Primitives and path independence

Try

For f(z)=2z, every piecewise smooth path from 1 to i has integral i²−1²=−2. The path may bend or loop; the primitive F(z)=z² records the entire integral through its endpoints. This is a stronger statement than computing one convenient path.

Now compare f(z)=1/z on the punctured plane. Locally it has logarithm primitives, but its unit-circle integral is 2πi. Local primitives and a single global primitive are different mathematical objects.

Predict

Predict which statement follows from having a primitive: zero integral along every path, or zero integral along every closed path. Only the second is correct. On an open path, the primitive’s endpoint values need not agree.

Predict also what information is needed to define a function by integrating from a base point. If two allowed routes give different values, the proposed function is not well-defined.

Experiment

Compare three integrands and domains.

Integrand and domain	Primitive	Closed integrals
2z on C	z²	Zero
1/z on the slit plane	Log z	Zero
1/z on C without zero	None	Unit circle gives 2πi

The same formula 1/z appears in two rows. The difference is the domain, which determines what loops and branches are available.

Understand

Suppose F is holomorphic and F′=f. Along each smooth contour piece, the real-parameter chain rule gives dF(γ(t))/dt=F′(γ(t))γ′(t). Integrating and summing gives ∫γ f dz=F(b)−F(a). In particular every closed integral vanishes, and two paths with the same endpoints have equal integrals.

Conversely, let f be continuous on a domain D, and suppose every closed piecewise C¹ contour in D has zero integral. Fix a base point a. Since an open connected subset of the plane is polygonally connected, choose a path to each z and define F(z) by its integral. Two choices differ by a closed contour formed from one path and the reverse of the other. Their integrals agree, so F is well-defined.

To differentiate F, choose a disk around z contained in D. For small h, the segment from z to z+h remains in that disk. Path independence gives [F(z+h)−F(z)]/h=∫₀¹ f(z+th)dt. Continuity at z bounds the difference from f(z) uniformly in t, so the quotient tends to f(z). Thus F is a primitive.

This proves the equivalence of a global primitive, endpoint path independence, and vanishing of every closed integral for continuous functions on a domain. It does not assume that holomorphy alone supplies a global primitive.

Apply

Repair a false global argument: “Every point of the punctured plane has a logarithm neighborhood; therefore choose those logarithms to make one global primitive.” On overlapping neighborhoods, the choices can differ by multiples of 2πi. Following them around zero forces a nonzero accumulated change.

Use the unit-circle integral as a decisive obstruction: if a global primitive existed, endpoint evaluation would give zero, contradicting 2πi. This argument is shorter and more reliable than trying to draw a branch cut through every possible route.

Continue

The next theorem proves that holomorphic functions have zero integrals around triangles and, with appropriate domain topology, around larger contours. Optional depth: prove that two primitives differ by a constant on a domain. Their derivative difference is zero, and path integration propagates equal values between every pair of points.