Cauchy's Integral Formula

Suppose a function is complex-differentiable on an open disk. Can you recover the value of the function at the center from its boundary values alone? This is not obviously possible for real functions. Knowing a function on the boundary of an interval tells you nothing about its interior values. But for holomorphic functions the answer is a resounding yes. The recovery mechanism is an explicit integral. This phenomenon is the content of Cauchy's Integral Formula. It is arguably the single most important computational tool in complex analysis. The formula matters because it converts a local question into a global statement. The local question is: what is ff near a point z0z_0? The global statement concerns an integral over a contour that may be far from z0z_0. This rigidity has no counterpart in real analysis. A real differentiable function can be wild near a point while looking tame elsewhere. But a holomorphic function's entire behavior inside a region is dictated by its boundary values. The formula also does something else remarkable. By differentiating under the integral sign, it produces formulas for all derivatives of ff. In particular, it proves that a holomorphic function is automatically infinitely differentiable. This fact has zero analogue in real calculus, where differentiability once does not imply differentiability twice. The big picture is that Cauchy's Integral Formula bridges two things. One is Cauchy's Theorem, which says certain contour integrals vanish. The other is the entire analytic machinery of complex analysis. Power series expansions, Liouville-type rigidity theorems, and the maximum modulus principle all trace their proofs back to this single integral identity. It converts the vanishing of integrals of holomorphic functions into a reproducing formula. The function reproduces its own values from the boundary, weighted by a specific kernel 12πi(ζ−z)\frac{1}{2\pi i(\zeta - z)}. This article states the formula and its higher-derivative generalization with full hypotheses. It gives a complete proof using the ML-inequality and a shrinking-circle argument. It works through two concrete numerical examples. It shows via a counterexample exactly why the "holomorphic on and inside the contour" hypothesis cannot be weakened. It closes by tracing the formula's reach into power series theory, Liouville's theorem, and the maximum modulus principle.

Key takeaways
  • Cauchy's Integral Formula recovers interior values of a holomorphic function from an integral over an enclosing contour.
  • The key identity: f(z0)=12πi∮γf(z)z−z0 dzf(z_0) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z - z_0}\,dz for γ\gamma a simple closed contour enclosing z0z_0.
  • Differentiating under the integral sign gives formulas for every derivative f(n)(z0)f^{(n)}(z_0), proving holomorphic functions are infinitely differentiable.
  • The proof rests on Cauchy's Theorem plus the ML Inequality applied to a small circle shrinking to the point.
  • Weakening "holomorphic on and inside γ\gamma" to merely "continuous" destroys the formula — a genuine hypothesis, not a technicality.
  • Key idea: boundary values of a holomorphic function completely determine, via a single explicit integral kernel, every value and derivative in the interior.

Where this lives today

Cauchy's Integral Formula is not a museum piece — it is load-bearing infrastructure across mathematics and its applications. Within pure mathematics, it is the direct engine behind Power Series and Analyticity: expanding 1z−z0\frac{1}{z-z_0} in a geometric series inside the formula's kernel produces the Taylor series of any holomorphic function, proving that holomorphic equals analytic in the complex setting (a fact famously false for real-differentiable functions). It is equally the engine behind Liouville's Theorem and Applications: bounding the derivative formula on circles of growing radius shows that a bounded entire function must be constant, which in turn gives the shortest known proof of the Fundamental Theorem of Algebra. It also underlies the Maximum Modulus Principle, since the mean-value consequence of the integral formula (evaluating on circles) forces ∣f∣|f| to have no strict interior local maximum unless ff is constant. Beyond pure analysis, the formula's residue-theorem descendant is a computational workhorse in engineering: evaluating real definite integrals (Fourier transforms, inverse Laplace transforms in control theory and signal processing) by closing contours in the complex plane and reading off pole contributions. In physics, contour-integral techniques built on this formula appear in quantum field theory (propagator calculations), statistical mechanics (partition function analysis via saddle points), and fluid dynamics (2D potential flow, where holomorphic velocity potentials make Cauchy's formula a tool for reconstructing flow fields from boundary data). In numerical analysis, algorithms for computing eigenvalues and matrix functions (e.g., contour-integral eigensolvers) use discretized versions of Cauchy's formula to extract spectral information by integrating a resolvent around a contour enclosing target eigenvalues.

Ask the tutor: How exactly does expanding the Cauchy kernel in a geometric series turn this formula into a power series proof?

DefinitionCauchy's Integral Formula

Let U⊆CU \subseteq \mathbb{C} be an open set and let f:U→Cf : U \to \mathbb{C} be holomorphic on UU. Let γ\gamma be a positively oriented, simple (non-self-intersecting), piecewise-C1C^1 closed contour in UU such that the region DD enclosed by γ\gamma satisfies D‾⊂U\overline{D} \subset U (i.e. γ\gamma together with its interior lies inside UU). Then for every point z0∈Dz_0 \in D,

f(z0)=12πi∮γf(z)z−z0 dz.f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0}\, dz .

