Cauchy's Integral Theorem

Complex integration hands you two objects — a contour $C$ and a function $f$ — and asks for the line integral $\int_C f(z)\,\mathrm{d}z$ . For a generic function this number depends on the path: bend the contour and the answer changes. Cauchy's integral theorem is the single statement that says when it does not. If $f$ is analytic on and inside a closed loop, and the region it bounds has no holes, then the loop integral is exactly zero — no approximation, no leading term, exactly $0$ . That one equation, $\oint_C f\,\mathrm{d}z = 0$ , is the hinge on which complex analysis turns. The reason it holds is that analyticity is an unusually rigid demand. A complex-differentiable function obeys the Cauchy–Riemann equations, which lash its real and imaginary parts together so tightly that the planar "circulation" and "flux" of the associated fields both vanish at every interior point. Green's theorem sums those vanishing local quantities over the area, and the boundary integral collapses. Read the other way, $f(z)\,\mathrm{d}z$ is locally an exact differential $\mathrm{d}F$ with $F' = f$ , and the integral of an exact differential around any closed loop is zero — path-independence in disguise. Édouard Goursat sharpened the hypothesis to its irreducible core. Cauchy's own argument and the Green's-theorem route quietly assume that $f'$ is continuous; Goursat's triangle-bisection proof shows that mere existence of $f'$ suffices — continuity is then a theorem, not an assumption. That refinement is what lets one prove analytic functions are infinitely differentiable without arguing in a circle. Extended by homotopy to domains with holes, the theorem becomes the deformation principle behind the residue calculus, the most powerful evaluation tool in mathematical physics.

Key takeaways

Cauchy's integral theorem states that if $f$ is analytic throughout a simply connected domain $D$ , then $\oint_C f(z)\,\mathrm{d}z = 0$ for every closed contour $C$ lying in $D$ .
It follows from the Cauchy–Riemann equations: split $f = u + iv$ and Green's theorem turns the loop integral into area integrals whose integrands vanish identically.
Goursat's proof drops the assumption that $f'$ is continuous — it proves the theorem for a triangle by repeated bisection, making analyticity alone the hypothesis.
Equivalent readings: $f\,\mathrm{d}z$ is a closed/exact one-form; a single-valued antiderivative $F$ exists; and integrals between two points are path-independent.
The benchmark $\oint_{|z|=1}\mathrm{d}z/z = 2\pi i \neq 0$ shows the hypothesis is essential — an enclosed singularity breaks the theorem and seeds the residue.
Key idea: For $f$ analytic on a simply connected domain $D$ , the one-form $f(z)\,\mathrm{d}z$ is closed and exact, so its integral around any closed loop in $D$ is zero — and integrals between two points depend only on the endpoints, not the path.

Where this lives today

Cauchy's theorem is used, unchanged, every time a physicist deforms a contour. In quantum field theory the propagator $\int \frac{\mathrm{d}^4 k}{(2\pi)^4}\frac{i}{k^2 - m^2 + i\epsilon}$ is evaluated by closing the energy contour, where the $i\epsilon$ prescription decides which poles are enclosed and hence encodes causality; the closing arc contributes nothing because the integrand is analytic there. Dispersion relations — the Kramers–Kronig relations of optics and their descendants for scattering amplitudes — are Cauchy's theorem applied to a function analytic in the upper half-plane, tying dispersive and absorptive parts together as a necessity of analyticity plus causality. Matsubara frequency sums in statistical mechanics collapse onto poles by the same deformation logic. The open frontier is that these methods increasingly meet functions with subtle analytic structure — Landau singularities of multi-loop amplitudes, branch cuts in effective field theories — where knowing exactly where analyticity fails is the whole game, and Cauchy's theorem is the ruler for those failures.

The physical picture

Picture $f(z)\,\mathrm{d}z$ as the work done by a planar flow as you march once around a loop. The Cauchy–Riemann equations force both the curl and the divergence of the relevant fields to vanish at every interior point, so there is nothing inside to accumulate: you return having gained exactly zero. It is the complex-analytic version of a conservative force field, whose potential is the antiderivative $F$ . The first figure makes this local-to-global logic visible — countless microscopic loops, each with zero circulation, tile the region so that all interior edges cancel and only the outer boundary survives, itself zero.

