Complex Analysis 27 | Cauchy's Integral Formula

Q: Why does splitting f(ζ) into (f(ζ)−f(z))+f(z) help?

The split creates two integrals. The part with f(z) is constant, so ∮ f(z)/(ζ−z) dζ evaluates directly to 2πi·f(z). The remaining part involves (f(ζ)−f(z))/(ζ−z). Because f is complex differentiable at z, this difference quotient stays well-behaved and the integral over the small circle becomes small: its magnitude is bounded by (maximum of the integrand) times the circle length 2π·ε, which forces the term to vanish as ε→0.

Q: What role does the limit ε→0 play in the proof?

The limit controls the “error” integral coming from (f(ζ)−f(z))/(ζ−z). As ε shrinks, the contour length scales like 2π·ε, so the integral’s magnitude is forced toward 0. Meanwhile, the constant term does not vanish; it yields the exact 2πi·f(z) contribution. Taking ε→0 leaves only the constant term, producing Cauchy’s formula.

TL;DR

Cauchy’s integral formula reconstructs f(z) for any interior point z from an integral of f(ζ)/(ζ−z) over the boundary circle ∂B_r.

Briefing Cornell Notes

Briefing

Cauchy’s integral formula turns the “zero integral” behavior of holomorphic functions into a precise reconstruction rule: for a holomorphic function on a domain containing a closed disk, the value at any interior point is determined entirely by the function’s values on the boundary circle. That matters because it sharply limits how much freedom holomorphic functions can have—knowing them on a curve fixes them everywhere inside.

The setup starts with a holomorphic function f on an open domain D. Choose a disk B_r(0) whose closure sits inside D, so the boundary circle stays away from the domain’s boundary. Let γ be the positively oriented boundary of that disk (the circle), which winds once around any point z inside the disk. Cauchy’s formula then states that for every point z in B_r(0), the function value f(z) can be computed by an integral over the circle:

f(z) = (1/(2πi)) ∮_γ f(ζ)/(ζ − z) dζ.

Equivalently, using the common shorthand ∮_{∂B_r} for the boundary circle, the same rule applies: the integral of f(ζ)/(ζ−z) around the circle produces exactly 2πi times f(z).

The proof strategy leans on a key tool from earlier work: constructing an auxiliary function with a removable singularity and then applying the “integral around closed curves is zero” principle. A slightly larger radius R̃ is chosen so that the closed disk of radius R̃ still lies inside D. Then an auxiliary function G is defined by

G(ζ) = f(ζ)/(ζ − z),

where the only problematic point is ζ = z. With the domain enlarged, the contour integrals can be handled cleanly.

Next comes the contour manipulation. Instead of integrating around the large circle centered at 0, the argument uses the earlier “keyhole contour” idea: the integral around the big circle can be related to integrals around a smaller circle that encloses the singularity at ζ = z, with a small radius ε (chosen less than r). This reduces the problem to analyzing what happens near the singularity.

To make the integral computable, the numerator is split by adding and subtracting f(z):

f(ζ) = (f(ζ) − f(z)) + f(z).

This yields two integrals. The second one is straightforward because f(z) is constant: its integral gives 2πi·f(z). The first integral involves the difference quotient (f(ζ) − f(z))/(ζ − z). Since f is complex differentiable at z, that quotient stays controlled, and the standard estimate for contour integrals shows the contribution from this term shrinks to 0 as ε → 0 (the circle’s length scales like 2π·ε). With the “error” term vanishing, only the constant term remains.

The result is Cauchy’s integral formula itself, derived with minimal extra machinery because the earlier theorem about holomorphic functions and closed-curve integrals already did most of the heavy lifting. The payoff is substantial: once f(z) is fixed by boundary data, holomorphic functions become rigid—setting the stage for powerful consequences in the next topic.

Cornell Notes

Cauchy’s integral formula says that if f is holomorphic on a domain D containing a closed disk B_r(0), then every interior value f(z) is determined by an integral over the boundary circle. For z inside the disk, f(z) equals (1/(2πi)) times the contour integral of f(ζ)/(ζ−z) around the positively oriented circle ∂B_r. The proof introduces an auxiliary function G(ζ)=f(ζ)/(ζ−z), which is holomorphic everywhere except at ζ=z. By shrinking the contour around the singularity (using a keyhole-contour idea) and splitting f(ζ) into (f(ζ)−f(z))+f(z), one term becomes 2πi·f(z) while the remaining term goes to 0 as the small radius ε→0. This rigidity is why holomorphic functions have so little freedom.

What does Cauchy’s integral formula actually compute, and what input data does it require?

It computes the value f(z) at any point z inside a disk B_r(0) using only the boundary values of f on the circle ∂B_r. The formula is f(z) = (1/(2πi)) ∮_{∂B_r} f(ζ)/(ζ−z) dζ, where ζ runs along the positively oriented boundary circle. The domain requirement is that f is holomorphic on an open set D that contains the closed disk (so the contour lies entirely inside D, not on its boundary).

Why is the winding number around z relevant?

The circle ∂B_r winds once around any interior point z, meaning the contour integral behaves like a “single wrap” around the singularity at ζ=z. In the proof, this ensures the contour integral of the constant term f(z)/(ζ−z) produces exactly the factor 2πi·f(z), matching the single winding. If the winding number changed, the resulting multiple of 2πi would change accordingly.

How does the proof use an auxiliary function, and where is the singularity?

An auxiliary function G(ζ)=f(ζ)/(ζ−z) is defined. It is holomorphic everywhere except at ζ=z, where the denominator vanishes. The earlier result about holomorphic functions and closed-curve integrals (integral equals 0 when there are no singularities inside) is then applied by modifying the contour so the singularity is handled separately.

What contour change makes the calculation easier?

Instead of integrating around a large circle, the argument shrinks to a smaller circle centered at z with radius ε (with ε<r). This uses the keyhole-contour idea: the integral around the big circle can be related to integrals around contours that isolate the singularity at ζ=z. The smaller circle makes the singular behavior local and allows estimates as ε→0.

Why does splitting f(ζ) into (f(ζ)−f(z))+f(z) help?