Complex Analysis 21 | Closed curves and antiderivatives
Based on The Bright Side of Mathematics's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.
A holomorphic function f on a path-connected open set D has an antiderivative if every closed contour integral ∮γ f(z) dz equals 0.
Briefing
A holomorphic function on a path-connected open set has an antiderivative exactly when every closed contour integral of that function vanishes. That “converse” turns the earlier one-way link between primitives and zero integrals into a full equivalence—an essential stepping stone toward Cauchy’s integral theorem.
The setup starts with a holomorphic function f on an open domain D ⊂ ℂ, where “domain” is taken to mean not just open, but also path-connected: any two points z and w in D can be joined by a curve γ lying entirely inside D. The key condition is that for every closed curve γ in D, the contour integral ∮γ f(z) dz equals 0. The central question is whether this condition forces f to have a primitive (an antiderivative) on D.
To build that antiderivative, the argument fixes a base point z0 in D. For any other point z in D, path-connectedness guarantees the existence of a curve γz from z0 to z. Using that curve, it defines a new function F by F(z) = ∫γz f(ζ) dζ, where ζ is used as the integration variable to avoid confusion with the point z. The proof then checks that F(z) does not depend on which particular path from z0 to z was chosen. If two different curves from z0 to z are used, combining one forward with the other reversed produces a closed curve. The assumed condition makes the integral over that closed curve zero, and splitting it into two pieces shows the two candidate values of F(z) must match. So F is well-defined.
The remaining task is to show that F is actually an antiderivative: F′(z) = f(z) for every z in D. Because complex differentiation is pointwise, it suffices to work near a fixed z. Since D is open, there is an ε-ball around z contained in D, allowing the use of nearby points z̃ within that ball. The derivative is obtained from the difference quotient (F(z̃) − F(z)) / (z̃ − z). The proof rewrites F(z̃) and F(z) as contour integrals from z0, then combines them into a single contour integral by again using the idea of “gluing” paths into a closed curve. The closed-curve integral vanishes by assumption, leaving an expression that can be estimated.
A standard bound for contour integrals is invoked: the magnitude of an integral is at most the maximum value of the integrand times the length of the contour. With holomorphicity implying continuity, the maximum of |f(ζ) − f(z)| over a shrinking neighborhood tends to 0 as ε → 0. As z̃ approaches z, the contour length also shrinks, forcing the difference quotient to converge to f(z). This completes the equivalence: zero integrals over all closed curves guarantee the existence of an antiderivative, and thus the groundwork for Cauchy’s integral theorem is in place.
Cornell Notes
For a holomorphic function f on a path-connected open set D ⊂ ℂ, the condition that ∮γ f(z) dz = 0 for every closed curve γ in D is enough to guarantee an antiderivative. The construction fixes a base point z0 and defines F(z) = ∫γz f(ζ) dζ using any path γz from z0 to z. Path-independence follows by combining two different paths into a closed curve, whose integral is zero by assumption. Finally, F′(z) = f(z) is proved using a difference quotient, rewriting it as a contour integral over a small path and applying an estimate that uses continuity of holomorphic functions. This yields the converse link needed for Cauchy’s integral theorem.
Why does path-connectedness matter for defining an antiderivative candidate F(z)?
How does the assumption “all closed contour integrals are zero” make F(z) independent of the chosen path?
What is the strategy for proving F′(z) = f(z) using contour integrals?
Where does continuity enter the derivative proof?
What role does the contour-length estimate play?
Review Questions
- Suppose f is holomorphic on a path-connected open set D and ∮γ f(z)dz = 0 for every closed curve γ in D. How would you construct an antiderivative F and why is it well-defined?
- In the proof that F′(z) = f(z), what happens to the integral over a closed curve, and how does that simplify the difference quotient?
- Why does the estimate for contour integrals require both a bound on the integrand and control of the contour’s length?
Key Points
- 1
A holomorphic function f on a path-connected open set D has an antiderivative if every closed contour integral ∮γ f(z) dz equals 0.
- 2
Path-connectedness guarantees that for each z ∈ D there exists a curve γz from a fixed base point z0 to z inside D.
- 3
Defining F(z) = ∫γz f(ζ) dζ produces a candidate antiderivative, but only the closed-curve condition ensures F is independent of the chosen path.
- 4
Independence of path follows by combining two different z0-to-z paths into a closed curve and using the zero-integral assumption.
- 5
To show F′(z) = f(z), the difference quotient is rewritten using contour integrals and then reduced via the vanishing of closed-curve integrals.
- 6
A contour-integral bound (maximum of the integrand times contour length) plus continuity of holomorphic functions forces the difference quotient to converge to f(z).