Figure

Analytic ⇒ zero circulation everywhere inside, so the boundary integral vanishes: every interior cell has ∮ f dz = 0 by Cauchy–Riemann, shared edges cancel in opposite directions, and only the outer loop C remains — giving ∮_C f dz = 0

The moment a hole appears — a point where $f$ stops being analytic — the summation breaks. There the "curl" concentrates into a delta-like source, and the loop integral no longer cancels: it picks up a fixed quantum, $2\pi i$ times the residue. This is not a failure but the theorem's most useful corollary, because it means a large contour can be shrunk onto the singularities it encloses without changing its value. The second figure carries the canonical number of the whole subject — the integral of $1/z$ around the unit circle — and shows the deformation that turns it into a computational engine.

Figure

The benchmark that fixes the scale: 1/z has a pole at z = 0, and ∮ over the unit circle gives ∮ dz/z = 2πi — a value invariant under deforming the loop to any circle around the origin, while a loop enclosing no singularity gives ∮ dz/z = 0

The core equations

The theorem itself is a single line: the integral of an analytic function around any closed contour in a hole-free region is zero. Everything downstream — the integral formula, series expansions, the residue theorem — follows from this one equation.

\oint_C f(z)\,\mathrm{d}z = 0 \qquad (1)

where: $f$ — a function analytic on and inside the closed contour $C$ ; $C$ — a piecewise-smooth simple closed contour in a simply connected domain $D$ ; $z$ — the complex integration variable. Read it as: no analytic circulation, no accumulation. The two hypotheses are inseparable — analyticity supplies the local vanishing, simple connectivity guarantees no interior holes. To prove it we split $f$ into real and imaginary parts and treat $\mathrm{d}z$ likewise, turning a complex line integral into two real ones.

f(z) = u(x,y) + i\,v(x,y), \qquad \mathrm{d}z = \mathrm{d}x + i\,\mathrm{d}y \qquad (2)

where: $u,v$ — the real and imaginary parts of $f$ ; $x,y$ — real coordinates with $z = x + iy$ . Pure bookkeeping, but the bridge that lets Green's theorem act on a complex integral. The engine that makes the real integrands vanish is the pair of Cauchy–Riemann equations, the defining condition of analyticity.

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \qquad (3)

where: $\partial_x, \partial_y$ — partial derivatives with respect to $x$ and $y$ . These couple $u$ and $v$ so rigidly that the associated fields are simultaneously irrotational and divergence-free; that double vanishing is what kills both area integrals in the proof below. When singularities are present, the estimation (ML) inequality bounds a contour integral by the size of the integrand times the length of the path — the tool that squeezes Goursat's remainder to zero and controls closing arcs at infinity.

\left|\int_C f(z)\,\mathrm{d}z\right| \le M L, \qquad M = \max_{z \in C}|f(z)| \qquad (4)

where: $M$ — the supremum of $|f|$ on the contour; $L$ — the arc length of $C$ . A worst-case bound: a small integrand or a short path cannot produce a large integral. The deepest reformulation is that $f$ possesses a single-valued antiderivative on a simply connected domain, so $f\,\mathrm{d}z$ is an exact differential.

F(z) = \int_{z_0}^{z} f(\zeta)\,\mathrm{d}\zeta, \qquad F'(z) = f(z) \qquad (5)

where: $F$ — the complex primitive, well defined because the integral is path-independent; $z_0$ — a fixed base point in $D$ . Once $F$ exists, $\oint_C f\,\mathrm{d}z = \oint_C \mathrm{d}F$ is the net change of $F$ around a loop, which is zero — the whole theorem in one sentence. Finally, the deformation theorem carries the result into multiply connected domains — the direct ancestor of the residue theorem.

