Residues
Objective. Calculate residues by multiple methods.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
The residue of f at an isolated singularity z₀, written Res(f,z₀), is the coefficient a₋₁ in the Laurent expansion of f about z₀, and it can be computed by several methods depending on the singularity type. For a simple pole, Res(f,z₀)=lim_{z→z₀}(z−z₀)f(z); for a pole of order m, Res(f,z₀)=(1/(m−1)!)lim_{z→z₀}d^{m-1}/dz^{m-1}[(z−z₀)^m f(z)]. For quotients f=g/h with h having a simple zero at z₀, Res(f,z₀)=g(z₀)/h'(z₀). For example, f(z)=e^z/(z−1) has a simple pole at z=1 with residue e^1/1=e.
Worked example
Find the residue of f(z) = z/((z−2)(z+1)) at its pole z=2.
- 1z=2 is a simple pole since (z−2) appears to first power
- 2Res = lim_{z→2}(z−2)f(z) = lim_{z→2} z/(z+1)
- 3Substitute z=2: 2/(2+1)
- 4Res = 2/3
Answer: 2/3
Concept mastery
Based on your past attempts at this lesson — the question types to practise again come first.
Complete a practice set to see your concept breakdown here.
Practice
Practice happens on its own screen, one question at a time. Answers stay hidden until you submit yours, and you can stop and pick up at the same question later.
Start practisingCommon mistake. Students apply the simple-pole formula Res=lim(z−z₀)f(z) to higher-order poles without differentiating, giving wrong answers; for f(z)=1/(z−1)², the naive limit gives 0, but the correct residue using the derivative formula is also 0 here — yet for 1/z(z−1)² misapplying the simple formula at z=1 gives the wrong finite value instead of using the derivative method.
Your progress
Tick each step as you finish it. Your exact reading position is saved automatically, so Continue drops you back on the same line.
0 of 5 steps complete
Your place saves automatically as you read.