The task is to find the singularities of the function $$f(z) = \frac{z^2}{1-e^{z^3}}$$ and decide their type.
So I started by finding the singularities. I got to $z^3 = 2\pi ki$ for every $k \in \mathbb{Z}$.
So:
$$z_0 = \sqrt[3]{2\pi k} e^{\frac{\pi}{6}i}, z_1 = \sqrt[3]{2\pi k} e^{\frac{5\pi}{6}i}, z_2 = \sqrt[3]{2\pi k} e^{-\frac{\pi}{2}i}, z_3 = \sqrt[3]{2\pi k} e^{-\frac{\pi}{6}i}, z_4 = \sqrt[3]{2\pi k} e^{-\frac{5\pi}{6}i}, z_5 = \sqrt[3]{2\pi k} e^{\frac{\pi}{2}i}$$
for every $k \in \mathbb{Z}^+$ and also the point $z=0$.
My problem is to determine their type. I though that $z=0$ is a removable singularity, but calculating the limit at $0$ of the functions using l'hopital's rule gave out the the limit is not finite.
About the other singularities I think that they are poles, since the limit at them is equal to $\infty$. But does it really enough to prove it?
I was not asked to find the order of the poles, but it would be nice if someone can show me how to do it here.
Hep would be appreciated.
For $z \to 0$, $e^{z^3} = 1 + z^3 + O(z^6)$ so $f(z) = z^2/(-z^3 + O(z^6)) = -z^{-1} + O(z^2)$, thus $0$ is a pole of order $1$.
The other singularities are also poles of order $1$, because they are simple zeros of the denominator:
$$ \dfrac{d}{dz}(1 - e^{z^3}) = - 3 z^2 e^{z^3} \ne 0 \ \text{when}\ z \ne 0$$