\oint_{\Gamma} f\,\mathrm{d}z = \sum_{k=1}^{n} \oint_{\gamma_k} f\,\mathrm{d}z \qquad (6)

where: $\Gamma$ — the outer, positively oriented boundary; $\gamma_k$ — small loops each encircling one hole, oriented the same way. Read it as the big loop equals the sum of the little ones: the analytic region between them contributes nothing, so a contour collapses onto the singularities it surrounds.

Derivation AThe Green's theorem proof via Cauchy Riemann

The goal of this first proof is to convert the complex loop integral $\oint_C f\,\mathrm{d}z$ into a pair of real area integrals using Green's theorem, and then show the Cauchy–Riemann equations make both integrands vanish identically at every interior point. This is the historically original route — it assumes $f'$ is continuous, an assumption Goursat later removed — and it is the one that exposes why analyticity, and nothing weaker, forces the answer to zero: the two CR equations are exactly the two conditions needed to annihilate the two area integrals.

Step 1. Substitute $f = u + iv$ and $\mathrm{d}z = \mathrm{d}x + i\,\mathrm{d}y$ into the loop integral and multiply out $(u+iv)(\mathrm{d}x + i\,\mathrm{d}y)$ , grouping the real and imaginary parts:

\oint_C f\,\mathrm{d}z = \oint_C (u\,\mathrm{d}x - v\,\mathrm{d}y) + i\oint_C (v\,\mathrm{d}x + u\,\mathrm{d}y)

Why: a complex line integral is nothing but two real line integrals bundled together, and separating them lets a real-variable theorem act on each.

Step 2. Apply Green's theorem $\oint(P\,\mathrm{d}x + Q\,\mathrm{d}y) = \iint_R (\partial_x Q - \partial_y P)\,\mathrm{d}A$ to each real integral over the enclosed region $R$ , with $(P,Q) = (u,-v)$ and $(v,u)$ respectively:

\oint_C f\,\mathrm{d}z = \iint_R (-\partial_x v - \partial_y u)\,\mathrm{d}A + i\iint_R (\partial_x u - \partial_y v)\,\mathrm{d}A

Why: Green's theorem trades the boundary integral for an area integral of derivatives, moving the whole question inside the loop, where analyticity is a pointwise statement.

Step 3. Invoke the Cauchy–Riemann equations $\partial_x u = \partial_y v$ and $\partial_y u = -\partial_x v$ . The first area integrand is $-\partial_x v - \partial_y u = -\partial_x v + \partial_x v = 0$ , and the second is $\partial_x u - \partial_y v = \partial_y v - \partial_y v = 0$ :

-\partial_x v - \partial_y u = 0, \qquad \partial_x u - \partial_y v = 0

Why: each CR equation cancels exactly one integrand — this is the precise sense in which analyticity is "two conditions that kill two integrals," and dropping either one destroys the result.

Step 4. With both integrands identically zero over $R$ , both area integrals vanish, so $\oint_C f\,\mathrm{d}z = 0$ . The result used only continuity of $u, v$ and their first partials plus the CR equations — continuous differentiability of $f$ . Why: this is the full theorem under the classical hypothesis; Goursat's job is to lift the crutch of assuming $f'$ continuous. *Verified: the CR cancellations reproduce zero integrands; matches Arfken–Weber §11.3 and Ahlfors §4.2.*

Go deeper · Derivation B · 2 minGoursats triangle proof without assuming continuity of the derivative

Derivation BGoursats triangle proof without assuming continuity of the derivative

The goal here is to prove $\oint_{\partial T} f\,\mathrm{d}z = 0$ for the boundary of a triangle $T$ using only the existence of $f'$ — never its continuity. The strategy is bisection and contradiction: suppose $|I| > 0$ , quarter the triangle repeatedly to trap a point where differentiability holds, and show the surviving integral shrinks in exact balance with its geometric lower bound, forcing zero. This is the logically essential upgrade, because the continuity of $f'$ and the smoothness of analytic functions are themselves consequences of this theorem, so assuming them would be circular.

Step 1. Let $I = \oint_{\partial T} f\,\mathrm{d}z$ . Join the midpoints of the three sides to split $T$ into four congruent sub-triangles $T_1,\dots,T_4$ . Interior edges are traversed twice in opposite senses and cancel, so

I = \sum_{j=1}^{4} I_j, \qquad I_j = \oint_{\partial T_j} f\,\mathrm{d}z

Why: subdivision preserves the total integral exactly while shrinking the geometry, setting up an infinite descent.

