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?