Characterizing the following function's singularities: $\frac{z^{1/3}+1}{z+1}$ I'm tasked with characterizing the singularities (pole, essential, branch point, cluster, ...) of
$$\frac{z^{1/3}+1}{z+1}$$
There is singularity at $z=-1$, a result of the denominator, and also branch point singularities at $z=0,\infty$ resulting from $z^{1/3}$. I would typically approach such a problem by expanding $\frac{1}{z+1}$ as a Laurent series, however, I'm unsure how to approach this with the addition of the multivaluedness. How should I proceed?
 A: You can't get Laurent series of $f(z) = \dfrac{z^{1/3}+1}{z+1}$ on $z=-1$.
This is because $z^{1/3}$ isn't continuous for $z\in\mathbb{R}, z<0$. Even worse, for those values of $z$, Taylor series of $z^{1/3}$ doesn't exist.
An interesting way to prove the non-continuity of $z^{1/3}$ is testing what is called directional limit or limit oriented. This concept is very similar to the concept of One-Sided Limit, but now extends this concept to any complex direction. I've talked a bit about this in this answer, with an interesting example application, too.
Thus, while one-side limits:
$$
\begin{align}
\lim_{z^{+}\to -1} f(z) &= 0 &&& \lim_{z^{-}\to -1} f(z) &= 0
\end{align}
$$
You might suppose that $f(z)$ is continuous on $z=-1$. However, when taking directional limit (one-side limits, but now from the imaginary values):
$$
\begin{align}
\lim_{z^{+i}\to -1} f(z) &= \lim_{x^{+}\to 0} f(-1+ix) = 0.5 + i 0.86605...   \\
\lim_{z^{-i}\to -1} f(z) &= \lim_{x^{-}\to 0} f(-1+ix) = 0.5 - i 0.86605...
\end{align}
$$
Therefore, $f(z)$ isn't continuous at $z=-1$. Same for $z\in\mathbb{R}, z<0$.
