Cauchy integral formula and derivatives

A closed contour integral looks like it should depend on every value $f(z)$ takes all along the curve $C$ . Cauchy's Integral Formula says something far stronger: if $f$ is analytic inside and on $C$ , the integral $\oint_C f(z)/(z-a)\,dz$ collapses to a single number, $2\pi i\,f(a)$ — the value at one interior point. Differentiate that formula and you get every derivative $f^{(n)}(a)$ from the same boundary data, for free. This turns contour integration from a parametrize-and-grind exercise into instant lookup: spot the pole, read off $f$ , $a$ , $n$ , and write the answer. It also delivers one of complex analysis's most surprising facts — an analytic function is automatically infinitely differentiable, with no real-variable counterpart. This article builds the formula from Cauchy's Integral Theorem, states the derivative generalization, and drills the pattern-matching skill on two fully worked contour integrals.

The core insight

Cauchy's Integral Formula reconstructs an analytic function's value (and, by differentiating, all its derivatives) at an interior point purely from its values on the boundary contour, via $f^{(n)}(a) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-a)^{n+1}}dz$ .

The core idea

Setup. Let $f$ be analytic inside and on a simple closed contour $C$ , traversed counter-clockwise, and let $a$ be any point strictly inside $C$ . Then $f(a)$ is fully determined by the boundary values of $f$ :

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

where: $C$ is the closed contour, $a$ is a fixed interior point, $z$ is the integration variable running along $C$ , and the $2\pi i$ factor is the same constant that appears in $\oint dz/(z-a) = 2\pi i$ .

Why it is plausible. The integrand $f(z)/(z-a)$ is analytic everywhere inside $C$ except at the single point $z=a$ . By the deformation principle (a direct consequence of Cauchy's Integral Theorem), the integral over $C$ equals the integral over a tiny circle around $a$ — and on that tiny circle $f(z)\approx f(a)$ , since $f$ is continuous. Shrinking the circle isolates exactly one copy of $f(a)$ times the standard $2\pi i$ .

Differentiate to get every derivative. Formally differentiating the formula with respect to $a$ , repeatedly, produces derivatives of all orders from the same boundary integral:

f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}}\, dz, \qquad n = 0, 1, 2, \dots

where: $n$ is the derivative order ( $n=0$ recovers the original formula), and $n!$ is the usual factorial that must not be dropped. The striking consequence: an analytic function is automatically infinitely differentiable, and every one of its derivatives is itself analytic — a guarantee real-variable calculus never offers, where being differentiable once says nothing about being differentiable twice.

How it is used in practice. The formulas are almost always run in reverse: given $\oint_C \dfrac{f(z)}{(z-a)^{n+1}}\,dz$ with $f$ analytic inside $C$ and $a$ inside $C$ , the answer is read off instantly as $\dfrac{2\pi i}{n!}f^{(n)}(a)$ , with no parametrization of $C$ ever needed.

Figure

Diagram showing a contour C enclosing a point a with a shrinking circle around a illustrating the deformation argument

Derivation ADeriving Cauchy's Integral Formula from Cauchy's Integral Theorem

The goal is to show, starting only from Cauchy's Integral Theorem and the ML-inequality — the standard bound that a contour integral's size is at most (arc length) times (maximum integrand size) — that $\oint_C f(z)/(z-a)\,dz$ collapses to the single value $2\pi i\,f(a)$ , no matter how complicated $f$ or how large $C$ is, provided $f$ is analytic inside and on $C$ and $a$ lies strictly inside $C$ .

Step 1. Isolate the singularity by deformation. The function $g(z) = f(z)/(z-a)$ is analytic inside and on $C$ except at the single point $z=a$ . Cauchy's Integral Theorem (applied to the region between $C$ and a small circle $C_\varepsilon: |z-a|=\varepsilon$ drawn around $a$ , where $g$ is analytic throughout that annular region) gives $\oint_C g(z)\,dz = \oint_{C_\varepsilon} g(z)\,dz$ : the original contour can be shrunk down to the tiny circle without changing the integral's value.

Step 2. Split off the constant piece $f(a)$ . Write $f(z) = f(a) + [f(z)-f(a)]$ inside the integral over $C_\varepsilon$ :

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

The first term is a pure constant $f(a)$ times a standard integral.

Step 3. Evaluate the standard integral. Parametrizing $C_\varepsilon$ as $z = a + \varepsilon e^{i\theta}$ , $\theta$ from $0$ to $2\pi$ , gives $\oint_{C_\varepsilon} dz/(z-a) = 2\pi i$ exactly, independent of $\varepsilon$ — the shifted version of the familiar $1/z$ result. So the first term is $f(a)\cdot 2\pi i$ .

Step 4. Show the second term vanishes as $\varepsilon \to 0$ . Since $f$ is continuous at $a$ , $|f(z)-f(a)| \to 0$ as $z \to a$ ; call this maximum value $M(\varepsilon)$ on $C_\varepsilon$ , with $M(\varepsilon)\to 0$ . The ML-inequality bounds the second integral by $M(\varepsilon) \cdot (\text{length of } C_\varepsilon) = M(\varepsilon)\cdot 2\pi\varepsilon$ , which shrinks to $0$ because $M(\varepsilon) \to 0$ while $2\pi\varepsilon$ stays bounded. Combining Steps 1 to 4, $\oint_C g(z)\,dz = 2\pi i\,f(a) + 0 = 2\pi i\,f(a)$ , which is exactly Cauchy's Integral Formula after dividing both sides by $2\pi i$ .

