Complex Analysis 27 | Cauchy's Integral Formula [dark version]

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

The setup starts with a holomorphic function f on an open domain D. Choose a disk centered at 0 with radius r such that the closed disk lies inside D (so the circle never hits the boundary of the domain). Let γ be the positively oriented circle (the boundary of that disk). For any point z inside the disk, Cauchy’s integral formula gives an explicit expression for f(z) as a contour integral over γ:

f(z) = (1/(2πi)) ∮_γ f(ζ)/(ζ − z) dζ.

Equivalently, many texts write the same circle integral using ∂B_r instead of γ, and the formula relies on the fact that the winding number of γ around z is 1 (the contour goes once around the point).

The proof uses a standard trick: build an auxiliary function G that is holomorphic everywhere except at one chosen point. Here, the point of exception is not fixed at the disk’s center; it is the interior point z where the value f(z) is sought. The construction is:

G(ζ) = f(ζ)/(ζ − z).

Because f is holomorphic and the only singularity comes from the denominator, G is holomorphic on the punctured disk (holomorphic with one exception at ζ = z). The argument then compares integrals over two contours: a large circle around the origin and a smaller “keyhole” circle around the singularity at ζ = z. Since Cauchy’s theorem applies to holomorphic functions away from the exception, the contour integral can be reduced to the contribution coming from the small circle around ζ = z.

To compute that contribution, the proof splits the integrand by adding and subtracting f(z) in the numerator:

f(ζ)/(ζ − z) = (f(ζ) − f(z))/(ζ − z) + f(z)/(ζ − z).

The second term is straightforward: its integral equals 2πi times the constant f(z). The first term involves a difference quotient. As the small radius ε of the inner circle shrinks to 0, the difference quotient approaches f′(z), and the contour length shrinks proportionally to ε, forcing the integral of that first part to vanish in the limit. What remains is exactly the 2πi f(z) term, yielding the formula.

In short, the formula is strong not because it’s complicated, but because it’s rigid: holomorphic functions are fully controlled by their boundary values on circles contained in the domain. The next logical step—hinted at for later—is what this rigidity implies for analyticity and further properties of holomorphic functions.