Recipe

Find the residue at a pole of finite order

Use when: you've confirmed z_0 is a pole of finite order m and need \operatorname{Res}_{z=z_0}f

Goal: the residue \operatorname{Res}_{z=z_0}f, computed from a finite number of derivatives instead of a full Laurent expansion

Taught in The pole order and residue shortcut formula

1. 1

   Use the classification recipe from the previous section, or given facts, to show that z_0 is a pole. Then find its order m.

   Example

   For f(z)=\dfrac{z^2+1}{(z-2)^2} at z_0=2: the denominator has (z-2)^2 exactly. The numerator z^2+1 is analytic and nonzero there, since 2^2+1=5. So the pole has order m=2.

   Why: every later step needs m exactly. It sets the power to multiply by and the number of derivatives to take.

2. 2

   Multiply by (z-z_0)^m to build \phi(z)=(z-z_0)^mf(z). It should now be analytic and nonzero at z_0.

   Example

   \phi(z)=(z-2)^2\cdot\dfrac{z^2+1}{(z-2)^2}=z^2+1, analytic everywhere, with \phi(2)=5\ne0.

   Why: this is exactly the algebraic move that clears every negative-power term of the principal part at once, leaving an ordinary Taylor series.

3. 3

   When m=1, the residue is just \phi(z_0). No differentiation is needed.

   • m=1 — the residue is \phi(z_0). Skip ahead to Step 5.
   • m>1 — continue to Step 4 and differentiate.

   ⚡ Shortcut

   Say f=p/q and q has a simple zero at z_0. Then skip building \phi and use \operatorname{Res}_{z=z_0}f=\dfrac{p(z_0)}{q'(z_0)} directly.

   Why: for m=1, the value of \phi at z_0 already is the coefficient the definition asks for. No derivative sits between them.

4. 4

   Differentiate \phi exactly m-1 times. Evaluate the result at z_0. Then divide by (m-1)!.

   Example

   Take \phi(z)=z^2+1 and m=2. Then \phi'(z)=2z, so \phi'(2)=4, and \operatorname{Res}_{z=2}f=\dfrac{4}{1!}=4.

   Why: the residue c_{-1} sits at power (z-z_0)^{m-1} in the Taylor series of \phi. The ordinary coefficient formula for that power is \phi^{(m-1)}(z_0)/(m-1)!.

5. 5

   Expand f directly in powers of (z-z_0). Check that the coefficient of (z-z_0)^{-1} matches the answer from Step 3 or Step 4.

   Example

   Writing w=z-2: f=\dfrac{w^2+4w+5}{w^2}=1+\dfrac4w+\dfrac{5}{w^2}, so c_{-1}=4, matching Step 4 exactly.

   Why: the shortcut and the full expansion compute the same coefficient. So agreement between them is a real check.