an essential singularity
An essential singularity is the wild child of the family. At a removable point the function is secretly fine; at a pole it blows up cleanly to infinity. At an essential singularity the function does neither — it has no limit at all, behaving so erratically near the point that it comes arbitrarily close to EVERY complex value, over and over, in any tiny neighbourhood. The standard example is e^(1/z) at 0.
In Laurent terms the diagnosis is immediate: an isolated singularity is essential precisely when the principal part has INFINITELY many nonzero negative terms — the series of negative powers never stops. For e^(1/z), substitute w = 1/z into e^w = 1 + w + w^2/2! + ... to get 1 + 1/z + 1/(2! z^2) + 1/(3! z^3) + ..., an endless string of negative powers. Because there is no smallest negative power, there is no finite order, and the controlled blow-up of a pole is replaced by genuine chaos: along z = 1/t with t real and positive, e^(1/z) shoots to infinity, but along z = -1/t it shrinks to 0, and along the imaginary axis it stays on the unit circle.
Essential singularities matter because they show the limits of how nicely complex functions can behave. The Casorati-Weierstrass theorem makes the chaos precise (the image of any neighbourhood is dense in the plane), and the great Picard theorem sharpens it astonishingly (every value, with at most one exception, is actually attained infinitely often). Recognizing one warns you that no naive limit, and no finite-order residue trick, will tame the point — though the residue a_{-1} still exists and still governs integration.
f(z) = e^(1/z) at z_0 = 0 has Laurent series 1 + 1/z + 1/(2 z^2) + 1/(6 z^3) + ..., with infinitely many negative powers — an essential singularity. To hit the value 1, solve e^(1/z) = 1: any z = 1/(2 pi i n) works, and these cluster at 0, so f equals 1 infinitely often in every neighbourhood of 0.
Infinitely many negative powers, and the value 1 attained infinitely often near 0 — the hallmark of an essential singularity.
Infinitely many negative terms must be a genuine feature of the function, not an artifact of expanding in the wrong annulus — use the Laurent series on the innermost punctured disk 0 < |z - z_0| < r to classify.