Cauchy's Integral Theorem and Formula

⚡ Key Takeaways

One-line statement: If $f$ is holomorphic on and inside a simple closed contour $\gamma$ , then $\oint_\gamma f(z)\,dz = 0$ , and the value of $f$ at any interior point is completely determined by its boundary values via $f(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-a}\,dz$ . Key result in words: Differentiability of a complex function in an open region forces an extraordinary rigidity — contour integrals vanish, and a function's interior is dictated by its boundary. Where it applies: Complex analysis, contour integration, residue calculus, evaluation of real integrals, physics (potential theory, fluid flow, electromagnetism, quantum field theory). Who/when: Augustin-Louis Cauchy, 1814–1825; later refined by Goursat (1900) and Riemann.

💡 The Core Insight

Holomorphy is a global constraint masquerading as a local one. Because the Cauchy–Riemann equations link the real and imaginary parts of $f$ , the differential form $f(z)\,dz$ is closed. By Green's theorem (or Stokes' theorem), the integral of a closed form around the boundary of a region depends only on topology, not on the path — and over a simply connected region with no singularities, it vanishes entirely. The Integral Formula then arises by isolating a single singularity $1/(z-a)$ and shrinking the contour around it.

DefinitionHolomorphic Function and Contour

A function $f:\Omega\to\mathbb{C}$ defined on an open set $\Omega\subseteq\mathbb{C}$ is holomorphic (analytic) at $z_0$ if the limit

1

\quad f'(z_0) = \lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}

exists, where $h\in\mathbb{C}$ and the limit is independent of the direction of approach. Derivation hint: This is the complex difference quotient; direction-independence is the strong condition that yields the Cauchy–Riemann equations. A contour (or path) $\gamma$ is a piecewise smooth curve $\gamma:[a,b]\to\mathbb{C}$ . The contour integral of $f$ along $\gamma$ is

2

\quad \int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\,\gamma'(t)\,dt.

Derivation hint: Pull the integral back to the real parameter interval via the chain rule; $dz = \gamma'(t)\,dt$ . Here $\Omega$ is the domain, $z_0$ a point, $h$ a complex increment, and $\gamma'(t)$ the velocity of the parametrization.

DefinitionCauchy–Riemann Equations

Writing $f = u + iv$ with $z = x+iy$ , holomorphy is equivalent to the partial differential equations

3

\quad \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},

together with continuity of the partials. Derivation hint: Compute $f'(z_0)$ from [1] approaching along the real axis ( $h$ real) and the imaginary axis ( $h = ik$ ); equate the two expressions. Here $u(x,y)$ and $v(x,y)$ are the real and imaginary parts.

TheoremCauchy's Integral Theorem (Cauchy–Goursat)

Let $\Omega\subseteq\mathbb{C}$ be a simply connected open set, and let $f$ be holomorphic on $\Omega$ . Then for any closed, piecewise-smooth contour $\gamma$ lying in $\Omega$ ,

4

\quad \oint_\gamma f(z)\,dz = 0.

Derivation hint: Apply Green's theorem to the real and imaginary parts, then substitute the Cauchy–Riemann equations [3] to kill both area integrands.

TheoremCauchy's Integral Formula

Let $f$ be holomorphic on and inside a positively oriented (counterclockwise) simple closed contour $\gamma$ , and let $a$ be any point in the interior. Then

5

\quad f(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-a}\,dz.

Derivation hint: Replace $\gamma$ by a tiny circle around $a$ (deforming via [4]), parametrize $z = a + \varepsilon e^{i\theta}$ , and let $\varepsilon\to 0$ using continuity of $f$ . More generally, $f$ has derivatives of all orders, given by

6

\quad f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\,dz.

Derivation hint: Differentiate [5] $n$ times under the integral sign with respect to $a$ ; each differentiation pulls down a factor and raises the power of $(z-a)$ .

ProofCauchy's Integral Theorem via Green's Theorem

Step 1 — Expand the integrand. Write $f = u+iv$ and $dz = dx + i\,dy$ . Then

7

\quad \oint_\gamma f\,dz = \oint_\gamma (u\,dx - v\,dy) + i\oint_\gamma (v\,dx + u\,dy).

Derivation hint: Multiply $(u+iv)(dx+i\,dy)$ and group real and imaginary parts.

Step 2 — Apply Green's theorem. For the region $R$ enclosed by $\gamma$ , Green's theorem $\oint_\gamma (P\,dx + Q\,dy) = \iint_R (Q_x - P_y)\,dA$ gives

8

\quad \oint_\gamma f\,dz = \iint_R \left(-v_x - u_y\right)dA + i\iint_R \left(u_x - v_y\right)dA.

Derivation hint: Take $P=u, Q=-v$ for the real part and $P=v, Q=u$ for the imaginary part.

Step 3 — Invoke Cauchy–Riemann. By [3], $u_x = v_y$ and $u_y = -v_x$ . Hence $u_x - v_y = 0$ and $-v_x - u_y = 0$ . Both area integrands vanish identically, so $\oint_\gamma f\,dz = 0$ . $\blacksquare$ (Goursat's refinement (1900) removes the assumption that $f'$ is continuous, deriving the result purely from differentiability by a bisection argument on triangles — the rigorous modern foundation.)

