Removable Singularity Problems at $z = 0$

Question

My problem is to find the radius of convergence $R$ of the function $f(z) = \frac{z}{\sin(z)}$ around the the point $z_{0} = \pi i$. I find the singularities of $f(z)$ are $z = n \pi$ ($n$ is integer). But why $z = 0$ is a removable singularity? If considering another function $g(z) = \frac{z^{2}}{(\sin(z))^{3}}$. At $z = 0$ is a pole. Can you please explain to me how to determine the difference between these two types of singularities.?

Answer 1

The point $0$ is a removable singularity of $f$ because the limit $\lim_{z\to0}f(z)$ exists in $\mathbb C$. In the case of $g$, the limit also exits, but it is equal to $\infty$. Therefore, $g$ has a pole at $0$, instead of a removable singularity.

And there is no such thing as the radius of convergence of a function at a point.