose $B$ as the branch cut, this corresponds to $[-i,i]$. This is where I'm slightly confused, from thinking about what happens to specific regions geometrically, it seems there is also a requirement for a branch cut along $(-\infty,\infty)$.

I was under the impression that branch cuts should be between branch points, but isn't the branch points of this function only $\pm i$? In this case why is there also one along the entire real axis? Is this just a demonstration of why the $A$ cut is more natural?

Answer 1

Having thought about this a lot more I think I have an explanation.

If one considers the behaviour of $\lim_{|z|\to\infty} f(z) \sim z^\frac{2}{3}$. Here we clearly have a branch point at $\infty$ as well as the ones previously found in the question.

In branch cut $A$, our choice of cut going through infinity is important as this then includes our new branch point. Originally it was thought that the path through infinity was similar to in the case of $g(z)=(1+z^2)^\frac{1}{2}$ where the principle branch cut 'happens' to go through infinity but isn't required to . It turns out this is not the case¹.

When we then move to branch cut $B$, we no longer have this branch point at infinity on our branch cut. This is why it is necessary to include the $(-\infty, \infty)$ branch cut. Presumably any additional branch cut through infinity would also suffice.

Hopefully this answer clears up this question if anyone else has it. If I am wrong or unclear about anything then please let me know and I'll have another look and hopefully understand it better afterwards.

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^{1. This can be seen by noticing that the asymptotic behaviour of $g$ is clearly different $\lim_{|z|\to\infty} g(z) \sim z$.}