Among all the Laurent coefficients ..., a₋₂, a₋₁, a₀, a₁, ... of a function near an isolated singularity, only a₋₁ is called the 'residue' and given special treatment. Why is this coefficient uniquely important?
AIt is always the largest coefficient in absolute value, so it dominates the behavior near the singularity
BIt is the only Laurent coefficient that is nonzero for poles — the others vanish for meromorphic functions
CIt is the only term in the Laurent series whose contour integral around the singularity is nonzero
DIt determines the order of the pole — a pole of order m has a₋₁ ≠ 0 but a₋m = 0
The key fact is that ∮ (z − z₀)ⁿ dz = 0 for every integer n ≠ −1 (because (z−z₀)ⁿ has an antiderivative), but ∮ 1/(z−z₀) dz = 2πi. So when you integrate the Laurent series term by term, every term except the n = −1 term vanishes. The contour integral of f picks out exactly 2πi · a₋₁ — the residue is the piece of f that survives integration. This is why it deserves a special name: it is the 'residue' left after all other terms integrate to zero.
Question 2 Multiple Choice
What is Res(f, 2) for f(z) = 1 / ((z − 2)(z + 3))?
A−1/5, because the formula gives 1/(2 + 3) with a sign error from the factored form
B1/5, because lim_{z→2} (z − 2) · f(z) = lim_{z→2} 1/(z + 3) = 1/5
C1/2, because the residue at a simple pole is the reciprocal of the value at the pole
DThe residue is undefined because the formula for simple poles only applies when the denominator has a simple zero
z = 2 is a simple pole (the denominator has a simple zero there). Applying the simple-pole formula: Res(f, 2) = lim_{z→2} (z − 2) · 1/((z − 2)(z + 3)) = lim_{z→2} 1/(z + 3) = 1/5. Option A gets the arithmetic right but invents a sign error. Option C confuses the residue formula with the 1/f'(z₀) formula that applies when f = g/h with h(z₀) = 0 and g(z₀) ≠ 0 (which gives the same answer here: g(2)/h'(2) = 1/(2·2+1) = 1/5, consistent). Option D is wrong — the formula applies here.
Question 3 True / False
For a pole of order 2 at z₀, you can compute the residue without finding the full Laurent series by multiplying f(z) by (z − z₀)², differentiating once with respect to z, and then evaluating at z₀.
TTrue
FFalse
Answer: True
This is the order-m residue formula with m = 2: Res(f, z₀) = (1/(2−1)!) · d/dz [(z−z₀)² f(z)] evaluated at z₀. Multiplying by (z−z₀)² clears the double pole, giving a function that is analytic near z₀. Differentiating once (and dividing by (m−1)! = 1! = 1) extracts the a₋₁ coefficient directly. This avoids computing the full Laurent expansion — you only need the leading behavior of the product near z₀.
Question 4 True / False
For an essential singularity, the shortcut formula Res(f, z₀) = lim_{z→z₀} (z − z₀) f(z) still gives the correct a₋₁ coefficient, though the limit may be harder to evaluate than for a simple pole.
TTrue
FFalse
Answer: False
This shortcut only works for simple poles. For an essential singularity, the Laurent series has infinitely many negative-power terms, and no finite power of (z − z₀) can clear all of them. The limit lim_{z→z₀} (z − z₀) f(z) does not isolate a₋₁ — near an essential singularity, f(z) oscillates wildly (by the Casorati-Weierstrass theorem), and this limit typically does not exist or gives zero, not a₋₁. For essential singularities, you must compute enough terms of the actual Laurent expansion to identify a₋₁ directly.
Question 5 Short Answer
Why does the coefficient a₋₁ in the Laurent expansion deserve the special name 'residue,' and why is it uniquely important compared to all other Laurent coefficients?
Think about your answer, then reveal below.
Model answer: The residue is the only Laurent coefficient that survives contour integration. When you integrate f(z) = Σ aₙ(z − z₀)ⁿ term by term around a small circle enclosing z₀, every term with n ≠ −1 integrates to zero (because (z − z₀)ⁿ has an antiderivative for n ≠ −1). Only the n = −1 term contributes: ∮ a₋₁/(z − z₀) dz = 2πi · a₋₁. So ∮_γ f(z) dz = 2πi · Res(f, z₀). The residue is literally what 'remains' — the residue — after all other terms cancel. This is why the residue theorem, which sums residues to evaluate contour integrals, is so powerful.
The naming is exact: residue comes from 'what remains.' All other Laurent terms integrate to zero, leaving behind only the a₋₁ contribution. This single number encodes everything about how f behaves under integration around the singularity, which is why residues are the computational workhorse of complex analysis.