Worked example

Problem. Evaluate $\displaystyle\oint_C \frac{\cos z}{z^4}\,dz$ where $C: |z|=1$ (counter-clockwise), and compare with the related integral $\displaystyle\oint_C \frac{\cos z}{z^3}\,dz$ over the same contour.

Step 1. Match the pattern for the first integral. Write the integrand as $\dfrac{f(z)}{(z-0)^{3+1}}$ , so $f(z) = \cos z$ (entire, hence analytic inside $C$ ), $a = 0$ , and $n=3$ . Check $a$ is inside $C$ : $|0| = 0 < 1$ , so it is. This is the derivative form of the Cauchy Integral Formula with $n=3$ :

\oint_C \frac{\cos z}{z^4}\,dz = \frac{2\pi i}{3!}\, f^{(3)}(0).

Step 2. Compute the needed derivative. $f'(z) = -\sin z$ , $f''(z) = -\cos z$ , $f'''(z) = \sin z$ , so $f^{(3)}(0) = \sin 0 = 0$ .

Step 3. Substitute back. $\oint_C \dfrac{\cos z}{z^4}\,dz = \dfrac{2\pi i}{6}\cdot 0 = \mathbf{0}$ .

Step 4. Do the companion integral $\oint_C \cos z/z^3\,dz$ the same way, to see the pattern shift by one power. Here the denominator is $(z-0)^{2+1}$ , so $n=2$ this time, and the formula gives $\dfrac{2\pi i}{2!}f''(0) = \pi i\cdot(-\cos 0) = \boldsymbol{-\pi i}$ , since $f''(z) = -\cos z$ and $\cos 0 = 1$ . Lining the two answers up side by side — $0$ for the $z^4$ case, $-\pi i$ for the $z^3$ case — is the fastest way to check the power of $(z-a)$ in the denominator was correctly matched to $n+1$ before reading off which derivative of $f$ is needed.

Common traps

Applying the formula when $a$ is outside $C$ . Then $f(z)/(z-a)$ is analytic throughout $C$ , and Cauchy's Integral Theorem (not the formula) gives $0$ — a different theorem for a different situation; always locate $a$ relative to $C$ first.
Misreading the power of $(z-a)$ . For $\cos z/z^4$ , it is tempting to peel off $\cos z/z^3$ as " $f(z)$ " with $n=1$ ; instead match the exact power in the denominator to $n+1$ and keep the rest as $f(z)$ , provided that remainder is analytic inside $C$ .
Dropping the $n!$ . $f^{(n)}(a) = \dfrac{n!}{2\pi i}\oint_C \dfrac{f(z)}{(z-a)^{n+1}}dz$ — omitting or misplacing $n!$ is a common slip for $n \ge 2$ , as seen in the $3! = 6$ above.
Using the formula when $f$ itself has a hidden singularity inside $C$ . If $f(z)$ is not analytic everywhere inside $C$ (a singularity distinct from $a$ ), the formula does not directly apply; such cases need residue calculus, covered later.

Ask the tutor: If distinguishing when the answer is zero versus a derivative value still feels shaky, ask the AI tutor for a quick mixed set of five contour integrals to test the inside-or-outside judgment call.

Why does Cauchy's Integral Formula work even though f(z)/(z-a) blows up at z=a?

The singularity at $z=a$ is exactly what makes the formula nontrivial, and the derivation handles it by shrinking the contour down to a tiny circle around $a$ . On that tiny circle $f(z)$ barely differs from $f(a)$ because $f$ is continuous, so the integral behaves like $f(a)$ times the standard result $\oint dz/(z-a) = 2\pi i$ . The blow-up is real, but it is exactly cancelled by the shrinking arc length of the tiny circle, which is why the ML-inequality step shows the leftover piece vanishes.

How do I quickly tell whether to use Cauchy's Theorem (answer zero) or Cauchy's Integral Formula (answer a derivative value)?

Check only one thing first: is the point $a$ where the denominator vanishes located inside the contour $C$ or outside it? If $a$ is outside $C$ , the whole integrand is analytic inside $C$ and Cauchy's Integral Theorem gives $0$ immediately, no further work needed. If $a$ is inside $C$ , the integrand has a genuine singularity inside the region and you switch to the Cauchy Integral Formula (or its derivative form), matching the power of $(z-a)$ to find $n$ .

Does this formula work for any function f, or only special ones like polynomials and cos z?

It works for any $f$ that is analytic inside and on the contour $C$ — polynomials, $e^z$ , $\sin z$ , $\cos z$ , and much more exotic analytic functions all qualify, since the derivation never used any special property of $f$ beyond analyticity and continuity. The one requirement is that $f$ itself must not have its own singularity inside $C$ ; if it does, this formula as stated does not apply and residue calculus is needed instead.

Cauchy integral formula and derivatives

The core idea

Worked example

Common traps

Before this

Where this leads