Consider $$ f(z) = \frac{z}{\sin{z}}. $$ I know that $\sin{z}$ has zeros at $ n\pi, n \in \mathbb{Z} $, so $f$ has singularities at those points. For $ z=0 $, I may use expansion $\sin{z} = z + \mathcal{O}(z^3)$, and obtain $ f(z) = \frac{z}{z + \mathcal{O}(z^3)} \to 1 $ as $ z \to 0 $. Therefore, $z=0$ is a removable singularity. For other points of the form $n\pi, n \in \mathbb{Z}$, we may use the fact that $|f(z)| \to \infty$ as $z \to \pm n\pi $, $n \neq 0,$ and find out that at these points, $f$ has simple poles.
If we are to write Laurent Series around the point $z = \pi i,$ $\sum_{n=-\infty}^{\infty} a_n (z-\pi i)^n $ we get that the radius of convergence is the distance from $\pi i$ to the nearest singularity, beyond which is impossible to extend the function analytically. In this case that is, $ R = |\pi i \pm \pi| = \sqrt{2}\pi.$
Now consider the case where $$f(z) = \frac{z^3}{\sin^3{z}}.$$ I would assume the same as above holds in this case, but solution says that in this case $z=0$ is a simple pole and so series have $R = \pi.$
Why isn't $z=0$ removable singularity in this example?
$\mathbf{Edit}:$ The solution in my book was wrong. $z=0$ is indeed a removable singularity in this example as well, but for a function $$\frac{z^2}{\sin^3{z}},$$ $z=0$ will be a simple pole.
$\dfrac{z^3}{\sin^3 z}$ does have a removable singularity at $z=0$. Either the solution is wrong or the problem is misquoted: maybe it should be $\dfrac{z^2}{\sin^3 z}$?