∎

ProofCauchy's Integral Formula

Step 1 — Subtract a known integral. Consider $g(z) = \dfrac{f(z) - f(a)}{z-a}$ . Since $f$ is holomorphic and the singularity at $a$ is removable in the bounded sense, write

9

\quad \oint_\gamma \frac{f(z)}{z-a}\,dz = f(a)\oint_\gamma \frac{dz}{z-a} + \oint_\gamma \frac{f(z)-f(a)}{z-a}\,dz.

Derivation hint: Add and subtract $f(a)$ in the numerator and split the integral linearly.

Step 2 — Evaluate the first integral. Deform $\gamma$ to a circle of radius $\varepsilon$ about $a$ (legitimate by [4] in the annular region where the integrand is holomorphic). With $z = a + \varepsilon e^{i\theta}$ , $dz = i\varepsilon e^{i\theta}d\theta$ :

10

\quad \oint_\gamma \frac{dz}{z-a} = \int_0^{2\pi}\frac{i\varepsilon e^{i\theta}}{\varepsilon e^{i\theta}}\,d\theta = \int_0^{2\pi} i\,d\theta = 2\pi i.

Derivation hint: The $\varepsilon e^{i\theta}$ factors cancel exactly, independent of radius.

Step 3 — Bound the second integral. Since $f$ is differentiable at $a$ , $\frac{f(z)-f(a)}{z-a}$ is bounded near $a$ by some $M$ . On the circle of radius $\varepsilon$ , the ML-estimate gives a bound $\le M\cdot 2\pi\varepsilon \to 0$ as $\varepsilon\to 0$ . Thus the second integral vanishes. Derivation hint: $\left|\oint h\,dz\right| \le (\max|h|)\cdot(\text{length})$ ; length $= 2\pi\varepsilon$ .

Step 4 — Combine. Equation [9] becomes $\oint_\gamma \frac{f(z)}{z-a}dz = 2\pi i\, f(a)$ , yielding [5]. $\blacksquare$

∎

Figure

Deforming the contour: a singularity at

a

enclosed by

\gamma

and a small circle

C_\varepsilon

mathematician · Augustin-Louis Cauchy | 1789–1857

A French mathematician of monumental output (nearly 800 papers), Cauchy laid the rigorous foundations of analysis, introducing precise definitions of limits, continuity, and convergence. He developed the residue calculus and the theory of complex integration in a series of memoirs beginning around 1814, motivated partly by problems in hydrodynamics and the evaluation of definite integrals. A devout Catholic and royalist, he spent years in exile after refusing allegiance to the new regime in 1830. His name attaches to dozens of results across analysis, and the integral theorem and formula remain his most celebrated contributions to complex analysis.

🧠 Intuition

Think of $f(z)\,dz$ as a vector field whose circulation you are measuring. The Cauchy–Riemann equations say this field is simultaneously curl-free and divergence-free — it is an ideal, source-free flow. Just as an incompressible, irrotational fluid in a region with no sources or sinks produces zero net circulation around any closed loop, a holomorphic function integrates to zero around any closed contour. The Integral Formula expresses a deeper "averaging" principle: because holomorphic functions are so rigid, the value at the center of a disk is exactly the average of the boundary values. The factor $1/(z-a)$ acts as a weight that concentrates all the information at the point $a$ , while $\oint dz/(z-a) = 2\pi i$ is the "fingerprint" of winding once around the singularity. Holomorphy means there is no slack: knowing the function on a closed curve pins down everything inside.

ExampleEvaluating a contour integral with a single pole

Given: Evaluate $\displaystyle I = \oint_{|z|=2} \frac{e^z}{z-1}\,dz$ , where $|z|=2$ is the circle of radius 2 traversed counterclockwise.

Step 1 — Identify $f$ and $a$ . Here $f(z) = e^z$ (entire, hence holomorphic everywhere) and $a = 1$ .

Step 2 — Check that $a$ is enclosed. Since $|1| = 1 < 2$ , the point $a=1$ lies inside the contour.

Step 3 — Apply Cauchy's Integral Formula [5].

11

\quad I = 2\pi i\, f(a) = 2\pi i\, e^{1}.

Derivation hint: Rearrange [5] to $\oint \frac{f(z)}{z-a}dz = 2\pi i f(a)$ and substitute.

Answer:

\boxed{I = 2\pi i\,e \approx 17.08\,i}

ExampleUsing the generalized formula for a higher-order pole

Given: Evaluate $\displaystyle J = \oint_{|z|=1} \frac{\cos z}{z^3}\,dz$ counterclockwise.

Step 1 — Match to formula [6]. Write the integrand as $\frac{f(z)}{(z-a)^{n+1}}$ with $f(z) = \cos z$ , $a = 0$ , and $n+1 = 3$ , so $n = 2$ .

Step 2 — Verify enclosure. $a = 0$ lies inside $|z|=1$ , and $\cos z$ is entire.

