Cauchy's integral formula

Theorem

Cauchy's integral formula

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Statement

Let \( \Omega \subseteq \mathbb{C} \) be open, let \( f : \Omega \to \mathbb{C} \) be holomorphic on \( \Omega \), and let \( \overline{D}(a,r) = \{ z \in \mathbb{C} : |z-a| \le r \} \) be a closed disc with \( \overline{D}(a,r) \subseteq \Omega \). Let \( \gamma \) be the boundary circle \( |z-a|=r \), traversed once counterclockwise. Then for every point \( w \) in the open disc \( D(a,r) \), \[ f(w) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-w}\,dz. \] More generally the same identity holds with \( \gamma \) replaced by the positively oriented boundary of any region whose closure lies in \( \Omega \) and for which \( w \) lies inside, provided \( f \) is holomorphic on \( \Omega \) and the boundary is a finite union of piecewise-\( C^1 \) closed curves null-homotopic in \( \Omega \).

Why it matters

This is the single most consequential formula in complex analysis: it converts a statement about the interior of a region into a statement about its boundary. Once you know \( f \) on a circle, you know \( f \) everywhere inside — a rigidity phenomenon with no counterpart for general differentiable functions of a real variable. Every subsequent pillar of the theory (power series expansions, Liouville's theorem, the maximum modulus principle, the residue calculus) is extracted from this one identity by differentiating or estimating under the integral sign.

It also explains, at a stroke, why holomorphic functions are infinitely differentiable and real-analytic: differentiating the formula under the integral sign produces convergent power series and derivative formulas for free, something that requires no extra hypothesis beyond the single complex derivative existing.

Hypotheses

\( f \) is holomorphic on an open set \( \Omega \) containing the closed disc \( \overline{D}(a,r) \). If \( f \) is only holomorphic on the open disc but singular somewhere on the boundary circle — e.g. \( f(z) = 1/(z-b) \) with \( |b-a|=r \) — the formula can fail for \( w \) near \( b \): the integrand develops a genuine singularity on the contour and the contour integral is not even defined without further interpretation (principal value), let alone equal to \( f(w) \).

\( w \) lies strictly inside the disc, \( |w-a| \lt r \). If \( w \) is outside, \( |w-a| \gt r \), the same integral \( \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-w}dz \) is not \( f(w) \) but \( 0 \), since \( z \mapsto f(z)/(z-w) \) is then holomorphic on and inside \( \gamma \) and Cauchy's theorem applies directly. Taking \( w \) outside and expecting the formula to reproduce \( f(w) \) is a standard error.

\( \gamma \) is a simple closed contour, traversed exactly once, with winding number \( 1 \) about every point of \( D(a,r) \) (here: counterclockwise). Reversing orientation flips the sign: the clockwise circle gives \( -f(w) \). Traversing \( \gamma \) twice doubles the answer to \( 2f(w) \), since the formula in its general form has an explicit winding-number factor \( n(\gamma,w) \) that must be tracked.

Complex differentiability of \( f \) (not merely real differentiability of \( f \) viewed as a map \( \mathbb{R}^2 \to \mathbb{R}^2 \)). A function that is \( C^1 \) as a map of two real variables but fails the Cauchy–Riemann equations — such as \( f(z) = \bar{z} \) — is not holomorphic, and the formula fails: for \( f(z)=\bar z \), \( a=0 \), \( r=1 \), \( w=0 \), one computes \( \frac{1}{2\pi i}\oint_{|z|=1} \frac{\bar z}{z}\,dz = \frac{1}{2\pi i}\oint \frac{1}{z^2}\,dz \) (using \( \bar z = 1/z \) on \( |z|=1 \)) \( = 0 \ne \bar 0 = f(0) \) trivially here, but more sharply the mean-value identity that underlies the proof (Goursat's theorem) requires the full complex derivative, not just partial derivatives in \( x,y \).

Proof

\text{Fix } w \in D(a,r). \text{ Define } g(z) = \dfrac{f(z)-f(w)}{z-w} \text{ for } z \ne w, \quad g(w) = f'(w).

Since \( f \) is holomorphic, the difference quotient extends continuously to \( z=w \) with value \( f'(w) \), by the very definition of the complex derivative \( f'(w) = \lim_{z\to w} \frac{f(z)-f(w)}{z-w} \). A

g \text{ is holomorphic on } \Omega \setminus \{w\} \text{ and continuous on all of } \Omega.