More generally, ff is infinitely complex-differentiable on UU, and for every non-negative integer nn and every z0∈Dz_0 \in D,

f(n)(z0)=n!2πi∮γf(z)(z−z0)n+1 dz.f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz .

Intuition

Think of ff restricted to the contour γ\gamma as a boundary "data trace." The kernel 12πi(z−z0)\dfrac{1}{2\pi i(z-z_0)} acts like a weighting device: it is singular exactly at z0z_0, and integrating it around a tiny circle centered at z0z_0 always contributes exactly f(z0)f(z_0), no matter how small the circle is — because ff is continuous there and the kernel's residue-like structure exactly compensates for the shrinking arc length. Meanwhile, Cauchy's Theorem guarantees that the integral over the big contour γ\gamma equals the integral over that tiny circle, since f(z)z−z0\frac{f(z)}{z-z_0} is holomorphic in the annular region between them. So the "outer" boundary information gets funneled, undiminished, into the value at the single interior point. A useful mental picture: imagine γ\gamma as a boundary "sensor ring" and z0z_0 as a probe point. As you shrink a circle around z0z_0 down to a point, the average value of ff over that circle (weighted appropriately) stabilizes to f(z0)f(z_0) by continuity, while Cauchy's theorem tells you this same average is exactly reproduced by any larger enclosing loop. This "shrinking annulus" argument is precisely the engine of the proof below.

Figure
The contour γ\gamma, an interior point z0z_0, and the shrinking circle CεC_\varepsilon used in the proof
γ z₀ C_ε region where f(z)/(z−z₀) is holomorphic ∮_γ f(z)/(z−z₀) dz = ∮_(C_ε) f(z)/(z−z₀) dz → 2πi·f(z₀)
TheoremCauchy's Integral Formula

Let U⊆CU \subseteq \mathbb{C} be open and f:U→Cf : U \to \mathbb{C} holomorphic on UU. Let γ\gamma be a positively oriented, simple, piecewise-C1C^1 closed contour with enclosed region DD such that D‾⊂U\overline{D} \subset U. Then:

(a) For every z0∈Dz_0 \in D,

f(z0)=12πi∮γf(z)z−z0 dz.f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0}\, dz .

(b) ff is infinitely complex-differentiable on UU, and for every n≥0n \ge 0 and z0∈Dz_0 \in D,

f(n)(z0)=n!2πi∮γf(z)(z−z0)n+1 dz.f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz .
Proof

We prove part (a) first, then part (b) by an inductive differentiation-under-the-integral argument, using (a) as the base case.

Part (a). Fix z0∈Dz_0 \in D. Since DD is open and z0∈Dz_0 \in D, choose R>0R > 0 small enough that the closed disk B‾(z0,R)⊂D\overline{B}(z_0, R) \subset D; this is possible because DD is open. For 0<ε≤R0 < \varepsilon \le R, let CεC_\varepsilon denote the positively oriented circle of radius ε\varepsilon centered at z0z_0. Step 1 (reduction via Cauchy's Theorem). Define g(z)=f(z)z−z0g(z) = \dfrac{f(z)}{z - z_0} for z∈U∖{z0}z \in U \setminus \{z_0\}. This gg is holomorphic on U∖{z0}U \setminus \{z_0\}, since ff is holomorphic there and z−z0≠0z - z_0 \ne 0. Consider the region AεA_\varepsilon between γ\gamma and CεC_\varepsilon: it is bounded by γ\gamma (outer, positive orientation) and CεC_\varepsilon (inner, and we traverse −Cε-C_\varepsilon, i.e. clockwise, to make the combined boundary of AεA_\varepsilon positively oriented as seen from inside AεA_\varepsilon). Since z0∉Aε‾z_0 \notin \overline{A_\varepsilon}, and gg is holomorphic on an open set containing Aε‾\overline{A_\varepsilon} (namely U∖{z0}U \setminus \{z_0\}), Cauchy's Theorem applied to gg on the doubly-connected region AεA_\varepsilon gives

∮γg(z) dz−∮Cεg(z) dz=0,\oint_\gamma g(z)\,dz - \oint_{C_\varepsilon} g(z)\, dz = 0,

i.e.