Step 2. By the triangle inequality at least one sub-triangle — relabel it $T^{(1)}$ — carries at least a quarter of the magnitude:

|I| \le \sum_{j}|I_j| \;\Rightarrow\; |I_{T^{(1)}}| \ge \tfrac{1}{4}|I|

Why: if all four were smaller than $|I|/4$ their sum could not reach $|I|$ ; picking the worst child keeps a definite fraction of the integral alive.

Step 3. Iterate the selection. After $n$ bisections the nested triangles $T^{(n)}$ shrink to a single point $a$ , and their perimeter and diameter halve each time:

|I_{T^{(n)}}| \ge 4^{-n}|I|, \qquad L^{(n)} = 2^{-n}L, \qquad d^{(n)} = 2^{-n}d

Why: the nested triangles share a single common point $a$ by completeness, and $a$ is where we cash in the one thing we may assume — differentiability.

Step 4. At $a$ , differentiability gives $f(z) = f(a) + f'(a)(z-a) + \eta(z)(z-a)$ with $\eta(z) \to 0$ as $z \to a$ . The constant and linear terms have exact antiderivatives, so they integrate to zero around any closed loop:

\oint_{\partial T^{(n)}}\!\big[f(a) + f'(a)(z-a)\big]\,\mathrm{d}z = 0

Why: only the remainder $\eta(z)(z-a)$ survives — anything with a primitive contributes nothing to a closed contour.

Step 5. Bound the surviving remainder with the ML inequality. Given $\varepsilon > 0$ , choose $n$ so large that $|\eta(z)| < \varepsilon$ throughout $T^{(n)}$ ; using $|z-a| \le d^{(n)}$ and length $L^{(n)}$ ,

|I_{T^{(n)}}| = \left|\oint_{\partial T^{(n)}}\eta(z)(z-a)\,\mathrm{d}z\right| \le \varepsilon\, d^{(n)} L^{(n)} = \varepsilon\,4^{-n}\,dL

Why: the product $d^{(n)}L^{(n)} = 4^{-n}dL$ decays at the same rate $4^{-n}$ as the lower bound — the argument hinges on this exact matching of powers.

Step 6. Combine the lower bound of Step 3 with the upper bound of Step 5. The $4^{-n}$ factors cancel:

4^{-n}|I| \le \varepsilon\,4^{-n}\,dL \;\Rightarrow\; |I| \le \varepsilon\,dL \to 0

Since $\varepsilon$ is arbitrary, $|I| = 0$ , hence $I = 0$ . Why: the geometric decay and the estimate balance precisely, leaving zero as the only self-consistent value. *Verified: standard Goursat argument; reproduced in Ahlfors §4.1 and Stone & Goldbart Ch. 17.*

Go deeper · Derivation C · 2 minFrom the triangle to a primitive and path independence

Derivation CFrom the triangle to a primitive and path independence

The goal of the third proof is to bootstrap the triangle result to arbitrary closed contours by constructing an antiderivative. Working first in a convex subdomain, we define $F(z) = \int_{z_0}^{z} f\,\mathrm{d}\zeta$ along straight segments, use the triangle theorem to show it is well defined, differentiate to prove $F' = f$ , and read off both the general theorem and path-independence as corollaries. This is the reading that connects Cauchy's theorem to the physics of a conservative field with a potential, and why the whole subject can be phrased as " $f\,\mathrm{d}z$ is exact."

Step 1. In a convex region fix a base point $z_0$ and define $F(z) = \int_{z_0}^{z} f(\zeta)\,\mathrm{d}\zeta$ integrated along the straight segment. Two such segments plus the segment between endpoints form a triangle, on whose boundary the Goursat theorem gives zero, so $F$ is independent of which vertex path is taken and is well defined:

F(z) = \int_{z_0}^{z} f(\zeta)\,\mathrm{d}\zeta

Why: the triangle theorem is exactly the statement needed to make the segment integral unambiguous, converting a proof about loops into a construction of a function.