Quotients of holomorphic functions are holomorphic away from zeros of the denominator; continuity at \( w \) was arranged in Step 1. A

\oint_\gamma g(z)\,dz = 0.

This is the crucial analytic input, and it needs Goursat's theorem (the version of Cauchy's theorem that assumes only complex differentiability, not continuity of \( f' \)) applied with the removable-singularity refinement: a function holomorphic on a disc minus one point but continuous at that point still has vanishing integral around any closed contour in the disc. Concretely, triangulate the disc; for triangles not containing \( w \), Goursat's theorem gives integral \( 0 \) directly; for a small triangle containing \( w \), continuity of \( g \) bounds the integral by \( O(\varepsilon) \) times its perimeter as the triangle is shrunk toward \( w \), which \( \to 0 \). Summing over the triangulation and using the standard "Cauchy's theorem for a disc" deformation argument (any closed contour in a convex region is homotopic to a point, and a holomorphic function with a primitive on that region has zero integral around it — the primitive being constructed by integrating \( g \) along paths from a fixed basepoint) gives the claim. C

\oint_\gamma \frac{f(z)}{z-w}\,dz - f(w)\oint_\gamma \frac{dz}{z-w} = \oint_\gamma g(z)\,dz = 0.

Rearrange the definition of \( g \): \( f(z) = g(z)(z-w) + f(w) \), so \( \dfrac{f(z)}{z-w} = g(z) + \dfrac{f(w)}{z-w} \); integrate termwise using linearity of the contour integral and Step 3. A

\oint_\gamma \frac{dz}{z-w} = 2\pi i.

Direct computation of the winding-number integral: parametrize \( \gamma(t) = a + re^{it} \), \( t \in [0,2\pi] \). Since \( w \in D(a,r) \), write \( \frac{1}{z-w} \) and integrate; the standard evaluation (e.g. via the antiderivative \( \log(z-w) \) tracked continuously around the loop, which increases by \( 2\pi i \) exactly once since \( \gamma \) has winding number \( 1 \) about \( w \)) gives \( \oint_\gamma \frac{dz}{z-w} = 2\pi i \, n(\gamma,w) = 2\pi i \cdot 1 \). This is the defining computation of the winding number / index of a curve about a point. B

\oint_\gamma \frac{f(z)}{z-w}\,dz = f(w)\cdot 2\pi i.

Substitute Step 5 into Step 4 and solve for the remaining integral. A

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

Divide both sides of Step 6 by \( 2\pi i \); this is the claimed formula, valid for the arbitrary \( w \in D(a,r) \) fixed at the start. A

\text{The formula extends to any positively oriented, piecewise-}C^1\text{, null-homotopic boundary curve } \gamma \text{ in } \Omega \text{ with } n(\gamma,w)=1.

Nothing in Steps 1–7 used circularity of \( \gamma \) except in the explicit evaluation of Step 5; replacing that computation with the general definition \( n(\gamma,w) = \frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-w} \) and invoking the homotopy-invariant form of Cauchy's theorem (integrals of holomorphic functions over null-homotopic closed curves in \( \Omega \) vanish) in place of Step 3's disc-specific argument reproduces the identity for any such \( \gamma \). C

Result

f(w) = \dfrac{1}{2\pi i}\oint_\gamma \dfrac{f(z)}{z-w}\,dz \qquad (w \text{ inside } \gamma)

Reading. The value of a holomorphic function at any interior point is a weighted average of its values on a surrounding contour, with weight \( \frac{1}{2\pi i(z-w)} \). Knowing \( f \) only on the boundary pins down \( f \) everywhere inside — there is no freedom left.

Scope. Applies to any \( f \) holomorphic on an open set containing a closed region bounded by a positively oriented, piecewise-\( C^1 \), simple (or null-homotopic) closed contour, for any point strictly inside that contour. It does not apply to points on or outside the contour, and requires genuine complex differentiability, not just smoothness in the real sense.

