= Solution
For $z\ne0$, the denominator zeroes are $z_n=1/[\pi(n+\tfrac12)]$. Its <derivative> is $\sin(1/z)/z^2$, nonzero at each such point. Thus
$$
\boxed{\text{simple poles at }z_n=\frac1{\pi(n+\tfrac12)},\quad n\in\mathbb Z.}
$$
These poles accumulate at zero. Consequently \b[zero is a <non-isolated singularity>], as described by <reciprocal cosine with accumulating poles>. It is not an isolated essential singularity: no punctured disc about zero is a domain of holomorphy for this reciprocal function. Although $\cos(1/z)$ itself has an <essential singularity> at zero, taking its reciprocal introduces this accumulation of poles.
Back to article page