Step 2. Form the difference quotient along the short segment from $z$ to $z+h$ , subtracting the exact integral of the constant $f(z)$ , which equals $h\,f(z)$ :

\frac{F(z+h) - F(z)}{h} - f(z) = \frac{1}{h}\int_z^{z+h}\big[f(\zeta) - f(z)\big]\,\mathrm{d}\zeta

Why: isolating $f(\zeta) - f(z)$ turns the derivative question into an average of small deviations that continuity can control.

Step 3. Bound the right side by $\max_{|\zeta - z| \le |h|}|f(\zeta) - f(z)|$ , which tends to $0$ as $h \to 0$ by continuity of $f$ :

\left|\frac{F(z+h)-F(z)}{h} - f(z)\right| \le \max_{|\zeta-z|\le|h|}|f(\zeta)-f(z)| \to 0

So $F'(z) = f(z)$ : a local primitive exists. Why: this is the fundamental theorem of calculus in the complex plane, and it is what makes $f\,\mathrm{d}z$ exact.

Step 4. Wherever a single-valued primitive exists, the closed-loop integral is the net change of $F$ , which is zero, and simple connectivity lets any contour be covered by overlapping convex patches with consistent primitives:

\oint_C f\,\mathrm{d}z = \oint_C \mathrm{d}F = 0

Consequently two contours from $a$ to $b$ give equal integrals — path-independence — since their difference is a closed loop. Why: this delivers the general theorem and its most-used corollary at once, the exact analogue of a conservative force doing zero net work around a loop. *Verified: matches the Cauchy–Goursat statement in Arfken–Weber §11.3 and Ahlfors §4.2.*

Limiting cases

The conclusion lives or dies on two things: analyticity of the integrand and topology of the region.

Case	Result	Meaning
No enclosed singularity	$\oint_{\|z\|=1}\dfrac{\mathrm{d}z}{z-2} = 0$	The pole at $z=2$ lies outside the unit circle; $f$ is analytic inside, so the theorem applies and the loop integral vanishes.
Enclosed singularity	$\oint_{\|z\|=1}\dfrac{\mathrm{d}z}{z} = 2\pi i$	Analyticity fails at the enclosed point $z=0$ ; the integral is nonzero and equals $2\pi i$ times the residue — the theorem's hypothesis is violated.
Homotopy / deformation	$\oint_{C_1} f\,\mathrm{d}z = \oint_{C_2} f\,\mathrm{d}z$	If $C_1$ and $C_2$ can be continuously deformed into one another within the analytic region, their integrals coincide — contours are free to move across holeless territory.
Multiply connected	$\oint_{\Gamma} f\,\mathrm{d}z = \sum_k \oint_{\gamma_k} f\,\mathrm{d}z$	An outer contour enclosing several holes equals the sum of small loops around each — the deformation theorem and the seed of the residue calculus.
Non-analytic integrand	$\oint_{\|z\|=1}\bar z\,\mathrm{d}z = 2\pi i \neq 0$	$\bar z$ fails Cauchy–Riemann everywhere, so no primitive exists; a closed loop need not vanish even with no "pole" in the naive sense.

Common trapThe theorem does not say every closed integral is zero

The vanishing is a statement about analytic integrands over hole-free regions — not a universal law for closed loops. The commonest error is to write $\oint_C f\,\mathrm{d}z = 0$ without checking that $f$ is analytic on and inside $C$ . The benchmark $\oint_{|z|=1}\mathrm{d}z/z = 2\pi i$ is nonzero because $1/z$ is singular at the enclosed origin; likewise $\oint_{|z|=1}\bar z\,\mathrm{d}z = 2\pi i$ because $\bar z$ is nowhere analytic. Both hypotheses must hold: analyticity supplies the local vanishing, and simple connectivity lets it sum to a global zero. Drop either and the theorem is silent — often the true answer is a nonzero multiple of $2\pi i$ .

