Cauchy's Theorem and Deforming a Contour
Prove Cauchy's theorem from Green's theorem plus the Cauchy-Riemann equations under explicit hypotheses, and use valid contour deformation with correct boundary orientations.
What a learner can do afterwards
- Derive Cauchy's theorem from Green's theorem plus the Cauchy-Riemann equations, with continuous partials on a neighbourhood of the filled region as an explicit proof hypothesis
- Deform contours through a holomorphic open domain with orientation and traversal preserved, including annulus inner-boundary orientation and multiple-hole boundary sums
- Explain what a singularity does: an interior pole removes the zero guarantee without forcing nonzero, a boundary pole leaves the ordinary integral undefined, and simply connected domains are sufficient but not necessary for a particular holomorphic function to have a primitive and path independence
1 · Read
Cauchy's theorem. Let γ be a simple closed piecewise C1 curve traversed once positively, and let f be holomorphic on an open set containing γ and its filled interior. Then the loop integral of f is 0. Holomorphic means complex differentiable at every point of an open connected domain, not merely at isolated points. The zero holds for either traversal direction.
The Green proof adds one hypothesis: u and v have continuous first partials on a neighbourhood of the filled region. Expand f dz = (u dx - v dy) + i(v dx + u dy). For the real component integral u dx - v dy, Green gives the double integral of -v_x - u_y. For the imaginary component integral v dx + u dy, Green gives the double integral of u_x - v_y. The Cauchy-Riemann equations u_x = v_y and u_y = -v_x cancel both curls, so both double integrals vanish. This proves the C1 case of the general theorem.
Deforming a contour. A deformation is a continuous family of admissible closed curves staying inside the holomorphic open domain, each traversed once with orientation preserved. For open paths, a continuous family of admissible paths inside the holomorphic domain with fixed endpoints and preserved orientation guarantees the same integral value. For two simple nested counterclockwise circles with f holomorphic on a neighbourhood of the closed annulus between them, outer CCW plus inner CW is 0, so outer CCW equals inner CCW. The inner boundary runs clockwise so the region stays on the left.
Several holes. Let f be holomorphic near a closed region with simple disjoint boundary curves. The outer boundary runs counterclockwise and each inner boundary clockwise, each once. The sum of all oriented boundary integrals is 0, and singularities inside the removed holes are allowed. Example: inner clockwise integrals -1 and -3 force the outer counterclockwise integral to be 4.
An interior pole removes the zero guarantee without forcing a nonzero value. For f(z) = 1/z² and γ(t) = R e^(it) with R > 0 and t in [0, 2π] traversed once counterclockwise, the integrand is (i/R) e^(-it). An antiderivative is -(1/R) e^(-it), and its values at 0 and 2π agree, so the integral is 0. The case R = 1 is the unit circle. Losing the guarantee is not the same as getting a nonzero answer.
Primitive and domains. A primitive is F with F' = f on a domain, and the endpoint rule ∫_γ f dz = F(end) - F(start) follows from the real-parameter chain rule. A domain is simply connected when every closed loop in it contracts to a point inside it. Holomorphic there gives a primitive and path independence. This is sufficient, not necessary for one function: on punctured C, 1/z² has primitive -1/z while 1/z does not. A boundary pole leaves the integrand undefined on the path, so the ordinary integral is not defined.
Worked derivative check. For F(z) = -1/z with z ≠ 0, the difference quotient [F(z+h) - F(z)]/h simplifies to 1/[z(z+h)], which tends to 1/z² as h tends to 0, so F' = 1/z². A global primitive gives 0 on every closed admissible path. Since the unit circle integral of 1/z is 2πi, no global primitive for 1/z exists on punctured C.
Holomorphic on a filled neighbourhood gives zero on the loop, valid deformations preserve the value with orientation kept, and a singularity inside removes the guarantee without forcing a nonzero answer.
All nine extra tasks and three readings are original authorship created for this topic.
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