How do I classify an isolated singularity, and what follows from each type?
Removable, pole, or essential — what the Laurent series' principal part tells you, and the shortcut (or lack of one) each type earns.
Reach for this the moment you've found a point where f isn't analytic but IS analytic on some punctured disk around it. Expand (or inspect) the Laurent series there and look at the principal part — its length is the whole diagnosis.
Start here — what does the principal part look like?
This is the entire decision variable below — empty, finite, or infinite.
Principal part is empty -> removable singularity
Example
f is bounded near z0, extends to an analytic function AT z0, and its residue is always 0 — nothing to compute.
Principal part is finite -> pole of order m
Term
Simple pole
z_0 is a pole of order m (a positive integer) of f if and only if f(z) = \frac{\phi(z)}{(z-z_0)^m} for some function \phi analytic at z_0 with \phi(z_0)\neq0. In that case the residue of f at z_0 is \operatorname{Res}_{z=z_0} f = \begin{cases} \phi(z_0) & m=1 \\[4pt] \dfrac{\phi^{(m-1)}(z_0)}{(m-1)!} & m>1. \end{cases}
The general shortcut once the order m is known — a derivative, not a full series expansion.
Show proof
By definition, z_0 is a pole of order m exactly when the Laurent series of f on a punctured disk around z_0 has c_{-m}\neq0 as its lowest-indexed nonzero coefficient, i.e. f(z) = \frac{c_{-m}}{(z-z_0)^m} + \frac{c_{-m+1}}{(z-z_0)^{m-1}} + \cdots + \frac{c_{-1}}{z-z_0} + c_0 + c_1(z-z_0)+\cdots with c_{-m}\neq0. Multiplying through by (z-z_0)^m gives \phi(z) := (z-z_0)^m f(z) = c_{-m} + c_{-m+1}(z-z_0) + \cdots + c_{-1}(z-z_0)^{m-1} + c_0(z-z_0)^m+\cdots, which is exactly a convergent power series in (z-z_0) on that same punctured disk, extendable to z_0 itself by the same series (a removable singularity), so \phi is analytic at z_0 with \phi(z_0)=c_{-m}\neq0. This proves the forward direction, and gives f(z)=\phi(z)/(z-z_0)^m. Conversely, if f(z)=\phi(z)/(z-z_0)^m with \phi analytic and nonzero at z_0, then \phi's own Taylor series \phi(z) = \sum_{k\ge0} \frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^k divided by (z-z_0)^m produces a Laurent series for f whose lowest term is \phi(z_0)/(z-z_0)^m, nonzero since \phi(z_0)\neq0, so z_0 is a pole of order exactly m.
For the residue, \operatorname{Res}_{z=z_0} f = c_{-1}, the coefficient of (z-z_0)^{-1} in the Laurent series above — equivalently the coefficient of (z-z_0)^{m-1} in the power series for \phi shown just above. By the ordinary Taylor coefficient formula, that coefficient is \phi^{(m-1)}(z_0)/(m-1)!. When m=1 this reads \phi^{(0)}(z_0)/0! = \phi(z_0), matching the stated case split (the m>1 formula already covers m=1 in this form, but the simple pole case is recorded separately since it is by far the most common).
Suppose f(z)=\dfrac{p(z)}{q(z)} with p and q analytic at z_0, q(z_0)=0, q'(z_0)\neq0, and p(z_0)\neq0. Then f has a simple pole at z_0, and \operatorname{Res}_{z=z_0} f = \frac{p(z_0)}{q'(z_0)}.
Example
An even faster shortcut for the common case f = p/q with a simple zero of q at z0.
Show proof
Since q(z_0)=0 and q'(z_0)\neq0, z_0 is a simple (order-1) zero of q, so q factors as q(z)=(z-z_0)h(z) for some function h analytic at z_0 — divide q(z) by z-z_0 using q's own Taylor series at z_0, whose constant term is q(z_0)=0, so z-z_0 divides it exactly. Comparing derivatives at z_0: q'(z) = h(z) + (z-z_0)h'(z), so q'(z_0)=h(z_0).
Substituting, f(z) = \dfrac{p(z)}{(z-z_0)h(z)} = \dfrac{\phi(z)}{z-z_0} where \phi(z):=p(z)/h(z) is analytic at z_0 (since h(z_0)=q'(z_0)\neq0) and \phi(z_0)=p(z_0)/h(z_0)\neq0 (since p(z_0)\neq0). This is exactly the pole-order-and-residue formula's m=1 case, with residue \phi(z_0): \operatorname{Res}_{z=z_0} f = \phi(z_0) = \frac{p(z_0)}{h(z_0)} = \frac{p(z_0)}{q'(z_0)}.
Principal part never terminates -> essential singularity
Example
Wild, dense behavior in every punctured neighborhood — no shortcut formula exists; the residue is whatever the c_{-1} coefficient of the full Laurent series happens to be.