Singularities of $ {1}/{\cos(\frac{1}{z})}$ I would like to determine the singularities of $f$, given by $$f(z) = \frac{1}{\cos(\frac{1}{z})}.$$
It is clear to me that $z = 0$ and $z = \frac{2}{(1+2k)\pi}$ for $k\in\mathbb Z$ are singularities. However, I don't know how to handle by
$$ \cos(\frac{1}{z}) = \sum_{n=0}^\infty \frac{(-1) z^{-2n}}{(2n)!} \;\forall z\neq 0 $$
which are the types of those singularities. I would be very pleased if you give me any hint or explanation. 
 A: The Taylor series for secant is
$$
\sec(z) = \sum_{n = 0}^{\infty}(-1)^n\frac{E_{2n}z^{2n}}{(2n)!}
$$
where $E_{2n}$ is the Euler number. Then 
$$
\sec(1/z) = \sum_{n = 0}^{\infty}(-1)^n\frac{E_{2n}}{z^{2n}(2n)!} = 1 + \frac{1}{2z^2} + \frac{5}{24z^4} + \frac{61}{720z^6} + \cdots
$$
Now, there are three types of singularities. We have removable, poles, and essential. What do we have here?
A removable singularity is defined as: 


*

*If $f$ is bounded in some neighborhood of $z_0$, then one can define
$f(z_0)$ in a unique way such that the function is also analytic at
$z_0$


A pole is defined as: 


*

*$f$ has a pole of finite order $m$ at $z_0$ if and only if $f(z)(z-z_0)^m$
is holomorphic at $z_0$ and has no zero at $z_0$.


Finally, an essential singularity is defined as: 


*

*If the Laurent series has an infinite number of negative terms, then
we say that $z_0$ is an essential singularity of $f$.

A: The answer depends on how you define essential singularities. Usually, the classification in removable singularities, poles, and essential singularities requires isolated singularities. However, since the zeros of the denominator $cos(1/z)$ are, as you pointed out, located at $z_0=\frac{1}{\pi (k + \frac{1}{2})}$, we actually have isolated poles arbitrarily close to $z=0$. Therefore, there's no isolated singularity at $z=0$ and thus a cluster point instead of an essential singularity.
See also Finding the singularity type at $z=0$ of $\frac{1}{\cos(\frac{1}{z})}$.