Worked examples

1 · The master computation ∮ zⁿ dz on the unit circle (fully worked)

Evaluate $\oint_{|z|=1} z^n\,\mathrm{d}z$ for every integer $n$ , and read off both the theorem and its failure from the single result.

Step 1. Parametrise the unit circle by $z = e^{i\theta}$ with $\theta$ running from $0$ to $2\pi$ , so $\mathrm{d}z = i e^{i\theta}\,\mathrm{d}\theta$ . Substituting,

\oint_{|z|=1} z^n\,\mathrm{d}z = \int_0^{2\pi} e^{in\theta}\,i e^{i\theta}\,\mathrm{d}\theta = i\int_0^{2\pi} e^{i(n+1)\theta}\,\mathrm{d}\theta.

Step 2. For any integer $n \neq -1$ the exponent $n+1$ is a nonzero integer, and $\int_0^{2\pi} e^{im\theta}\,\mathrm{d}\theta = 0$ for integer $m \neq 0$ because the exponential completes whole cycles. Hence for all $n \neq -1$ ,

\oint_{|z|=1} z^n\,\mathrm{d}z = i\cdot\frac{e^{i(n+1)2\pi} - 1}{i(n+1)} = \frac{1 - 1}{n+1} = \mathbf{0}.

For $n \ge 0$ , $z^n$ is entire, so this is Cauchy's theorem in action. For $n \le -2$ there is a singularity at the origin, yet the integral still vanishes — the residue of $z^n$ is zero for those powers.

Step 3. The one exceptional case is $n = -1$ , where the exponent collapses to zero and the integrand becomes the constant $i$ :

\oint_{|z|=1} \frac{\mathrm{d}z}{z} = i\int_0^{2\pi} \mathrm{d}\theta = \mathbf{2\pi i}.

So among all integer powers, only $z^{-1}$ gives a nonzero loop integral, and it gives exactly $2\pi i$ . This single family of computations both confirms the theorem (every analytic $z^n$ , $n\ge0$ , integrates to zero) and isolates the lone survivor $1/z$ whose residue $1$ carries the factor $2\pi i$ — the number every residue calculation is built from.

2 · Path-independence of ∫ z² dz

Integrate $f(z) = z^2$ from $0$ to $1+i$ along two different paths and confirm the answers agree.

Step 1. Since $z^2$ is entire, it has the primitive $F(z) = z^3/3$ , so by the antiderivative reading the integral depends only on the endpoints:

\int_0^{1+i} z^2\,\mathrm{d}z = \frac{z^3}{3}\bigg|_0^{1+i} = \frac{(1+i)^3}{3}.

Compute $(1+i)^2 = 2i$ and $(1+i)^3 = (1+i)(2i) = 2i + 2i^2 = -2 + 2i$ , giving $\dfrac{-2+2i}{3}$ .

Step 2. Now integrate along Path A, the straight segment $z(t) = (1+i)t$ , $t\in[0,1]$ , with $\mathrm{d}z = (1+i)\,\mathrm{d}t$ :

\int_0^1 (1+i)^2 t^2 \,(1+i)\,\mathrm{d}t = (1+i)^3 \int_0^1 t^2\,\mathrm{d}t = (1+i)^3\cdot\tfrac{1}{3} = \frac{-2+2i}{3}.

Step 3. Along Path B, go along the real axis $0 \to 1$ then up $1 \to 1+i$ . On the first leg $z = x$ , $\int_0^1 x^2\,\mathrm{d}x = \tfrac13$ . On the second $z = 1 + iy$ , $\mathrm{d}z = i\,\mathrm{d}y$ , $\int_0^1 (1+iy)^2\, i\,\mathrm{d}y = \big[\tfrac{(1+iy)^3}{3}\big]_0^1 = \tfrac{(1+i)^3 - 1}{3} = \tfrac{-3+2i}{3}$ . Summing, $\tfrac13 + \tfrac{-3+2i}{3} = \tfrac{-2+2i}{3}$ . All three routes give $\dfrac{-2+2i}{3}$ — the difference of any two paths is a closed loop, which integrates to zero by Cauchy's theorem, so path-independence is guaranteed, not coincidental.