Step 3 — Apply [6]. We need $f''(0)$ . Since $f(z) = \cos z$ , $f''(z) = -\cos z$ , so $f''(0) = -1$ .

12

\quad J = \frac{2\pi i}{2!}\,f''(0) = \frac{2\pi i}{2}(-1) = -\pi i.

Derivation hint: Solve [6] for the integral: $\oint \frac{f(z)}{(z-a)^{n+1}}dz = \frac{2\pi i}{n!}f^{(n)}(a)$ .

Answer:

\boxed{J = -\pi i}

⚠️ Common mistake

The theorem requires holomorphy everywhere inside the contour, not just on it. A function may be perfectly well-behaved on $\gamma$ yet have a singularity in the interior, in which case $\oint_\gamma f\,dz \neq 0$ . The classic example: $\oint_{|z|=1} \frac{1}{z}\,dz = 2\pi i \neq 0$ , because $1/z$ blows up at $z=0$ inside.

Simple connectivity matters. On a domain with holes (like an annulus), Cauchy's theorem fails for loops that enclose a hole. Always check that the region bounded by your contour contains no singularities.

Orientation flips the sign. The formula $f(a)=\frac{1}{2\pi i}\oint$ assumes counterclockwise (positive) orientation. A clockwise contour introduces a factor of $-1$ .

Counterexample

Consider $f(z) = \bar{z}$ (complex conjugate). On the unit circle $z = e^{i\theta}$ , $\bar z = e^{-i\theta}$ and $dz = ie^{i\theta}d\theta$ , so

13

\quad \oint_{|z|=1}\bar z\,dz = \int_0^{2\pi} e^{-i\theta}\,i e^{i\theta}\,d\theta = \int_0^{2\pi} i\,d\theta = 2\pi i \neq 0.

Derivation hint: The $e^{\pm i\theta}$ cancel; integrate the constant $i$ . This does not contradict Cauchy's theorem: $f(z)=\bar z$ is nowhere holomorphic because it violates the Cauchy–Riemann equations (with $u=x$ , $v=-y$ , we get $u_x = 1 \neq v_y = -1$ ). The nonzero integral is exactly the signature of failed holomorphy.

connection

Cauchy's results sit at the heart of a web of complex-analytic theorems. The Residue Theorem generalizes the Integral Formula: instead of one singularity, it sums residues over all enclosed poles, turning contour integration into algebra. Liouville's Theorem (a bounded entire function is constant) follows from the derivative formula [6] by bounding $|f'(a)|$ as the radius grows. The Maximum Modulus Principle is a direct consequence of the averaging interpretation of [5]. The theorem rests on the Cauchy–Riemann Equations and on Green's Theorem from vector calculus, while Power Series and Taylor Expansion of holomorphic functions is derived by expanding the kernel $1/(z-a)$ inside [5] as a geometric series. The vanishing-circulation idea connects directly to Conservative Vector Fields and Path Independence.

🔬 Where this appears

⏳ History

- 1814 — Cauchy presents his first memoir on definite integrals to the French Academy, containing the seeds of contour integration and integration between complex limits.

- 1821 — Cauchy publishes Cours d'analyse, establishing rigorous limits and continuity that underpin the later theory.

- 1825 — Cauchy's landmark memoir Mémoire sur les intégrales définies prises entre des limites imaginaires states the integral theorem (assuming continuous derivative) and introduces residues.

- 1831 — Cauchy formulates the integral formula and the resulting power-series expansion during his exile in Turin.

- 1851 — Bernhard Riemann's dissertation reframes complex functions geometrically, deriving the Cauchy–Riemann equations and conformal mapping from first principles.

- 1855 — Cauchy dies; by this time his integral theory is central to European analysis.

- 1883 — Édouard Goursat begins refining the hypotheses of the theorem.

- 1900 — Goursat proves the theorem without assuming continuity of

f'

, giving the modern Cauchy–Goursat form via triangle bisection.

- 1900s onward — Pringsheim, Moore, and others streamline the proof; the homology and homotopy versions (general cycles, winding numbers) crystallize in 20th-century textbooks (Ahlfors).

open-problem

The classical theorem assumes a simply connected domain, but the precise topological formulation — the homology version — replaces "simple closed contour" with a cycle $\gamma$ that is null-homologous in $\Omega$ (winds zero times around every point outside $\Omega$ ). Stating and proving this cleanly requires the notion of winding number and considerable care; Ahlfors' formulation is the standard reference. Subtleties that trip up students:

Removable vs. essential singularities. The formula handles poles, but functions with essential singularities (like $e^{1/z}$ ) require the full Laurent-series / residue machinery, and Picard-type phenomena lie beyond elementary treatment.
Boundary regularity. Extending Cauchy's formula to contours that are merely rectifiable (not piecewise smooth), or to functions only continuous up to the boundary, leads into delicate questions of $H^p$ Hardy-space theory.
Several complex variables. The naive generalization fails; the correct multidimensional analogues (the Bochner–Martinelli and Cauchy–Fantappiè formulas) are an active and intricate field, and the geometry of domains of holomorphy is far richer.

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also