# Classifying essential singularities

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?

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 $$f$$.

• Thanks. I am still confused on why you choose $z_n=\frac{1}{1-n \pi}$ to be the sequence you take the limit to approach zero as?
– user764658
Commented Apr 14, 2020 at 19:53
• Is it because $z_n \rightarrow 0$ as $n \rightarrow \infty$?So the sequence of points is approaching the value of interest?Also in general should you show two different sequences approach two different values as they approach $0$?But in this case it isn't necessary because the one sequence does not exists?
– user764658
Commented Apr 14, 2020 at 19:59

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

• I need to find ways to show it without using laurent series but thank you.
– user764658
Commented Apr 14, 2020 at 4:48

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?