I'm trying to find what kind of singularity is $z=0$ in $\dfrac{z\cos\left(\frac{1}{z}\right)}{\cos(z)-1}$
But the Laurent's serie of that function is pretty hard to calculate. Exist other way to see what kind of singularity is $z=0$?. I think that is a Essential singularity, but i don't know how to justificate that.
Since $z/[\cos(z)-1]$ is analytic at $z=0$, only the $\cos(z^{-1})$ part matters. You can show that $\cos(z^{-1})$ has an essential singularity at $z=0$ from the fact that $z^n \cos(z^{-1})$ has a singularity there for all $n\ge 0$.