Corollaries & converses

Differentiating under the integral sign (justified by uniform convergence of difference quotients on compact subsets of \( D(a,r) \)) gives the generalized Cauchy integral formula \( f^{(n)}(w) = \dfrac{n!}{2\pi i}\oint_\gamma \dfrac{f(z)}{(z-w)^{n+1}}\,dz \); hence holomorphic functions are automatically infinitely differentiable, with all derivatives themselves holomorphic.
Cauchy's inequalities and Liouville's theorem (bounded entire functions are constant) follow immediately by estimating the derivative formula on circles of growing radius.
The Mean Value Property for holomorphic functions is the special case \( w=a \): \( f(a) = \frac{1}{2\pi}\int_0^{2\pi} f(a+re^{i\theta})\,d\theta \), the average of \( f \) over the circle.
Existence of local power series (Taylor series) expansions of holomorphic functions, convergent on any disc contained in the domain, follows by expanding \( \frac{1}{z-w} \) as a geometric series in \( \frac{w-a}{z-a} \) inside the integral.
Converse: there is no useful converse in the sense of "if the averaging formula holds for one contour then \( f \) is holomorphic" being a separate free-standing theorem — rather, Morera's theorem gives a genuine converse to Cauchy's theorem (vanishing of all contour integrals of a continuous function implies holomorphy), and combined with Cauchy's formula this characterizes holomorphic functions completely. The identity itself is an equivalence only in the sense that a continuous function satisfying the mean-value/circle-average property for all discs in its domain can be shown to be holomorphic (this is a genuine and non-trivial converse, sometimes attributed to Morera's approach).

Fails without

Drop holomorphy (keep only real-differentiability): for \( f(z) = \bar z \) on \( \Omega = \mathbb{C} \), \( \gamma = \{|z|=1\} \), \( w=0 \): direct computation gives \( \frac{1}{2\pi i}\oint_{|z|=1} \bar z \, dz = \frac{1}{2\pi i}\int_0^{2\pi} e^{-it} \cdot ie^{it}\,dt = \frac{1}{2\pi i}\int_0^{2\pi} i\,dt = 1 \ne 0 = f(0)\); the formula gives the wrong value entirely because \( \bar z \) is nowhere holomorphic.
Drop "\( w \) inside \( \gamma \)" (take \( w \) outside instead): for \( f(z)=z \), \( \gamma=\{|z|=1\} \), \( w=2 \): \( \frac{1}{2\pi i}\oint_{|z|=1} \frac{z}{z-2}\,dz = 0 \) by Cauchy's theorem (integrand holomorphic inside \( |z|\le 1 \)), while \( f(2) = 2 \ne 0 \); the formula's right-hand side silently switches to computing \( 0 \) rather than \( f(w) \) once \( w \) leaves the disc.
Drop "\( \Omega \) contains the closed disc" (allow a puncture on the boundary): for \( f(z) = \dfrac{1}{z-1} \), \( \Omega = \mathbb{C}\setminus\{1\} \), \( \gamma = \{|z|=1\} \), \( w=0 \) inside: \( f \) is not holomorphic on any open set containing the whole closed disc \( \overline{D}(0,1) \) since it blows up exactly on the boundary at \( z=1 \); the contour integral \( \oint_{|z|=1} \frac{dz}{(z-1)(z-w)} \) is not even a well-posed Riemann/complex line integral along that exact contour, since the integrand is unbounded at the point \( z=1\) on the path itself.

Common errors

Applying the formula with \( w \) outside the contour and expecting to recover \( f(w) \), rather than realizing the integral is simply \( 0 \) there (by Cauchy's theorem, since the integrand is then holomorphic throughout the enclosed region).
Forgetting the \( \frac{1}{2\pi i} \) prefactor, or dropping the orientation sign when the contour is traversed clockwise (giving \( -f(w) \) instead of \( f(w) \)).
Confusing this formula with Cauchy's theorem (\( \oint_\gamma f(z)\,dz = 0\) for \( f \) holomorphic with no pole inside) — the integral formula has an extra factor \( \frac{1}{z-w} \) creating a genuine pole at \( w \), which is precisely the point of the theorem, not an oversight to be "cancelled".
Using the formula for a contour that winds around \( w \) more than once (or not at all) without inserting the winding number \( n(\gamma,w) \), producing an answer off by an integer factor.
Trying to apply the formula to functions that are only piecewise holomorphic or holomorphic except at isolated singularities inside the contour, without switching to the residue theorem (the correct generalization once poles are allowed inside \( \gamma \)).

Discussion

Cauchy's integral formula, proved by Augustin-Louis Cauchy in the 1820s as part of his foundational work on complex analysis, is often described as the point where complex analysis definitively parts ways with real analysis. In the real case, knowing a smooth function on the boundary of an interval says essentially nothing about its values or derivatives in the interior — one can prescribe boundary values almost arbitrarily and interpolate. In the complex case, a single complex derivative existing everywhere on an open set is such a strong constraint that the entire function is baked into its boundary values, a phenomenon with no real-variable analogue.

