Cauchy’s integral formula and derivatives

Try

Suppose a function is holomorphic throughout a disk. Cauchy’s formula says its values inside are already encoded by its values on a surrounding circle. At the center, the formula becomes an average over that circle. This is much stronger than ordinary continuity of a real function.

For f(z)=z²+1 on the unit circle, the average is one: the rotating z² term cancels while the constant remains. The cancellation is exact, not a statement that positive and negative complex numbers can be ordered.

Predict

Predict whether changing a holomorphic function at one interior point while leaving all boundary values fixed is possible. The integral formula will force that point’s value to stay unchanged.

Now predict how a denominator squared rather than to the first power changes the contour information. It extracts a derivative. Keep the factorial visible when extending this observation to higher powers; forgetting it produces incorrect residue and derivative calculations.

Experiment

On a positive circle enclosing a, compare the kernel powers.

Kernel denominator	Contour integral
z−a	2πi f(a)
(z−a)²	2πi f′(a)
(z−a)³	πi f″(a)

All rows assume f is holomorphic on a neighborhood of the closed interior disk. They do not apply when the evaluation point lies on the contour.

Understand

Here is the value formula for a disk: f(a)=(1/(2πi))∫ f(z)/(z−a)dz, with positive orientation. Subtract f(a) from the numerator. The quotient g(z)=(f(z)−f(a))/(z−a) is holomorphic away from a and bounded near a because it tends to f′(a). Apply Cauchy’s theorem on the region between the outer circle and a small circle about a. The outer integral of g equals the small-circle integral. The latter tends to zero by its boundedness and shrinking length. Thus the subtracted part contributes nothing, while f(a)/(z−a) contributes 2πi f(a).

This proof avoids assuming that the quotient is already holomorphic at a; that stronger removable-singularity conclusion comes later. The bounded estimate is enough.

To differentiate with respect to a, restrict a to a smaller disk. The contour denominator is then uniformly separated from zero. The kernel’s difference quotients converge uniformly to its derivative, so the limit passes through the finite-length integral. Iteration gives f^(n)(a)=n!/(2πi)∫ f(z)/(z−a)^(n+1)dz. Consequently a holomorphic function has derivatives of every order, despite the original definition asking only for one derivative at each point.

A finite contour sampling can approximate these integrals, but the exact formula follows from holomorphy and contour hypotheses, not from numerical agreement at a few points.

Apply

Calculate ∫ exp(2z)/(z−1)² dz around |z|=2. The derivative extracted is 2exp(2), so the result is 4πi exp(2). Reverse the circle and explain the sign change.

Diagnose an incorrect application with the point a=3 outside the radius-two circle. There is no enclosed singularity, and the entire integrand is holomorphic inside. Cauchy’s theorem gives zero; evaluating a derivative at three would use the wrong formula hypotheses.

Continue

Cauchy estimates turn these identities into bounds. They will prove Liouville’s theorem and the fundamental theorem of algebra. Optional depth: show directly that the center formula is f(a)=(1/(2π))∫ from 0 to 2π of f(a+r exp(it))dt, explaining how the parametrization cancels the reciprocal kernel.