Complex Analysis 21 | Closed curves and antiderivatives [dark version]

TL;DR

If f is holomorphic on a path-connected open set D and ∮γ f(z) dz = 0 for every closed curve γ in D, then f has an antiderivative on D.

Briefing Cornell Notes

Briefing

A holomorphic function on a path-connected open set has an antiderivative exactly when every closed contour integral of that function is zero. That “converse” turns a one-way fact from real analysis into a complex-analytic equivalence, setting the stage for Cauchy’s integral theorem.

The starting point is the familiar direction: if a complex function f admits a primitive (an antiderivative) on a domain, then the integral of f around any closed curve γ in that domain must vanish. The new question flips the logic. Suppose f is holomorphic on an open set D ⊂ ℂ, and suppose that for every closed curve γ lying in D, the contour integral ∮γ f(z) dz equals 0. Is that condition strong enough to guarantee an antiderivative exists? The answer is yes—provided D is not split into disconnected pieces.

To make that precise, the domain D is required to be path-connected: any two points z and w in D can be joined by a curve γ lying entirely in D. Holes are allowed; what matters is that there is always a path inside D connecting points. With this assumption, the proof fixes a base point z0 in D. For any other point z in D, path-connectedness supplies a curve γz from z0 to z (parameterized on [0,1]). Using that curve, an antiderivative candidate is defined by

F(z) = ∫γz f(ζ) dζ.

A key issue is whether F(z) depends on which path from z0 to z is chosen. It does not: if two different curves connect z0 to z, combining one forward with the other reversed produces a closed contour. By the hypothesis that all closed contour integrals are zero, the two path integrals must match, so F is well-defined.

The remaining task is to show that F really differentiates back to f. The argument uses the complex difference quotient. For a point z and a nearby point z̃ within a small ε-ball, one studies (F(z̃) − F(z)) / (z̃ − z). Each term F(z̃) and F(z) is expressed as a contour integral along a path from z0 to the relevant endpoint. Subtracting these integrals is converted into a single contour integral around a closed loop formed by following one path and then returning along another. The integral is then estimated using standard bounds for contour integrals: the magnitude is controlled by the maximum of |f| on a small neighborhood times the length of the contour. Because f is holomorphic, it is continuous, so as ε shrinks the maximum value on that neighborhood becomes small, forcing the difference quotient to converge to f(z).

With that, the existence of an antiderivative follows from the vanishing of all closed contour integrals, and the logic runs in both directions on path-connected open sets. That equivalence is positioned as a direct bridge toward Cauchy’s integral theorem in the next step.

Cornell Notes

For a holomorphic function f on a path-connected open set D, the condition “∮γ f(z) dz = 0 for every closed curve γ in D” is enough to guarantee an antiderivative exists on D. The construction fixes a base point z0 and, for each z, chooses any path γz from z0 to z, then defines F(z)=∫γz f(ζ)dζ. The value of F(z) is independent of the chosen path because two different paths form a closed contour whose integral is zero. Finally, a difference-quotient estimate using contour-integral bounds shows that F′(z)=f(z). This creates an equivalence between primitives and vanishing closed contour integrals, a key ingredient for Cauchy’s integral theorem.

Why is path-connectedness of D essential for defining F(z)=∫γz f(ζ)dζ?

Path-connectedness ensures that for every z in D there exists at least one curve γz lying entirely in D that starts at the fixed base point z0 and ends at z. Without such a path, the integral defining F(z) might not be possible for points in a disconnected component.

How does the “closed contour integral equals zero” assumption guarantee that F(z) does not depend on the chosen path?

Take two paths γ1 and γ2 from z0 to z. Traversing γ1 forward and then γ2 backward forms a closed curve. The integral over that closed curve is zero by assumption, so the integrals along γ1 and γ2 must be equal after accounting for the sign change from reversing direction. Therefore both choices produce the same F(z).

What is the core idea behind proving F′(z)=f(z)?

Compute the complex difference quotient (F(z̃)−F(z))/(z̃−z) for z̃ near z. Express F(z̃) and F(z) as contour integrals from z0 to each endpoint, then combine them into a single contour integral around a closed loop. Use an estimate that bounds the contour integral by (maximum of |f| on a small neighborhood) × (length of the contour). As the neighborhood shrinks, continuity of f forces the bound to go to 0, yielding the limit f(z).

Why does reversing a curve change the sign of a contour integral?

If a contour is traversed in the opposite direction, the parameterization runs backward. This reverses the orientation, so ∫γ f(ζ)dζ becomes −∫γback f(ζ)dζ. That sign flip is used when forming closed curves from two different paths.

What role does choosing a small ε-ball around z play in the derivative argument?

It localizes the estimate. The contour connecting z to z̃ is kept inside a neighborhood where |f| is controlled. Since f is holomorphic, it is continuous, so as ε→0 the maximum of |f| on that neighborhood decreases appropriately. This makes the difference-quotient error term vanish in the limit.

Review Questions

State the exact conditions on D and f under which vanishing closed contour integrals imply the existence of an antiderivative.
Explain why F(z) defined via path integrals is well-defined (path-independent).
Outline how the difference quotient is converted into a contour integral and then estimated to show F′(z)=f(z).

Key Points

1
If f is holomorphic on a path-connected open set D and ∮γ f(z) dz = 0 for every closed curve γ in D, then f has an antiderivative on D.
2
Fix a base point z0 in D and define F(z)=∫γz f(ζ)dζ using any curve γz from z0 to z.
3
F(z) is path-independent because two different paths from z0 to z form a closed contour whose integral is zero.
4
To prove F′(z)=f(z), use the complex difference quotient (F(z̃)−F(z))/(z̃−z) with z̃ near z.
5
The difference of two path integrals is rewritten as a single contour integral around a closed loop.
6
A contour-integral bound (maximum of |f| on a neighborhood times contour length) plus continuity of f forces the error term to vanish as the neighborhood shrinks.
7
On path-connected open sets, “having a primitive” and “all closed contour integrals vanish” become equivalent conditions.

Highlights

Vanishing integrals over every closed contour are strong enough to force the existence of an antiderivative for holomorphic functions.

The antiderivative candidate F(z)=∫γz f(ζ)dζ is well-defined because any two choices of γz differ by a closed contour integral.

The derivative step reduces to estimating a contour integral over a small loop and letting the loop shrink to a point.

Path-connectedness allows every point z to be reached from a fixed base point z0 by a curve staying inside D.

Topics

Closed Curves
Antiderivatives
Path-Connected Domains
Contour Integrals
Complex Differentiation