Contour Deformation & the Cauchy Integral Formula

Lesson 4 — Cauchy’s theorem & the integral formula. Contour integral: ∫γf(z) dz := ∫abf(γ(t))γ′(t) dt

∮γ dz/z  (singularity at z = 0)

winding number around 0:  
∮γ dz/z = 2πi × n =  
Drag the control points. Pull the loop across the × at z = 0 and watch the value jump.

Why deformation invariance holds

If f is holomorphic on the region between two closed contours γ1, γ2 (no singularities in between), then ∮γ1f dz = ∮γ2f dz.
Reason: apply Cauchy–Goursat to the boundary of the in-between region — γ1 together with −γ2 bounds a region where f is holomorphic, so the integral over that whole boundary is 0.
The integral only changes when the contour crosses the singularity — each crossing changes the winding number by ±1, and the value by ±2πi.

Preset (today’s worked examples)

Shrink Cε

Continuity in action: f sampled on Cε

The integral (computed numerically)

∮Cε g dz = i∫₀²π f(z₀+εeiθ) dθ =  
2πi · f(z₀) =  
These agree at every ε — deformation invariance demonstrated, not just asserted.

Unwinding back to f(z₀) — the whole chain

∮γ f(z)z − z0 dz  =  ∮γ g(z) dz
↓substitute g’s definition: g(z) = f(z)/(z−z0) — same integral, new name
∮γ g dz  =  ∮Cε g dz  =  2πi f(z0)
↓shrink to Cε (Cauchy–Goursat), then let ε→0 (continuity of f at z0) — then divide both sides by 2πi
f(z0)  =  12πi ∮γ f(z)z − z0 dz
The integral recovers the value of f at z₀ from f’s values on a curve that never touches z₀. That’s the CIF.