The formula is best understood as an averaging or reproducing-kernel identity: the Cauchy kernel \( \frac{1}{2\pi i(z-w)} \) reproduces holomorphic functions from their boundary data, in exactly the sense that the Poisson kernel reproduces harmonic functions from boundary data (indeed, taking real parts of the Cauchy kernel on a circle recovers the Poisson kernel, linking this theorem directly to the Dirichlet problem and potential theory).

Historically and pedagogically, the formula is the engine that converts Cauchy's theorem (a statement that certain integrals vanish) into a source of new information (a formula for function values). This is a recurring pattern in the subject: vanishing-integral theorems become computational tools once a controlled singularity, here \( \frac{1}{z-w} \), is deliberately introduced into the integrand.

A subtler point, easy to miss on first exposure: the proof above depends critically on Goursat's theorem holding under the weak hypothesis of mere complex differentiability, without assuming a priori that \( f' \) is continuous. Historically, Cauchy's own proof assumed continuity of \( f' \) (via Green's theorem), and it was Goursat who later showed this extra hypothesis is unnecessary. This matters because the corollary that holomorphic functions are automatically \( C^\infty \) would otherwise be circular — one cannot assume continuity of \( f' \) to prove that \( f' \) is continuous.

Common misconception: students sometimes think the formula "assumes \( f \) is analytic (given by a convergent power series)" and is therefore somewhat trivial. In fact the logical order is the reverse — holomorphy (mere existence of a complex derivative) is the hypothesis, and the existence of a local power series expansion is a corollary extracted from this formula, not an input to it.

Worked examples

\text{Compute } I = \oint_{|z|=2} \frac{e^z}{z-1}\,dz.

Identify the pattern of the formula: \( f(z)=e^z \) is entire, hence holomorphic on an open set containing \( \overline{D}(0,2) \); the pole of the integrand is at \( w=1 \), which lies inside \( |z|=2 \). A

I = 2\pi i \, f(1) = 2\pi i\, e^1.

Direct application of Cauchy's integral formula with \( a=0, r=2, w=1 \): \( f(w) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-w}dz \) rearranges to \( \oint_\gamma \frac{f(z)}{z-w}dz = 2\pi i\, f(w) \). A

\oint_{|z|=2} \dfrac{e^z}{z-1}\,dz = 2\pi i\, e

\text{Find } I = \oint_{|z-2|=3} \frac{\cos(\pi z)}{z^2 - 1}\,dz.

Factor the denominator: \( z^2-1=(z-1)(z+1) \), giving simple poles at \( z=1 \) and \( z=-1 \). Both lie inside \( |z-2|=3 \) since \( |1-2|=1\lt3 \) and \( |-1-2|=3 \) is on the boundary — adjust: check \( |-1-2|=3\), which lies exactly on the contour, so instead take the contour \( |z-2|=3.5 \) to keep both poles strictly interior (or equivalently note the problem is set up so both are interior; here we use radius \( 3.5\) for a well-posed contour). B

\frac{\cos(\pi z)}{z^2-1} = \frac{\cos(\pi z)}{(z+1)}\cdot\frac{1}{z-1} = \frac{\cos(\pi z)}{(z-1)}\cdot\frac{1}{z+1}.

Split into two Cauchy-formula shapes by partial-fraction-style regrouping around each pole separately, since the contour formula as stated handles one singularity \( w \) at a time; use linearity of the contour integral to treat the two poles independently via small circles \( \gamma_1 \) about \( 1 \) and \( \gamma_2 \) about \( -1 \), each null-homotopic to the big contour minus the other pole by the deformation form of Cauchy's theorem. B

\oint_{\gamma_1} \frac{\cos(\pi z)/(z+1)}{z-1}\,dz = 2\pi i \cdot \frac{\cos(\pi)}{1+1} = 2\pi i \cdot \frac{-1}{2} = -\pi i.

Apply Cauchy's formula with \( f(z) = \cos(\pi z)/(z+1) \) (holomorphic near \( z=1 \), since its only singularity at \( z=-1 \) is excluded) and \( w=1 \). A

\oint_{\gamma_2} \frac{\cos(\pi z)/(z-1)}{z+1}\,dz = 2\pi i \cdot \frac{\cos(-\pi)}{-1-1} = 2\pi i \cdot \frac{-1}{-2} = \pi i.

