At $z=0$ the function $f(z)=\exp({z\over 1-\cos z})$ has
$1$. A removable singualrity
$2$. A pole
$3$. An essential singularity
$4$. Laurent series around $z=0$ has infinitely man positive and negative power of $z$
I see $\lim_{z\to\infty}f(z)=\infty\cup -\infty$ if you approach to $0$ from left and right side of the real line, so $f$ can not be bounded near $0$ so it has $f$ essential singularity?
Best Answer
How do you know it isn't a pole? For small $z$, $\frac{z}{1-\cos z}\approx \frac{z}{z^2}=\frac{1}{z}$, so that for small $z$ your function is $\displaystyle e^{\frac{1}{z}}=\Sigma \frac{1}{n!z^n}$, and this clearly goes to infinity faster than any polynomial goes to $0$, ie. there is no $m$ such that $\,\,\displaystyle z^m\Sigma \frac{1}{n!z^n}$ is bounded near $0$, hence it is an essential singularity.