Singularities of $ {1}/{\cos(\frac{1}{z})}$

Question

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.

Answer 1

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 $f$.

Answer 2

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 $\frac{1}{\cos(\frac{1}{z})}$.