So, I'm trying to classify the singular points of the following function:
$$ f(z)=e^{\cot(\frac {1}{z})} $$
Obviously, when z is zero, the function tends to approach infinity, so that must be a singularity. To characterize the singularity, I think that I must expand it out into a Laurent series and see what it looks like. Here's what I did:
$$\cot(z)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\cdots$$
$$\cot\left(\frac{1}{z}\right)=z-\frac{1}{3z}-\frac{1}{45z^3}+\cdots$$
Let's call this function above A for simplicity!
$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots$$
$$e^A=1+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots$$
Since A consists of an infinite number of z's with negative powers increasing each term, we notice two things. First, $z=0$ is indeed a singularity. Second, does this allow us to conclude that the singularity $z=0$ is an isolated, essential singularity? Is there any other singularity, such as infinity?
EDIT: So, to consider $z=infinity$, I considered $f(\frac{1}{z})$ at the point $z=0$. Expanding out this, we end up with:
$$\cot(z)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\cdots$$
Let's call this expansion B!
$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots$$
$$f(\frac{1}{z})=e^{\cot(z)}=1+B+\frac{B^2}{2!}+\frac{B^3}{3!}+\frac{B^4}{4!}+\cdots$$
Thus, that $\frac{1}{z}$ term in B will continually increase in power as the terms go further out. For this reason, I think that it's also an isolated,essential singularity. What are your thoughts; am I thinking of this correctly? Are there any other singularities?
Some comments on the part of your work studying $\cot$.
The series for $\cot z$ that you write down is not defined everywhere: that Laurent series only converges for the annulus $0 < |z| < \pi$. Correspondingly, the Laurent series for $\cot(1/z)$ you wrote is only valid for $\frac{1}{\pi} < |z| < \infty$. In particular, it's not defined near $0$, and thus can't tell you if $\cot(1/z)$ has any singularities at $z=0$ or if so, what type they are.
In fact, we can see that $\cot(1/z)$ doesn't even have a Laurent series in any neighborhood of $0$: any annulus $0 < |z| < R$ must include a pole of $\cot(1/z)$, and thus cannot be the annulus of convergence of a Laurent series for $\cot(1/z)$. This fact let us conclude $0$ is an essential singularity of $\cot(1/z)$.
Note that any neighborhood of $z=0$ also contains infinitely many zeroes of $\cot(1/z)$ as well: you can't say that this function converges to $+\infty$ as $z \to 0$. In fact, if it did converge to $+\infty$, then $z=0$ would have to be a pole!