Determine the type of each singularity.

Question

I got the following question, and I think I got the correct answer with justification but I am note sure. Can anyone verify if this is correct?

For each of the following functions, find all of it's singularities and determine the type of each singularity. If the singularity is a pole, also find the order of this pole:

1. $\cos(1/z)$

2. $\sin(z)/(1-\cos(z))$

3. $z/\sin(z)$

Here's my attempt at the problem:

1. The following is the Laurent Expansion of $\cos(1/z)$: $$\cos(1/z) = \sum_{n = 0}^\infty (-1)^n\frac{z^{-2n}}{(2n)!}$$ We find that there is a singularity at $z = 0$. Since there are infinitely many terms of negative powers, we find that the singularity at $z = 0$ is an essential singularity.

2. Consider the following function: $$\frac{\sin(z)}{1-\cos(z)}$$ There are singularities whenever $z = 2\pi n$, $n \in \mathbb Z$. Consider the following limit: $$\lim_{z \rightarrow 2\pi n}\Big|\frac{\sin(z)}{1-\cos(z)}\Big| = \infty$$ Therefore, all of these singularities are poles. Consider the following limit: $$\lim_{z \rightarrow 2\pi n} \Big|\frac{z\sin(z)}{1-\cos(z)}\Big| = 2$$ Therefore, each of these poles is a pole of order $1$.

3. There is a removable singularity at $z = 0$ and essential singularities at $z = \pi n$, $n \in \mathbb Z \backslash \{0\}$. For $z = 0$, we can make $f(z) = 1$ to remove the singularity. For $z = n\pi$, there exists no $n$ such that $\lim_{z \rightarrow n\pi}z^nf(z) = C$. Therefore, $z = \pi n$ are essential singularities.