Classifying essential singularities

Question

I want to figure out the process for showing why the function $\cos(1-\frac{1}{z})$ has an essential singularity at $z=0$ without using knowledge of the Laurent expansion. I know the process should be to rule out the possibility of removable singularities or poles, but do not know how to do this for this function.

Attempt I was thinking I would show since $$\lim_{z\to 0} |\cos(1-\frac{1}{z})| \text{ DNE } $$ since the function oscillates between $1$ and $-1$ for $z$ near zero for positive values, this rules out the possibility of a pole since the limit is not $\infty$ and the singularity is not removable since the limit is not finite.Is this the correct approach? What are some other ways, to show that zero is an essential singularity?

Answer 1

Let $z_n:= \frac{1}{1-n \pi}$ for $n \in \mathbb N$ and $f(z):= \cos(1-1/z).$

Then $z_n \to 0$ as $n \to \infty$ and

$$ f(z_n)= (-1)^n$$

for all $n$. This shows that $z=0$ is not a pole of $f$ and not a removable singularity of $f$.

Answer 2

$$\cos\left(1-\frac1z\right)=\cos1\cos\frac1z+\sin1\sin\frac1z =\cos1\sum_{n=0}^\infty\frac{(-1)^nz^{-2n}}{(2n)!} +\sin1\sum_{n=0}^\infty\frac{(-1)^nz^{-2n-1}}{(2n+1)!}.$$ This is a Laurent series expansion with infinitely many terms with negative powers, so the function has an essential singularity at $z=0$.

Answer 3

Here is an attempt: $\lim_{z\to 0}f(z) =\lim_{z\to 0}\cos(1-1/z)$ and $\lim_{z\to 0}1/f(z)=\lim_{z\to 0}1/\cos(1-1/z)$

both do not exist. Can we conclude now the singularity $0$ is an essential singularity?