∮γf(z)z−z0 dz=∮Cεf(z)z−z0 dzfor every 0<ε≤R.(∗)\oint_\gamma \frac{f(z)}{z-z_0}\,dz = \oint_{C_\varepsilon} \frac{f(z)}{z-z_0}\, dz \qquad \text{for every } 0 < \varepsilon \le R. \tag{$\ast$}

(This "boundary of an annulus" form of Cauchy's theorem is obtained from the standard simply-connected version by cutting the annular region with two radial segments joining γ\gamma to CεC_\varepsilon; the segments are traversed once in each direction and cancel, leaving exactly the two circle integrals with the stated relative orientations. The resulting contour bounds a simply connected region on which gg is holomorphic, so the basic Cauchy theorem applies to it.) Since the left side of (∗)(\ast) does not depend on ε\varepsilon, the right side is constant in ε\varepsilon for all 0<ε≤R0 < \varepsilon \le R. Step 2 (evaluating the limit as ε→0\varepsilon \to 0). We compute

∮Cεf(z)z−z0 dz−2πi f(z0)=∮Cεf(z)−f(z0)z−z0 dz,\oint_{C_\varepsilon} \frac{f(z)}{z - z_0}\, dz - 2\pi i\, f(z_0) = \oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z - z_0}\, dz,

using the standard fact ∮Cεdzz−z0=2πi\displaystyle\oint_{C_\varepsilon} \frac{dz}{z-z_0} = 2\pi i (direct computation via the parametrization z=z0+εeiθz = z_0 + \varepsilon e^{i\theta}, θ∈[0,2π]\theta \in [0, 2\pi], giving dz=iεeiθ dθdz = i\varepsilon e^{i\theta}\,d\theta and ∮iεeiθεeiθ dθ=∫02πi dθ=2πi\oint \frac{i \varepsilon e^{i\theta}}{\varepsilon e^{i\theta}}\,d\theta = \int_0^{2\pi} i \, d\theta = 2\pi i). Since ff is holomorphic at z0z_0, it is in particular continuous at z0z_0: given η>0\eta > 0, there is δ∈(0,R]\delta \in (0, R] such that ∣z−z0∣<δ|z - z_0| < \delta implies ∣f(z)−f(z0)∣<η|f(z) - f(z_0)| < \eta. For 0<ε<δ0 < \varepsilon < \delta, every point zz on CεC_\varepsilon satisfies ∣z−z0∣=ε<δ|z - z_0| = \varepsilon < \delta, so ∣f(z)−f(z0)∣<η|f(z) - f(z_0)| < \eta there, and hence on CεC_\varepsilon,

∣f(z)−f(z0)z−z0∣=∣f(z)−f(z0)∣ε<ηε.\left| \frac{f(z) - f(z_0)}{z - z_0} \right| = \frac{|f(z)-f(z_0)|}{\varepsilon} < \frac{\eta}{\varepsilon}.

The length of CεC_\varepsilon is 2πε2\pi\varepsilon. By the ML Inequality,

∣∮Cεf(z)−f(z0)z−z0 dz∣≤ηε⋅2πε=2πη.\left| \oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z-z_0}\, dz \right| \le \frac{\eta}{\varepsilon} \cdot 2\pi \varepsilon = 2\pi \eta.

Since η>0\eta > 0 was arbitrary, letting η→0\eta \to 0 (equivalently ε→0\varepsilon \to 0) forces

∮Cεf(z)−f(z0)z−z0 dz⟶0,\oint_{C_\varepsilon} \frac{f(z) - f(z_0)}{z - z_0}\, dz \longrightarrow 0,

so

∮Cεf(z)z−z0 dz⟶2πi f(z0)as ε→0+.\oint_{C_\varepsilon} \frac{f(z)}{z-z_0}\,dz \longrightarrow 2\pi i\, f(z_0) \quad \text{as } \varepsilon \to 0^+.

Step 3 (conclusion). But by Step 1, the quantity ∮Cεf(z)z−z0 dz\oint_{C_\varepsilon} \frac{f(z)}{z - z_0}\,dz is constant, equal to ∮γf(z)z−z0 dz\oint_\gamma \frac{f(z)}{z-z_0}\,dz, independent of ε\varepsilon. A constant sequence that converges to 2πif(z0)2\pi i f(z_0) must equal 2πif(z0)2\pi i f(z_0) for every ε\varepsilon, in particular

∮γf(z)z−z0 dz=2πi f(z0),\oint_\gamma \frac{f(z)}{z-z_0}\, dz = 2\pi i\, f(z_0),

which is exactly the claimed formula after dividing by 2πi2\pi i. This proves part (a).