Apply Cauchy's formula with \( f(z) = \cos(\pi z)/(z-1) \) (holomorphic near \( z=-1 \)) and \( w=-1 \). A

I = -\pi i + \pi i = 0.

Sum the two contributions, since the original large contour integral splits additively into the sum of small-circle integrals around each interior singularity (standard contour-deformation argument, the precursor to the residue theorem). B

\oint_{|z-2|=3.5} \dfrac{\cos(\pi z)}{z^2-1}\,dz = 0

Problems

Evaluate \( \displaystyle\oint_{|z|=1} \frac{\sin z}{z}\,dz \).
Solution
\( f(z)=\sin z\) is entire, \( w=0\) is inside \( |z|=1 \). By Cauchy's integral formula, the integral equals \( 2\pi i\, f(0) = 2\pi i \sin 0 = 0 \).
Evaluate \( \displaystyle\oint_{|z-i|=1} \frac{z^2+1}{z-i}\,dz \).
Solution
\( f(z)=z^2+1\) is entire, \( w=i\) is inside \( |z-i|=1\) (it is the center). By the formula the integral equals \( 2\pi i\, f(i) = 2\pi i\,(i^2+1) = 2\pi i\,(0) = 0 \).
Evaluate \( \displaystyle\oint_{|z|=3} \frac{e^{2z}}{(z-1)}\,dz \) and explain why \( |z|=3 \) versus \( |z|=1/2 \) would give different answers.
Solution
For \( |z|=3\): \( w=1\) is inside, \( f(z)=e^{2z}\) entire, so the integral is \( 2\pi i\, e^{2} \). For a contour \( |z|=1/2\): \( w=1\) lies outside, so \( z\mapsto e^{2z}/(z-1)\) is holomorphic throughout the enclosed disc and Cauchy's theorem gives integral \( 0\). The two answers differ because Cauchy's integral formula's right-hand side reproduces \( f(w)\) only when \( w\) is interior to the contour; outside, it collapses to \( 0\) by the plain (non-formula) Cauchy theorem.
Let \( f \) be holomorphic on all of \( \mathbb{C} \) (entire) and suppose \( |f(z)| \le M \) for all \( z \). Use Cauchy's integral formula (differentiated once) to show \( f'(0) \) satisfies \( |f'(0)| \le M/r \) for every \( r\gt0 \), and deduce \( f \) is constant.
Solution
Differentiating Cauchy's formula once gives \( f'(0) = \dfrac{1}{2\pi i}\oint_{|z|=r} \dfrac{f(z)}{z^2}\,dz \). Bounding the integral by (length of contour) \( \times \) (max modulus of integrand): \( |f'(0)| \le \dfrac{1}{2\pi}\cdot 2\pi r \cdot \dfrac{M}{r^2} = \dfrac{M}{r} \). Since \( r \) can be taken arbitrarily large while \( M\) stays fixed (as \(f\) is entire and bounded on the whole plane), letting \( r \to \infty \) forces \( f'(0)=0\). The same argument centered at any point \( a\) gives \( f'(a)=0\) for all \( a\), so \( f\) is constant. (This is Liouville's theorem, derived directly from the differentiated Cauchy formula — the estimate used is Cauchy's inequality.)
Explain, using the formula's proof, exactly which step would fail if \( \gamma \) were the boundary of a square instead of a circle, and why the theorem nonetheless still holds for a square contour.
Solution
The proof step that used circularity explicitly was Step 5, the direct computation \( \oint_\gamma \frac{dz}{z-w} = 2\pi i\) via the parametrization \( \gamma(t)=a+re^{it}\). For a square this exact parametrization and antiderivative bookkeeping changes, but the identity \( \oint_\gamma \frac{dz}{z-w} = 2\pi i \, n(\gamma,w)\) still holds because it depends only on \( \gamma \) being a simple closed piecewise-\(C^1\) curve winding once around \( w\), not on it being a circle — this is the content of Step 8 (the winding-number/homotopy-invariant generalization). All other steps (1–4, 6–7) use only that \( g \) is holomorphic away from \( w \) and continuous at \( w\), plus Cauchy's theorem for null-homotopic curves in \( \Omega\), neither of which references the shape of \( \gamma \) at all. Hence the formula holds verbatim for a square (or any simple closed contour) enclosing \( w \).