Questions: Contour Integration

Question 1

Let f(z) = 1/(z − 2). You integrate f counterclockwise around a circle of radius 3 centered at the origin. What is the value of ∮ f(z) dz?

• A. 0 — f has a singularity, so Cauchy's theorem does not apply and the integral cannot be computed
• B. 2πi — the pole at z = 2 lies inside the contour, contributing 2πi times the residue, which equals 1
• C. 6πi — the residue is multiplied by the radius of the contour
• D. πi — because the pole is not at the center of the contour, only half the residue contributes

Answer: (B)

The residue theorem: ∮ f(z) dz = 2πi × Σ(residues inside γ). The only singularity of f(z) = 1/(z−2) is at z = 2, which lies inside the circle of radius 3 centered at the origin. The residue of 1/(z−2) at z = 2 is lim(z→2)(z−2) · 1/(z−2) = 1. So the integral equals 2πi × 1 = 2πi. The radius of the contour is irrelevant — what matters is only which singularities are inside the contour and their residues. This is the power of the residue theorem: the value is determined entirely by what's inside.

Question 2

Let f(z) = 1/(z² + 1) = 1/((z − i)(z + i)). Evaluated counterclockwise around a circle of radius 2 centered at the origin, ∮ f(z) dz equals:

• A. 0 — f is holomorphic everywhere on and inside the circle since it has no singularities there
• B. 2πi — only the pole at z = i lies inside the contour
• C. 0 — both poles z = i and z = −i lie inside the contour, but their residues are equal and opposite, summing to zero
• D. 4πi — both poles lie inside the contour and each contributes 2πi

Answer: (C)

Both poles z = i and z = −i lie inside the circle of radius 2. Res(f, i) = lim(z→i)(z−i) · 1/((z−i)(z+i)) = 1/(2i). Res(f, −i) = lim(z→−i)(z+i) · 1/((z−i)(z+i)) = 1/(−2i) = −1/(2i). Sum of residues = 1/(2i) − 1/(2i) = 0. So ∮ f(z) dz = 2πi × 0 = 0. The integral is zero even though f has singularities inside the contour — because the residues cancel. This shows that zero contour integral does NOT necessarily mean f is holomorphic inside; the residues may simply cancel.

Question 3

For a holomorphic function f on a simply connected domain, the value of ∮_γ f(z) dz depends on the shape and size of the closed contour γ.

• T. True
• F. False

Answer: False

By Cauchy's theorem, if f is holomorphic throughout the region bounded by γ (and on γ itself), then ∮_γ f(z) dz = 0, regardless of the shape, size, or placement of γ. A large square, a tiny circle, a wiggly path — any closed contour in a region where f has no singularities gives the same answer: zero. The shape of the contour matters only in that it determines *which singularities lie inside* — and hence which residues contribute. For a holomorphic function, there are no singularities, so the answer is always zero.

Question 4

A singularity of f(z) located strictly inside a closed contour γ can affect the value of ∮_γ f(z) dz, even though γ never passes through the singularity.

• T. True
• F. False

Answer: True

This is the topological heart of contour integration. The path γ goes around the singularity, never touching it, yet the value of the integral reflects the singularity's residue. The singularity contributes 2πi × Res(f, z₀) to the integral, detectable by any closed path that winds around it once counterclockwise. This is a fundamentally non-local property with no analogue in real single-variable calculus, where the integral between two points depends only on the path, not on what lies in nearby regions.

Question 5

Explain in your own words why ∮_γ (1/z) dz = 2πi around the unit circle, even though the unit circle never passes through the singularity at z = 0.

Model answer: The function 1/z has a simple pole at z = 0 with residue 1. As the unit circle winds once counterclockwise around the origin, it 'sees' the singularity topologically — the path encloses z = 0. The residue theorem tells us the integral equals 2πi times the residue, so 2πi × 1 = 2πi. This can be verified directly: parameterize as z = e^(iθ), dz = ie^(iθ) dθ, and ∮ (1/z) dz = ∫₀^(2π) e^(−iθ) · ie^(iθ) dθ = ∫₀^(2π) i dθ = 2πi. The path doesn't touch z = 0, but by going around it, the integral accumulates a contribution from the singularity's local behavior — specifically, the 1/z term that doesn't cancel when integrated over a full loop.

Contrast with f(z) = z: ∮ z dz = 0 around any closed path, because z is holomorphic everywhere. The 2πi answer for 1/z is a signature of the singularity at the origin. The residue theorem systematizes this: any meromorphic function can be decomposed into a holomorphic part (contributing 0 to any closed integral) and polar parts near each singularity (contributing 2πi × residue for each enclosed singularity).