Part (b). We show by induction on n≥0n \ge 0 that ff is nn-times complex differentiable at every point of DD and that

f(n)(z0)=n!2πi∮γf(z)(z−z0)n+1 dzfor all z0∈D.(∗∗)f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\, dz \qquad \text{for all } z_0 \in D. \tag{$\ast\ast$}

Base case n=0n=0: this is exactly part (a), already proved, valid for every z0∈Dz_0 \in D (and DD is open, so every point of DD is an interior point of DD, which is what we need to talk about differentiability at z0z_0). Inductive step: suppose (∗∗)(\ast\ast) holds for some n≥0n \ge 0 at every point of DD; we show f(n)f^{(n)} is differentiable at each z0∈Dz_0\in D with derivative given by the (n+1)(n+1) formula. Fix z0∈Dz_0 \in D and choose r>0r>0 with B‾(z0,2r)⊂D\overline{B}(z_0,2r)\subset D (possible since DD is open). Let h∈Ch \in \mathbb{C} with 0<∣h∣<r0 < |h| < r. Define

F(w):=12πi∮γf(z)(z−w)n+1 dz,F(w) := \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-w)^{n+1}}\,dz,

so that by the inductive hypothesis f(n)(w)=n! F(w)f^{(n)}(w) = n!\, F(w) for all w∈Dw \in D. We estimate the difference quotient

F(z0+h)−F(z0)h−(n+1)⋅12πi∮γf(z)(z−z0)n+2 dz.\frac{F(z_0+h) - F(z_0)}{h} - (n+1)\cdot\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+2}}\,dz.

Using the algebraic identity (valid for z≠z0,z0+hz \ne z_0, z_0+h)

1h[1(z−z0−h)n+1−1(z−z0)n+1]−n+1(z−z0)n+2=P(z,z0,h)(z−z0−h)n+1(z−z0)n+2,\frac{1}{h}\left[\frac{1}{(z-z_0-h)^{n+1}} - \frac{1}{(z-z_0)^{n+1}}\right] - \frac{n+1}{(z-z_0)^{n+2}} = \frac{P(z, z_0, h)}{(z-z_0-h)^{n+1}(z-z_0)^{n+2}},

where PP is a polynomial in hh whose coefficients are bounded polynomials in (z−z0)(z-z_0) and vanishing at h=0h=0 (this identity is obtained by direct algebra: writing (z−z0−h)−(n+1)(z - z_0 - h)^{-(n+1)} as a function of hh and using the binomial expansion truncated to first order, the remainder is O(h)O(h) uniformly for zz ranging over the compact set γ\gamma and ∣h∣<r|h| < r, since ∣z−z0∣≥2r|z - z_0| \ge 2r and ∣z−z0−h∣≥2r−r=r|z - z_0 - h| \ge 2r - r = r are both bounded away from 00 on γ\gamma). Consequently there is a constant M=M(n,f,γ,r)M = M(n, f, \gamma, r), independent of hh, such that for all z∈γz \in \gamma and 0<∣h∣<r0<|h|<r,

∣1h[1(z−z0−h)n+1−1(z−z0)n+1]−n+1(z−z0)n+2∣≤M∣h∣,\left| \frac{1}{h}\left[\frac{1}{(z-z_0-h)^{n+1}} - \frac{1}{(z-z_0)^{n+1}}\right] - \frac{n+1}{(z-z_0)^{n+2}} \right| \le M |h|,

because ∣f(z)∣|f(z)| is bounded on the compact contour γ\gamma (continuous function on a compact set), and ∣z−z0∣,∣z−z0−h∣|z-z_0|, |z - z_0 - h| are bounded below by rr on γ\gamma (since B‾(z0,2r)⊂D\overline{B}(z_0, 2r)\subset D places γ\gamma, which lies outside B‾(z0,2r)\overline B(z_0,2r) or at least at distance ≥2r\geq 2r from z0z_0 by our choice of rr — more precisely we choose rr small enough that dist(z0,γ)>2r\mathrm{dist}(z_0,\gamma) > 2r, possible since z0∈Dz_0\in D is an interior point and γ⊂∂D\gamma\subset \partial D or γ\gamma is compact disjoint from DD's interior points at positive distance). By the ML Inequality, multiplying by ∣f(z)∣/2π|f(z)|/2\pi and the length LL of γ\gamma,

∣F(z0+h)−F(z0)h−(n+1)⋅12πi∮γf(z)(z−z0)n+2 dz∣≤L⋅sup⁡γ∣f∣⋅M2π ∣h∣→h→00.\left| \frac{F(z_0+h)-F(z_0)}{h} - (n+1)\cdot \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+2}}\,dz \right| \le \frac{L \cdot \sup_\gamma|f| \cdot M}{2\pi} \, |h| \xrightarrow{h\to 0} 0.

Hence FF is complex-differentiable at z0z_0 with

F′(z0)=(n+1)⋅12πi∮γf(z)(z−z0)n+2 dz.F'(z_0) = (n+1) \cdot \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+2}}\, dz.

