Complex line integrals and estimates

Try

Integrating z from zero to 1+i along the straight segment is not an area-under-a-graph problem. Parametrize the segment by z(t)=(1+i)t for 0≤t≤1. Then dz=(1+i)dt and z dz=2it dt, giving integral i.

Keep the parameter interval beside the substitution so that the starting point, endpoint and direction can be checked independently. Both the function value and the direction of motion contribute. Omitting the derivative of the parametrization changes the problem. A contour is an oriented path, possibly made of finitely many continuously differentiable pieces; its geometric trace alone does not specify its integral.

Predict

Predict the integral after reversing the segment. It should be −i, not the conjugate for any general reason. Predict also whether two paths with the same endpoints always give the same integral. That property depends on the integrand and its domain, not on the notation dz.

The conjugate function will supply a small example where the difference between paths is visible in exact arithmetic.

Experiment

For the unit circle z=exp(it), 0≤t≤2π, compare these products.

Integrand	Integrand times dz	Integral
1	i exp(it)dt	0
1/z	i dt	2πi
z	i exp(2it)dt	0

The constant integrand has zero integral because the endpoint displacement is zero. The reciprocal detects a turn around a missing point. Reversing orientation changes every answer’s sign.

Understand

For a piecewise C¹ parametrization γ:[a,b]→C and continuous f on its trace, define ∫γ f(z)dz as the sum of real-parameter integrals ∫ f(γ(t))γ′(t)dt over the smooth pieces. Complex integration means integrating the real and imaginary parts. The ordinary substitution rule proves invariance under orientation-preserving smooth reparametrization. Reversing the parameter gives the negative integral; concatenation adds integrals.

If the path has length L and |f|≤M on it, then |∫γ f dz|≤∫|f(γ)||γ′|dt≤ML. This estimate is often called the ML estimate. It bounds a complex modulus, not an ordering of complex numbers.

For a positively oriented circle of radius r about zero, set z=r exp(it). For integer m, z^m dz=i r^(m+1) exp(i(m+1)t)dt. The integral vanishes unless m=−1; in that exceptional case it equals 2πi independently of r. The missing primitive of 1/z on a full punctured neighborhood will explain the exception.

These calculations use smooth contours. They do not claim that every arbitrary continuous or fractal path has a line integral defined by this formula.

Apply

Compare conjugate(z) along two paths from zero to 1+i. On the diagonal the integral is one. On the horizontal segment followed by the vertical segment it is 1+i. Write the two parametrizations and verify the extra imaginary contribution.

Diagnose the tempting shortcut “replace dz by the path length.” The complex differential includes direction, while arc length is nonnegative. The ML inequality replaces it by length only after taking a modulus and accepting an inequality rather than an equality.

Continue

Keep a small contour ledger: parametrization, parameter interval, orientation, integrand after substitution, and final value. The next lesson identifies when a primitive reduces that ledger to endpoint evaluation. Optional depth: derive the decomposition ∫(u+iv)(dx+i dy)=∫(u dx−v dy)+i∫(v dx+u dy).