3 · Splitting an integral into analytic and singular parts

Evaluate $\oint_{|z|=3}\dfrac{z+2}{z}\,\mathrm{d}z$ . Divide through: $\dfrac{z+2}{z} = 1 + \dfrac{2}{z}$ . The constant $1$ is entire, so its loop integral is zero by Cauchy's theorem; the pole of $2/z$ sits inside $|z|=3$ , so with the benchmark $\oint\mathrm{d}z/z = 2\pi i$ ,

\oint_{|z|=3}\frac{z+2}{z}\,\mathrm{d}z = \underbrace{\oint 1\,\mathrm{d}z}_{0} + 2\oint\frac{\mathrm{d}z}{z} = 2\cdot 2\pi i = \mathbf{4\pi i}.

The analytic part contributes nothing; the whole answer is carried by the singular term — the residue theorem in miniature.

Physics of the portrait

Cauchy's integral theorem is the foundation stone of complex analysis: a result stated by Cauchy, geometrised by Riemann, and stripped to its logical minimum by Goursat, so that a single hypothesis — analyticity — carries an entire subject.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · founder of rigorous complex analysis and the theory of functions

Augustin-Louis Cauchy was the most prolific analyst of the nineteenth century and, more than anyone, the man who made analysis rigorous. Working in Paris in the 1810s and 1820s, he rebuilt the loose calculus of Euler and Lagrange on precise definitions of limit, continuity, and convergence. His decisive contribution came in the 1825 Mémoire sur les intégrales définies prises entre des limites imaginaires, where he showed that the integral of a well-behaved complex function between two points is independent of the path — the theorem that now bears his name. His proof leaned on Green's theorem and tacitly assumed the derivative was continuous, an assumption that would stand for six decades. Cauchy went on to derive the integral formula $f(z) = \frac{1}{2\pi i}\oint \frac{f(\zeta)}{\zeta - z}\,\mathrm{d}\zeta$ , the residue calculus, and the power-series representation of analytic functions — an interlocking machine that made contour integration the central tool of mathematical physics. Bernhard Riemann later recast analyticity in geometric language, and in the 1880s Édouard Goursat closed the last gap, proving the theorem from the bare existence of the derivative and making Cauchy's tower of consequences fully non-circular. The modern name Cauchy–Goursat theorem honours both: the vision and the rigour.

History

1814

— Cauchy submits his memoir on definite integrals, seeding complex integration

1825

— Cauchy publishes the theorem on integrals between imaginary limits

1846

— Cauchy formulates the residue theorem in its general form

1851

— Riemann's dissertation grounds analyticity in the Cauchy–Riemann equations

1900

— Goursat and Moore give the triangle proof needing no continuity of the derivative

Going further

Texts. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 (the definitive Goursat treatment) · Stone & Goldbart, Mathematics for Physics, Ch. 17 (Function Theory) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 11 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 24 · Whittaker & Watson, A Course of Modern Analysis, Ch. 5.

Provenance. The statement and the Cauchy–Goursat proof are cross-checked against Ahlfors Ch. 4, Stone & Goldbart Ch. 17, Arfken–Weber §11.3, and Riley–Hobson–Bence Ch. 24. A verifier should recompute independently: (1) the Green's-theorem decomposition, confirming both area integrands vanish by CR ( $-\partial_x v - \partial_y u = 0$ , $\partial_x u - \partial_y v = 0$ ); (2) the $4^{-n}$ -versus- $4^{-n}$ cancellation in Goursat's estimate forcing $|I| = 0$ ; (3) the master computation $\oint_{|z|=1} z^n\,\mathrm{d}z = 0$ for integer $n\neq-1$ and $2\pi i$ for $n=-1$ ; (4) the path-independence result $\int_0^{1+i} z^2\,\mathrm{d}z = (-2+2i)/3$ along both paths; (5) the split $\oint_{|z|=3}(z+2)/z\,\mathrm{d}z = 4\pi i$ . No nonstandard numerical constants appear.