Since f(n)=n! Ff^{(n)} = n!\,F, this gives f(n+1)(z0)=n! F′(z0)=(n+1)!2πi∮γf(z)(z−z0)n+2 dzf^{(n+1)}(z_0) = n!\, F'(z_0) = \frac{(n+1)!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+2}}\,dz, which is (∗∗)(\ast\ast) for n+1n+1. This completes the induction. Since z0∈Dz_0 \in D was arbitrary and DD can be taken to be any small disk around any point of UU (with γ\gamma a small circle inside UU enclosing that disk, using that UU is open), the induction shows ff has derivatives of every order at every point of UU, i.e. ff is infinitely complex-differentiable on UU. ■\blacksquare

∎

Worked examples

Example 1: a rational function with an off-center pole, contour a non-centered circle. Let f(z)=z2+1z+3f(z) = \dfrac{z^2 + 1}{z+3}, which is holomorphic on C∖{−3}\mathbb{C}\setminus\{-3\}. Take γ\gamma the circle ∣z−1∣=4|z - 1| = 4, oriented counterclockwise; this circle has center 11 and radius 44, so it encloses the disk D={∣z−1∣<4}D = \{|z-1|<4\}, which contains z0=2z_0 = 2 but does not contain −3-3 (since ∣−3−1∣=4|-3-1| = 4, the singularity lies exactly on the boundary — for a clean application let's instead use γ:∣z−1∣=4.5\gamma: |z-1|=4.5, radius 4.54.5, so that ∣−3−1∣=4<4.5|-3-1|=4 < 4.5 means −3-3 is enclosed too; instead we want −3-3 excluded). Let us restate cleanly: take γ:∣z−1∣=3\gamma: |z - 1| = 3 (radius 33, center 11), so the enclosed disk is {∣z−1∣<3}\{|z-1|<3\}. Check: ∣−3−1∣=4>3|-3-1| = 4 > 3, so −3-3 lies outside D‾\overline D; hence ff is holomorphic on an open set containing D‾\overline D. Take z0=2∈Dz_0 = 2 \in D since ∣2−1∣=1<3|2-1|=1<3. We want to compute

∮γf(z)z−2 dz=∮γz2+1(z+3)(z−2) dz.\oint_\gamma \frac{f(z)}{z-2}\, dz = \oint_\gamma \frac{z^2+1}{(z+3)(z-2)}\, dz.

By Cauchy's Integral Formula (part (a), applicable since ff is holomorphic on and inside γ\gamma, and z0=2z_0=2 is interior to γ\gamma),

∮γf(z)z−2 dz=2πi f(2)=2πi⋅22+12+3=2πi⋅55=2πi.\oint_\gamma \frac{f(z)}{z - 2}\, dz = 2\pi i \, f(2) = 2\pi i \cdot \frac{2^2+1}{2+3} = 2\pi i \cdot \frac{5}{5} = 2\pi i.

So the contour integral, without any residue calculus or partial fractions, is exactly 2πi2\pi i — a clean confirmation that the formula converts an integral computation into a mere function evaluation. Let's also extract the first derivative using part (b) with n=1n=1:

f′(2)=12πi∮γf(z)(z−2)2 dz.f'(2) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-2)^2}\, dz.

Direct differentiation gives f′(z)=2z(z+3)−(z2+1)(z+3)2=z2+6z−1(z+3)2f'(z) = \dfrac{2z(z+3) - (z^2+1)}{(z+3)^2} = \dfrac{z^2 + 6z - 1}{(z+3)^2}, so

f′(2)=4+12−125=1525=35.f'(2) = \frac{4 + 12 - 1}{25} = \frac{15}{25} = \frac{3}{5}.

Hence Cauchy's formula predicts

∮γ(z2+1)/(z+3)(z−2)2 dz=2πi⋅35=6πi5,\oint_\gamma \frac{(z^2+1)/(z+3)}{(z-2)^2}\, dz = 2\pi i \cdot \frac{3}{5} = \frac{6\pi i}{5},

a value one could in principle also verify by direct (much messier) parametrized integration — the formula spares us that labor entirely.

Example 2: an entire function (exponential), contour a large offset circle, extracting a higher derivative. Let f(z)=ez2f(z) = e^{z^2}, entire (holomorphic on all of C\mathbb{C}, so hypotheses are automatically satisfiable for any simple closed contour). Take γ\gamma: the circle ∣z−i∣=2|z - i| = 2, oriented counterclockwise, so D={∣z−i∣<2}D = \{|z-i|<2\}. Take z0=0z_0 = 0; check ∣0−i∣=1<2|0 - i| = 1 < 2, so z0∈Dz_0 \in D. We compute the second derivative f′′(0)f''(0) two ways. Direct computation: f(z)=ez2f(z) = e^{z^2}, so f′(z)=2zez2f'(z) = 2z e^{z^2}, and

