Path Independence

Complex Analysis · Axiom Academy

When an integral forgets the path it took — and depends only on where it started and stopped. Take f(z) = z and integrate from 0 to 1+i . Go straight, or bulge out along a curve — it makes no difference. Watch both running totals climb to the same complex number as each route reaches the endpoint. Same endpoints, same answer — the path drops out 2. Why It Happens: The Antiderivative Path independence is not luck. If f has an antiderivative F (a function with F'(z) = f(z) ), then the integral collapses to a difference of endpoint values — the contour version of the Fundamental Theorem of Calculus. The path wiggles; the two endpoints are all that survive. For f(z)=z take . Then the integral from 0 to 1+i is — exactly the value both paths reached, computed with no path at all. 3. What Breaks It: Singularities The whole guarantee rests on f being analytic everywhere inside . Puncture the region with a singularity — a point where f blows up — and path independence can fail. The signature example is : send a loop once around the origin and the integral does not return to zero. Free to deform the path; every closed loop gives 0 ; the integral is path independent. Crossing it changes the value; the loop need not be 0 ; deform only without crossing the bad point. You've seen when a contour integral forgets its path — and the one thing that brings the path back into play. Scroll up to revisit any step.

This is the written version of the interactive lesson above. See the full Complex Analysis course.