The function $f(z) = \frac{z^{1/3} - 1 }{z-1}$ has a branch point at $z = 0$. How do I tell what kind of singularities $f$ has at $z=1$ depending on the branch taken?
The explanation I have been given is that letting $z=re^{i\theta}$ for $0 \leq \theta<2\pi$, $z=1$ is a removable singularity; letting $z = re^{i\theta}$ for $2\pi \leq θ < 4\pi$, $z = 1$ is a simple pole; letting $z=re^{i\theta}$ for $4\pi\leq\theta<6\pi$, $z=1$ is a simple pole.
Why is the singularity in the last two cases a simple pole instead of a removable singularity?
To investigate the singularity at $z=1$, let's change variables $w=z-1$ and investigate the singularity at $w=0$.
$$ \frac{z^{1/3}-1}{z-1} = \frac{(1+w)^{1/3}-1}{w} $$ First case... Using the branch of the cube root with $1^{1/3}=1$, we get as $w \to 0$: \begin{align*} (1+w)^{1/3} &= 1 + \frac{1}{3}w + o(w^2)\\ (1+w)^{1/3} - 1 &= \frac{1}{3}w + o(w^2)\\ \frac{(1+w)^{1/3}-1}{w} &= \frac{1}{3} + o(w) \end{align*} Thus, the singularity at $w=0$ is removable.
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Here is the graph of $\displaystyle \frac{z^{1/3}-1}{z-1}$ from $z=0$ to $2$ ... See the value $\frac13$ at $z=1$.
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Second case: Using the branch with $1^{1/3}= \omega = \frac{-1+i\sqrt{3}}{2}$, we get \begin{align*} (1+w)^{1/3} &= \omega\left(1+\frac{1}{3}w+o(w^2)\right) =\omega + \frac{\omega}{3}\;w + o(w^2) \\ (1+w)^{1/3}-1 &= (\omega-1) + \frac{\omega}{3}\;w + o(w^2) \\ \frac{(1+w)^{1/3}-1}{w} &=\frac{\omega-1}{w} + \frac{\omega}{3} + o(w) \\ &= \frac{-3+i\sqrt{3}}{2}\;\frac{1}{w} + \frac{-1+i\sqrt{3}}{6} + o(w) \end{align*}
Third case: Using the branch with $1^{1/3}=\frac{-1-i\sqrt{3}}{2}$, we get $$ \frac{(1+w)^{1/3}-1}{w} = \frac{-3-i\sqrt{3}}{2}\;\frac{1}{w} + \frac{-1-i\sqrt{3}}{6} + o(w) $$
In these two cases we have poles at $z=1$ with residues that can be read here.