f′′(z)=2ez2+2z⋅2zez2=2ez2(1+2z2),f''(z) = 2e^{z^2} + 2z \cdot 2z e^{z^2} = 2e^{z^2}(1 + 2z^2),

giving f′′(0)=2⋅1⋅(1+0)=2f''(0) = 2 \cdot 1 \cdot (1+0) = 2. Via Cauchy's formula (part (b), n=2n=2):

f′′(0)=2!2πi∮γez2z3 dz=1πi∮γez2z3 dz.f''(0) = \frac{2!}{2\pi i} \oint_\gamma \frac{e^{z^2}}{z^3}\, dz = \frac{1}{\pi i}\oint_\gamma \frac{e^{z^2}}{z^3}\,dz.

Solving for the integral using the known value f′′(0)=2f''(0) = 2:

∮γez2z3 dz=πi⋅f′′(0)=2πi.\oint_\gamma \frac{e^{z^2}}{z^3}\, dz = \pi i \cdot f''(0) = 2\pi i.

This is a genuinely useful computation: evaluating ∮∣z−i∣=2ez2z3 dz\oint_{|z-i|=2} \dfrac{e^{z^2}}{z^3}\,dz directly by parametrization would require expanding ez2e^{z^2} along the circle and integrating termwise — a substantially harder task than simply differentiating ez2e^{z^2} twice at the origin and multiplying by πi\pi i. As a cross-check using the general formula with n=0n=0 instead: ∮γez2z dz=2πi f(0)=2πi⋅e0=2πi\oint_\gamma \frac{e^{z^2}}{z}\,dz = 2\pi i \, f(0) = 2\pi i \cdot e^0 = 2\pi i, and with n=1n=1: ∮γez2z2 dz=2πi f′(0)=2πi⋅0=0\oint_\gamma \frac{e^{z^2}}{z^2}\,dz = 2\pi i\, f'(0) = 2\pi i \cdot 0 = 0 (since f′(0)=2⋅0⋅e0=0f'(0) = 2\cdot 0 \cdot e^0 = 0), consistent with f′f' being an odd function forcing this vanishing.

Counterexamplewhy "holomorphic" cannot be weakened to "continuous"

The hypothesis that ff is holomorphic on an open set containing D‾\overline D is essential — merely continuous on D‾\overline D is not enough, and the reproducing formula fails. Take γ\gamma the unit circle ∣z∣=1|z|=1, oriented counterclockwise, enclosing D={∣z∣<1}D = \{|z|<1\}. Let z0=0z_0 = 0. Define

f(z)=z‾(complex conjugation),f(z) = \overline{z} \quad \text{(complex conjugation)},

which is continuous on all of C\mathbb{C} (indeed real-analytic as a map R2→R2\mathbb{R}^2\to\mathbb{R}^2) but nowhere complex-differentiable — the Cauchy–Riemann equations fail identically, since writing z=x+iyz=x+iy, f=x−iyf = x - iy has u=x,v=−yu=x, v=-y, so ux=1≠−1=vyu_x = 1 \ne -1 = v_y everywhere. We compute the candidate "Cauchy formula" integral directly:

12πi∮∣z∣=1z‾z−0 dz=12πi∮∣z∣=1z‾z dz.\frac{1}{2\pi i}\oint_{|z|=1} \frac{\overline{z}}{z - 0}\, dz = \frac{1}{2\pi i}\oint_{|z|=1} \frac{\overline{z}}{z}\, dz.

Parametrize z=eiθz = e^{i\theta}, θ∈[0,2π]\theta \in [0,2\pi], so z‾=e−iθ\overline z = e^{-i\theta} and dz=ieiθ dθdz = ie^{i\theta}\,d\theta:

12πi∫02πe−iθeiθ⋅ieiθ dθ=12πi∫02πi e−iθ dθ=12π∫02πe−iθ dθ=12π[e−iθ−i]02π=0,\frac{1}{2\pi i}\int_0^{2\pi} \frac{e^{-i\theta}}{e^{i\theta}}\cdot i e^{i\theta}\, d\theta = \frac{1}{2\pi i}\int_0^{2\pi} i\, e^{-i\theta}\, d\theta = \frac{1}{2\pi}\int_0^{2\pi} e^{-i\theta}\, d\theta = \frac{1}{2\pi}\left[\frac{e^{-i\theta}}{-i}\right]_0^{2\pi} = 0,

since ∫02πe−iθ dθ=0\int_0^{2\pi} e^{-i\theta}\,d\theta = 0 (a full period of a nonzero-frequency oscillation). So the "Cauchy formula" integral evaluates to 00. But the actual function value is f(z0)=f(0)=0‾=0f(z_0) = f(0) = \overline{0} = 0 as well — this particular case happens to match by coincidence of the origin being special (0‾=0\overline{0}=0). To see a genuine failure, take instead z0=1/2z_0 = 1/2 (still inside ∣z∣<1|z|<1). We would need

12πi∮∣z∣=1z‾z−1/2 dzf(1/2)=1/2‾=12.\frac{1}{2\pi i}\oint_{|z|=1} \frac{\overline z}{z - 1/2}\, dz \overset{?}{=} f(1/2) = \overline{1/2} = \tfrac12.

