Finding the singularity type at $z=0$ of $\frac{1}{\cos(\frac{1}{z})}$

Question

I have the following homework problem:

What kind of singular point does the function $\frac{1}{\cos(\frac{1}{z})}$ have at $z=0$ ?

What I tried:

We note (visually) that $z_{0}$ is the same type of singularity for both $f,f^{2}$ hence the type of singularity of $f(z)=\frac{1}{\cos(\frac{1}{z})}$ have at $z=0$ is the same type of singularity $f^{2}(z)=\frac{1}{\cos^{2}(\frac{1}{z})}$. We recall $$ \frac{1}{\cos^{2}(z)}=1+\tan^{2}(z) $$

We also note that for any constant $z_{0}\in\mathbb{C}$ the singularity of $f,f+z_{0}$ are the same, this can be proved by noting the Laurent expansions of both functions are the same, up to an additive constant.

It remains to determine the singularity type at the origin of $$ \tan(\frac{1}{z}) $$

This is where I'm stuck, we didn't study what the Taylor series of $\tan(z)$.

I also know that type of singularity $$\cos(\frac{1}{z})$$ have, but I don't know how to connect this with the singularity type of $$\frac{1}{\cos(\frac{1}{z})}$$

Can someone please hint me in the right direction ?

Answer 1

The singularity is not an isolated singularity, as $\cos(1/z)$ has a sequence of zeroes approaching $0$ (namely $z_n=1/(n\pi+\pi/2)$). In particular, $0$ cannot be an essential singularity of the function.

Answer 2

Notice that neither of the limits $\lim_{z \to 0} \frac{1}{\cos(1/z)}$ and $\lim_{z \to 0} \cos(1/z)$ exist. Therefore the singularity is of essential type.

Edit: The conclusion is wrong since $0$ is an accumulation point of poles. (See Ted Shifrin's answer below.)