Questions: Complex Line Integrals

Question 1

You want to compute ∫_γ z dz from 0 to 1+i. You try two paths: a straight line, and an L-shaped path via the point 1. What result should you expect?

• A. Different values, because the two paths trace different curves through the complex plane
• B. The same value, because f(z) = z is holomorphic and the domain ℂ is simply connected
• C. The same value only if both paths have the same arc length
• D. Different values, depending on the winding number of each path around the origin

Answer: (B)

f(z) = z is holomorphic everywhere (it's entire), and ℂ is simply connected, so by path-independence for holomorphic functions, the integral depends only on the endpoints. Winding number and arc length are irrelevant for holomorphic integrands on simply connected domains — they matter only when there are singularities. The key distinction: path-independence is a consequence of holomorphicity, not of the path's geometry.

Question 2

A student computes ∮ (1/z) dz along the unit circle (traversed once counterclockwise) and gets 2πi. Her classmate argues she must have made an error: 'The path is closed, so it returns to the starting point — the integral must be 0.' Who is correct?

• A. The classmate — by Cauchy's theorem, any integral along a closed path is 0
• B. The student — 1/z has a singularity at z = 0 inside the unit circle, so Cauchy's theorem does not apply, and the integral equals 2πi
• C. Neither — the integral of 1/z around a closed path is always πi regardless of the contour
• D. The classmate — the integral is 0 because 1/z is holomorphic on the punctured plane ℂ\{0}

Answer: (B)

Cauchy's theorem requires the function to be holomorphic on a *simply connected* domain containing the path. The unit circle encloses z = 0, where 1/z has a singularity — the domain ℂ\{0} is not simply connected (there's a hole at the origin). The integral ∮ (1/z) dz = 2πi is one of the foundational results of complex analysis; its value reflects the path winding once around the singularity. Option D confuses 'holomorphic on a punctured plane' with 'holomorphic on a simply connected domain.'

Question 3

Since a complex line integral along a closed path generally returns to the starting point in the complex plane, it typically evaluates to 0.

• T. True
• F. False

Answer: False

Returning to the starting point does not make the integral zero — this confuses the path being closed geometrically with the integral being zero analytically. The integral is 0 for closed paths only when the integrand is holomorphic on a simply connected region containing the path. For f(z) = 1/z integrated around the origin, the path is closed but the integral is 2πi. The value depends on whether (and how) the path encircles singularities.

Question 4

When integrating a holomorphic function f over a simply connected domain, the value of ∫_γ f(z) dz depends only on the endpoints of γ, not on the specific path taken between them.

• T. True
• F. False

Answer: True

Path-independence is exactly what holomorphicity buys you on simply connected domains. This is the complex analogue of gradient vector fields in real calculus: just as integrating a conservative field depends only on endpoints, integrating a holomorphic function on a simply connected domain depends only on where you start and end. The simply connected condition is essential — if the domain has holes (like ℂ\{0}), path-independence can fail because different paths may wind differently around the holes.

Question 5

Why does ∫_γ (1/z) dz around a circle enclosing the origin equal 2πi rather than 0, and what concept connects the integral's value to the geometry of the path?

Model answer: The function 1/z is not holomorphic at z = 0, and the circle encloses that singularity. The value 2πi comes from the winding number of the path around the singularity: a path that winds once counterclockwise around the origin gives 2πi · 1 = 2πi; a path winding twice would give 4πi; a path not enclosing the origin gives 0. The winding number connects the integral's value to the topological relationship between the path and the singularity — this is the first glimpse of why complex integration connects analysis to topology.

Computing the integral explicitly confirms this: parametrize γ(t) = e^{it} for 0 ≤ t ≤ 2π, so dz = ie^{it} dt and 1/z = e^{-it}. Then ∫ e^{-it} · ie^{it} dt = ∫ i dt = 2πi. The 'i times 1' factor comes directly from the winding — the path going all the way around.