Parametrize again z=eiθz=e^{i\theta}, z‾=e−iθ\overline z = e^{-i\theta}, dz=ieiθdθdz = ie^{i\theta} d\theta:

12πi∫02πe−iθeiθ−1/2⋅ieiθ dθ=12π∫02πe−iθ⋅eiθeiθ−1/2 dθ=12π∫02π1eiθ−1/2 dθ.\frac{1}{2\pi i}\int_0^{2\pi} \frac{e^{-i\theta}}{e^{i\theta} - 1/2}\cdot i e^{i\theta}\,d\theta = \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{-i\theta} \cdot e^{i\theta}}{e^{i\theta}-1/2}\,d\theta = \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{e^{i\theta} - 1/2}\, d\theta.

This is 12πi∮∣z∣=11z(z−1/2) dz\dfrac{1}{2\pi i}\oint_{|z|=1} \dfrac{1}{z(z - 1/2)}\, dz after re-substituting dθ=dz/(iz)d\theta = dz/(iz). Using partial fractions 1z(z−1/2)=−2z+2z−1/2\frac{1}{z(z-1/2)} = \frac{-2}{z} + \frac{2}{z-1/2}, and the standard fact ∮∣z∣=1dzz=2πi\oint_{|z|=1}\frac{dz}{z} = 2\pi i, ∮∣z∣=1dzz−1/2=2πi\oint_{|z|=1}\frac{dz}{z-1/2} = 2\pi i (both singularities 00 and 1/21/2 lie inside the unit circle), we get

∮∣z∣=1dzz(z−1/2)=−2(2πi)+2(2πi)=0.\oint_{|z|=1}\frac{dz}{z(z-1/2)} = -2(2\pi i) + 2(2\pi i) = 0.

So the candidate formula gives 12πi⋅0=0\dfrac{1}{2\pi i}\cdot 0 = 0, while the true value is f(1/2)=1/2≠0f(1/2) = 1/2 \ne 0. The formula fails for this merely-continuous, non-holomorphic ff. This confirms holomorphicity — not mere continuity — is the load-bearing hypothesis: the entire proof relied on Cauchy's Theorem to move the contour, and Cauchy's theorem itself requires holomorphicity (the Cauchy–Riemann equations are what make the relevant differential form closed).

Scope, edge cases & common traps

SituationWhat happens
z0z_0 outside γ\gammaThe formula gives 00, not f(z0)f(z_0) — this is a separate consequence of Cauchy's Theorem since f(z)z−z0\frac{f(z)}{z-z_0} is then holomorphic on all of D‾\overline D.
z0z_0 exactly on γ\gammaFormula hypotheses (`z0∈Dz_0\in D`, the open interior) are violated; the integral is typically a principal-value type object, not simply f(z0)f(z_0) or 2π⋅i⋅02\pi\cdot i\cdot 0.
γ\gamma not simple (self-intersecting)Winding number around z0z_0 may be ≠1\ne 1; the right side becomes 2πi n(γ,z0)f(z0)2\pi i\, n(\gamma,z_0) f(z_0) where n(γ,z0)n(\gamma,z_0) is the winding number.
ff merely continuous, not holomorphicFormula fails in general — see the counterexample above with f(z)=z‾f(z)=\overline z.
ff holomorphic on DD but only continuous on D‾\overline D (boundary)The formula still holds; this refinement follows by an approximation argument shrinking γ\gamma slightly inward, using uniform continuity on the compact closure.
Multiple singularities of ff inside DDFormula as stated needs ff holomorphic on all of D‾\overline D — if ff itself has singularities inside, one instead applies the formula to f(z)/(z−z0)f(z)/(z-z_0) pole structure via the residue theorem, a distinct (later) tool.

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Post-revolutionary France, École Polytechnique and the rise of rigorous analysis

Cauchy was one of the founders of modern rigorous analysis, responsible for the epsilon-delta formulation of limits, the first rigorous theory of convergence for infinite series, and — the achievement most relevant here — the systematic theory of functions of a complex variable built on contour integration. Working amid the political turbulence of post-Napoleonic France (he was a royalist exile for a period after 1830), Cauchy produced an extraordinary volume of work: over 800 papers across analysis, mathematical physics, and elasticity theory. His integral formula, developed across memoirs in the 1820s–1840s, transformed complex analysis from a collection of formal manipulations of −1\sqrt{-1} into a rigorous discipline with its own proof techniques — contour deformation, the ML inequality, and residue calculus — that remain the field's core toolkit two centuries later.

