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A point where a specified complex function fails to be holomorphic, but which is approached by points of its domain. An isolated singularity has a punctured neighbourhood on which the function is holomorphic; a non-isolated singularity has other singularities arbitrarily close to it. The classification as a removable singularity, a pole, or an essential singularity applies to isolated singularities.

Away from zero, singularities occur exactly when $e^{1/z}=-1$, namelyFor $D(z)=z(e^{1/z}+1)$, differentiation at $z_k$ gives $D^{\prime }(z_k)=1/z_k\ne 0$. Thus every $z_k$ is a simple pole, with residue $z_k$.

As $|k|\to \infty$, these poles accumulate at zero. Therefore zero is a non-isolated singularity. Every punctured disc around it contains poles, so the function is not holomorphic on any punctured disc there. In particular zero cannot be classified as an isolated essential singularity; the isolated-singularity classification does not apply. The displayed poles and their accumulation point zero exhaust the finite singularities.

For $z\ne 0$, the denominator zeroes are $z_n=1/[\pi (n+\frac{1}{2})]$. Its derivative is $\sin (1/z)/z^2$, nonzero at each such point. ThusThese poles accumulate at zero. Consequently 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.

At every $z_k$, the function $\tan (1/z)$ has a simple pole. The exponential of a pole is an essential singularity, soare essential singularities of $\exp (\tan (1/z))$.

These points accumulate at $0$. Therefore no punctured neighbourhood of zero is a domain of holomorphy for the function: zero is a non-isolated singularity, specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in $\mathbb{C}$.

The complex exponential never vanishes. The zeros of $\sinh z$ occur at $z=n\pi i$, $n\in \mathbb{Z}$, and they are simple because $\cosh (n\pi i)=(-1{)}^n\ne 0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. ThereforeFor example their residues are $1/\cosh (n\pi i)=(-1{)}^n$.

Infinity requires care: the poles themselves tend to infinity. Equivalently $f(1/w)$ has poles at $w=1/(n\pi i)$ accumulating at zero. Thus $\boxed{\infty \text{ is a non-isolated singularity}}$, and it is not a zero. In the usual classification of isolated singularities, it cannot be called a removable singularity, pole or isolated essential singularity. The non-isolated singularity distinction is necessary even though the function is meromorphic at every finite point.

The function $1/\cos (1/z)$ has simple poles at $z=1/[\pi (n+1/2)]$, accumulating at zero. Its origin is therefore a non-isolated singularity, despite the underlying cosine of $1/z$ having an isolated essential singularity.