How to analytically continue this function?

Question

I wish to show one solution of $w^{3}=z^{2}+1$ can be analytically continuation around $z=0,w=1$ to two different solutions. It seems the standard way is to consider the points $z=1,z=-1,z=\infty$ and connect them. Once the analytically continuation passed the base point, then we should expect a change of sign by $\omega=e^{\frac{2\pi i}{3}}$. But in practice I found it is quite difficult. For example the Taylor series expansion of $(1+z^{2})^{1/3}$ is quite difficult to compute. At $z=1$ we have $$f(z)=2^{1/3}+\frac{2^{1/3}}{3}(z-1)+\frac{2^{1/3}}{9*2}(z-1)^{2}-\frac{8}{27*6}2^{1/3}(z-1)^{3}+\frac{46}{81*24}2^{1/3}(z-1)^{4}-\frac{170}{243*120}2^{1/3}(z-1)^{5}-\frac{1070}{729*720}2^{1/3}(z-1)^{6}+\frac{38080}{2187*5140}2^{1/3}(z-1)^{7}...$$

Theoretically analytically continue this expression along $z=e^{it}$, passing through the point $\omega$ we expect a change of sign. But it seems very difficult to tell this fact from the "functional element" found above. Even the radius of convergence is not very clear unless we plead to Cauchy's inequality, etc to bound the n-th derivative. What is the routine way of doing this? Did I do something wrong?

Answer 1

There's no need to consider the Taylor expansion and it would be next to impossible to read off any continuation behaviour directly from its expansion around a single point. The only role for Taylor expansion here is to show (theoretically) that $w$ can be continued along any path that avoids the singular points $z \in \{i, -i, \infty\}$.

First a simpler but fundamental case. The function $z^{1/3} = e^{\log(z)/3}$ branches at $z=0$. Continuing a branch along a loop around $z = 0$ in counter clockwise direction picks up a factor $e^{2 \pi i/3}$. This is clearly seen in the explicit logarithmic form where you get $e^{\log(z)/3 + 2 \pi i/3}$.

This process can be applied in your example almost exactly the same. Note that the branch points of $w$ are the zeroes of $z^2+1$, i.e. $\pm i$. Let's look at the branch point $z=i$. There $w$ can be written as

$$ w(z) = (z-i)^{1/3}(z+i)^{1/3}. $$

Now note that any branch of $(z+i)^{1/3}$ is holomorphic on a neighbourhood of $z=i$ so the branch behaviour is entirely due to the $(z-i)^{1/3}$ factor. This factor behaves as described before: going around $z=i$ in counter clockwise direction picks up a factor $e^{2 \pi i/3}$. The point $z=-i$ can be treated similarly.

So the monodromy of $w(z)$ starting at a base point ($z=0$ say) is generated by two loops $\gamma_i$ and $\gamma_{-i}$ going around $i$ and $-i$ respectively. Continuing along either of these $w$ picks up a factor $e^{2 \pi i/3}$.

It is sometimes convenient to also include the loop $\gamma_{\infty}$ around $\infty$. It can be chosen such that it satisfies the relation $\gamma_i \gamma_{-i} \gamma_{\infty} = e$ in the fundamental group and therefore also in the monodromy of $w$. This shows that continuation along $\gamma_{\infty}$ picks up a factor $e^{-4\pi i/3} = e^{2 \pi i3}$ as well (in particular $w$ is singular at $\infty$).

Answer 2

Suppose we have an analytic function $w=w(z)$ defined on some open set $U\subseteq\mathbb C\setminus0$ such that $w(z)^3=z^2+1$ for all $z\in U$. Then $3w^2w'=2z$ and since $w^2=(z^2+1)/w$, we have $$\frac{w'}{w}=\frac{2}{3}\frac{z}{z^2+1}.$$ This is an ODE with its two variables separated which is satisfied by any branch of the function we are trying to continue, and, up to initial conditions, actually solves our problem because it determines its solutions completely —solving it as usual (proceeding formally) gives us the formula $$w(z)=\exp\frac{2}{3}\int_{z_0}^z\frac{\zeta}{\zeta^2+1}\,\mathrm d\zeta.$$ Now let us make sense of this. First, the integral has to start at a point $z_0\in\Omega:=\mathbb C\setminus\{\pm i\}$ and be taken over a curve $\gamma$ going from $z_0$ to $z$, so that the integral makes sense. Once this is taken care of, this formula gives us an analytical continuation of the solution to our equation. It is branched at $i$ and $-i$, because the integrand there has non-zero residues. Monodromy is not difficult to compute.