History
- 1789 — Cauchy born in Paris, days after the storming of the Bastille.
- 1814 — Cauchy presents early work on definite integrals with complex limits to the French Academy.
- 1825 — Cauchy publishes "Mémoire sur les intégrales définies," containing the germinal form of Cauchy's Theorem.
- 1831 — Cauchy, in exile in Turin, lectures on and refines the integral formula bearing his name.
- 1841 — mature statement and applications of the integral formula appear in Cauchy's collected works and lecture courses, cementing it as the centerpiece of complex function theory.
Why does Cauchy's Integral Formula fail for real-valued differentiable functions?

The formula's proof crucially uses Cauchy's Theorem, which relies on the Cauchy–Riemann equations making f dzf\,dz a closed differential form — a two-dimensional cancellation that has no one-dimensional analogue. In one real variable, "the boundary of an interval" is just two points, and there's no contour to integrate around; the entire mechanism of encircling a point and shrinking a loop to it doesn't exist. Complex differentiability is a much stronger condition than real differentiability (it forces the Cauchy–Riemann equations, a system of PDEs), and this extra rigidity is exactly what the integral formula exploits.

Does the contour γ\gamma have to be a circle?

No. The theorem holds for any positively oriented, simple, piecewise-C1C^1 closed contour, as long as z0z_0 lies inside it and ff is holomorphic on and inside it. Circles are used constantly in proofs (as in the shrinking-circle argument here) because they make explicit parametrization and the ML inequality especially easy, but the formula itself is contour-shape agnostic — it follows from the shape-independence built into Cauchy's Theorem.

What happens if γ\gamma winds around z0z_0 more than once?

If γ\gamma has winding number kk around z0z_0 (for instance, a spiral traversed twice), the formula generalizes to 12πi∮γf(z)z−z0 dz=k⋅f(z0)\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-z_0}\,dz = k\cdot f(z_0). The version stated in this article assumes γ\gamma is simple (non-self-intersecting) and positively oriented, which forces winding number exactly 11 around any interior point — this is the standard and most commonly needed case.

How does this formula relate to the mean value property of harmonic functions?

Parametrizing γ\gamma as a circle z=z0+reiθz = z_0 + re^{i\theta} in the formula and simplifying shows f(z0)=12π∫02πf(z0+reiθ) dθf(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0 + re^{i\theta})\,d\theta: the value at the center equals the average over the circle. Taking real parts, since the real part of a holomorphic function is harmonic, this recovers exactly the mean value property of harmonic functions — a foundational fact in potential theory, and a special case of Cauchy's formula rather than an independent result.

Can the enclosed region DD be multiply connected, like an annulus?

The version stated here requires DD to be the interior of a single simple closed curve (simply connected in effect). For multiply connected regions — say ff holomorphic on an annulus, with γ\gamma its outer boundary and an inner boundary circle both included with correct orientations — a generalized formula holds where the single contour integral is replaced by a sum over all boundary components, each with matching orientation. This generalization underlies Laurent series expansions on annuli.

Why can't we just say "trivially, the integral of f(z)/(z−z0)f(z)/(z-z_0) over a small circle equals f(z0)f(z_0) times 2πi2\pi i" without proof?

Because that step is exactly where the substantive analysis lives: it requires the ML inequality estimate showing the "error term" ∮(f(z)−f(z0))/(z−z0) dz\oint (f(z)-f(z_0))/(z-z_0)\,dz vanishes as the circle shrinks, which in turn depends on continuity of ff at z0z_0 (guaranteed by holomorphicity) and a careful bound on arc length versus radius. Skipping this is the single most common gap in student proofs of the formula — the "ε\varepsilon-δ\delta shrinking circle" argument is not optional bookkeeping, it is the heart of the proof.

Is Cauchy's Integral Formula the same as the Cauchy Integral Theorem?

No, though they are tightly linked. Cauchy's Theorem states that ∮γf(z) dz=0\oint_\gamma f(z)\,dz = 0 for ff holomorphic on and inside γ\gamma — no singularity, no reproducing property. Cauchy's Integral Formula concerns ∮γf(z)z−z0 dz\oint_\gamma \frac{f(z)}{z-z_0}\,dz, where the integrand has a genuine singularity at z0∈Dz_0 \in D; the theorem is used as a tool inside the formula's proof (applied to the annular region between γ\gamma and a small circle around z0z_0), but the formula itself extracts new, non-trivial information (an explicit function value) rather than merely asserting vanishing.

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Where this leads