the residue formula for a pole of order m
When a function blows up faster than 1 / (z - z_0) — like 1 / (z - z_0)^2 or 1 / (z - z_0)^3 — the simple-pole trick fails, because multiplying by a single (z - z_0) does not fully tame the singularity. For a pole of order m there is a slightly heavier but still mechanical formula, and the new ingredient is differentiation.
The rule: if f has a pole of order m at z_0, then Res(f, z_0) = (1 / (m - 1)!) times the limit as z approaches z_0 of the (m - 1)-th derivative, with respect to z, of (z - z_0)^m f(z). Here is the logic. Multiplying f by (z - z_0)^m clears the pole entirely, turning it into an ordinary holomorphic function g(z) = (z - z_0)^m f(z) whose Taylor series starts a_{-m} + a_{-m+1}(z - z_0) + .... We want a_{-1}, which sits in g as the coefficient of (z - z_0)^(m-1). Taylor's theorem says that coefficient equals g^(m-1)(z_0) / (m - 1)!. So you differentiate g exactly m - 1 times, evaluate at z_0, and divide by (m - 1)!. For m = 1 the derivative is the zeroth one — no differentiation — and the formula collapses to the simple-pole limit.
The formula is foolproof but the differentiation can get messy for large m, so in practice many people short-circuit it by expanding the relevant numerator in a Taylor series and reading a_{-1} directly — often quicker for m = 2 or 3. The two routes always agree; choose whichever has less algebra for the problem in front of you.
For f(z) = e^z / z^2, the pole at 0 has order m = 2. Res(f, 0) = (1 / 1!) times the limit as z -> 0 of d/dz of (z^2 f(z)) = d/dz of e^z evaluated at 0 = e^0 = 1.
Clear the pole with (z - z_0)^m, differentiate m - 1 times, divide by (m - 1)!.
You must use the correct order m. Guessing m too small leaves a residual blow-up; guessing m too large is harmless to the limit but adds needless differentiation. Determine m from the order of the pole first.