A function f(z) has isolated singularities at z = 0, z = 1+i, and z = 3. You integrate f around the circle |z−1| = 1.5. How many singularities contribute to the contour integral?
AAll 3 — the residue theorem sums over all singularities of f, regardless of position
B2 — both z=0 (distance 1 from center) and z=1+i (distance 1 from center) lie inside the contour
C1 — only z=1+i lies inside since it is closest to the center
D0 — none of the singularities lie exactly at the center z=1
The residue theorem sums only over singularities enclosed by the contour. |0−1|=1 < 1.5, so z=0 is inside. |1+i−1|=|i|=1 < 1.5, so z=1+i is inside. |3−1|=2 > 1.5, so z=3 is outside. Option 0 is the classic error: position relative to the contour, not mere existence, determines which singularities contribute.
Question 2 Multiple Choice
A student correctly computes Res(f, 1) + Res(f, −2) for a function with poles at z=1 and z=−2, both enclosed by the contour γ. The student announces this sum as the value of ∮_γ f(z) dz. What is missing?
AThe student must multiply the sum by 2πi to get the correct integral value
BNothing is missing — the sum of residues equals the contour integral
CThe student must divide by 2πi because the residues already include that factor
DThe student must subtract the contribution from the contour itself
The residue theorem states ∮_γ f(z) dz = 2πi × Σ Res(f, zₖ). The factor 2πi arises from the fundamental integral ∮ dz/(z−z₀) = 2πi and cannot be omitted. Forgetting this factor is the single most common computational error when applying the theorem. The sum of residues alone is not the integral.
Question 3 True / False
A function that is holomorphic everywhere inside and on a closed contour (no singularities enclosed) must have a contour integral equal to zero.
TTrue
FFalse
Answer: True
This is Cauchy's theorem, the foundation from which the Residue Theorem generalizes. If f is holomorphic (analytic) everywhere inside and on γ, there are no singularities to contribute residues, and the sum Σ Res = 0, giving ∮_γ f dz = 2πi · 0 = 0. The Residue Theorem extends this: isolated singularities are the sole sources of nonzero contour integrals.
Question 4 True / False
The Residue Theorem applies primarily to functions with simple poles; it cannot be used for functions with poles of order 2 or higher.
TTrue
FFalse
Answer: False
The Residue Theorem applies to any isolated singularity, including poles of any order and essential singularities. The residue is defined as the coefficient a₋₁ of the 1/(z−z₀) term in the Laurent series, regardless of pole order. For a pole of order m at z₀, the residue is computed via lim_{z→z₀} (1/(m−1)!) d^{m-1}/dz^{m-1} [(z−z₀)^m f(z)]. The theorem remains ∮_γ f dz = 2πi Σ Res(f, zₖ) in all cases.
Question 5 Short Answer
Why does the factor 2πi appear in the residue theorem? Where does it come from?
Think about your answer, then reveal below.
Model answer: It comes from the most fundamental contour integral in complex analysis: ∮_{|z−z₀|=r} 1/(z−z₀) dz = 2πi. Parametrizing with z = z₀ + re^{iθ} gives dz = ire^{iθ}dθ, and the integrand becomes (ire^{iθ})/(re^{iθ}) = i. Integrating i from 0 to 2π yields 2πi. Every pole's contribution to a contour integral ultimately reduces to a scalar multiple of this kernel integral, which is why 2πi appears universally — it is the winding number of the contour around the singularity multiplied by 2πi.
The 2πi factor is not an artifact of notation — it encodes the topology of the contour. The winding number of a positively oriented simple closed curve around an interior point is exactly 1, contributing one factor of 2πi. If the contour wound around a point twice, you would get 4πi. This topological interpretation is what makes the theorem so powerful: the integral depends only on which singularities are enclosed and how many times the contour winds around them.