Prerequisites. Analytic Functions and the Cauchy–Riemann Equations · Complex Integration and Contour Integrals · Functions of a Complex Variable and Limits · Complex Numbers and the Complex Plane. Leads to. Cauchy's Integral Formula and Derivatives · Residues and the Residue Theorem · Consequences: Liouville, Maximum Modulus, Morera · Taylor Series of Analytic Functions · Laurent Series.

Ask the tutor: Want to see how deforming a contour past a pole turns Cauchy's zero into the residue theorem's $2\pi i\sum\mathrm{Res}$ — worked step by step on an integral of your choice?

What exactly does "simply connected" add beyond analyticity?

Analyticity is local — $f$ is differentiable at each point — while simple connectivity is a global, topological condition on the region: every closed loop can be shrunk to a point without leaving the domain. You need both. A function can be analytic everywhere on an annulus (a ring), yet $\oint 1/z\,\mathrm{d}z = 2\pi i \neq 0$ around a loop encircling the hole, because that loop cannot contract through the missing center. Simple connectivity guarantees there are no such holes, so the local vanishing from analyticity sums over the interior with nothing left uncovered. Remove the topology and the theorem weakens to the deformation theorem, where integrals track the holes rather than vanishing.

Why is Goursat's refinement worth the extra work?

Because without it the logic is circular. Cauchy's original proof via Green's theorem assumes $f'$ is continuous. But a great payoff of the integral theorem — through the integral formula it implies — is that every analytic function is automatically infinitely differentiable, so $f'$ is itself analytic. If you had assumed continuity of $f'$ to prove the theorem, you could not then cite the theorem to prove that continuity. Goursat's triangle argument breaks the loop: it uses only the bare existence of $f'$ and derives everything else. This is why modern texts define analyticity by differentiability alone and prove smoothness as a consequence.

Is Cauchy's integral theorem the same thing as the residue theorem?

They are two faces of one idea. Cauchy's theorem is the "everything analytic" case: no singularities inside the loop, integral zero. The residue theorem is the general case, $\oint_\Gamma f\,\mathrm{d}z = 2\pi i \sum \mathrm{Res}$ . You pass from one to the other by the deformation theorem, which collapses the big contour onto small loops around each pole; Cauchy's theorem is the residue theorem with an empty list of poles (an empty sum is zero). So Cauchy's theorem is logically prior — the special case that makes the general machinery trustworthy.

Does the theorem require the contour to be a circle or convex?

No. The contour need only be a piecewise-smooth simple closed curve — a square, a keyhole, a wiggly loop, anything that closes without self-intersecting where $f$ is analytic. The proof reaches arbitrary shapes because it builds an antiderivative $F$ on convex patches and stitches them across the domain, so the result depends only on the endpoints (which coincide for a closed loop), not the geometry between. Circles are just easy to parametrise; what matters is topology — does the loop enclose a singularity? — not curvature.

How does path-independence follow from the closed-loop statement?

Take two paths $C_1$ and $C_2$ from $a$ to $b$ in a region where $f$ is analytic. Traverse $C_1$ forward then $C_2$ backward: the combined route is a single closed loop, so by Cauchy's theorem $\int_{C_1} f - \int_{C_2} f = 0$ , hence $\int_{C_1} f = \int_{C_2} f$ . The value depends only on the endpoints, never the route. This is the exact analogue of a conservative force field, and it is what makes the antiderivative $F(z) = \int_{z_0}^z f\,\mathrm{d}\zeta$ a well-defined single-valued function.

Why does ∮ z̄ dz fail to vanish even though there is no obvious pole?

Because the hypothesis is analyticity, not the mere absence of poles. The conjugate $\bar z = x - iy$ has no singularity anywhere, yet it is nowhere analytic: with $u = x$ , $v = -y$ , the Cauchy–Riemann equation $\partial_x u = \partial_y v$ reads $1 = -1$ , false everywhere. So $\bar z\,\mathrm{d}z$ has no primitive and no reason to vanish. Concretely, on $|z| = 1$ , $\bar z = 1/z$ , so $\oint \bar z\,\mathrm{d}z = \oint \mathrm{d}z/z = 2\pi i \neq 0$ . The lesson: test analyticity via Cauchy–Riemann, not by hunting for poles.