--- title: "Cauchy's Integral Formula and Derivatives of Every Order" description: "Recover a function's value inside a contour from its values on the contour, then differentiate under the integral to get every derivative at once." canonical: https://lightmysky.com/learn/mathematics/cauchys-integral-formula-and-derivatives-of-every-order-mt_mViFwrFNWQ source: https://lightmysky.com/learn/mathematics/cauchys-integral-formula-and-derivatives-of-every-order-mt_mViFwrFNWQ.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Cauchy's Integral Formula and Derivatives of Every Order Recover a function's value inside a contour from its values on the contour, then differentiate under the integral to get every derivative at once. Subject: Mathematics · Area: Complex Analysis · Ages 20 to 21 Page: https://lightmysky.com/learn/mathematics/cauchys-integral-formula-and-derivatives-of-every-order-mt_mViFwrFNWQ ## Ready when they can - State and apply the integral formula to evaluate a value or a derivative - Explain why holomorphy on an open neighbourhood forces differentiability infinitely often - Deduce Liouville's theorem and, from it, the fundamental theorem of algebra ## Lesson: One loop that rebuilds a whole function Cauchy's integral formula. Let γ be a simple closed piecewise C1 curve traversed once counterclockwise, f holomorphic on an open set containing γ and its filled interior, and a strictly inside. Then the loop integral of f(z)/(z - a) dz is 2πi f(a), and f(a) is that integral divided by 2πi. Worked example: for f(z) = z² with a = 0.5 strictly inside the unit circle, the integral is 2πi × 0.25 = πi/2. Clockwise traversal reverses the sign; traversing k times multiplies by k, the signed turn count. Why the formula holds. The quotient f(z)/(z - a) is holomorphic on the region between γ and a small circle C_ρ around a, so deformation preserves the integral I. Parametrising a + ρe^(it) gives I = i times the integral from 0 to 2π of f(a + ρe^(it)) dt. |I - 2πi f(a)| ≤ 2π times the supremum of |f(z) - f(a)| on C_ρ. Differentiability at a gives f(a + h) - f(a) = ((f(a + h) - f(a))/h) h, which tends to 0, so the supremum tends to 0 as ρ shrinks. Since I does not depend on ρ, I = 2πi f(a). Derivatives of every order. Differentiating under the integral sign n times gives f^(n)(a) = (n!/(2πi)) I_n, where I_n is the loop integral of f(z)/(z - a)^(n+1) dz. Solving for the integral, I_n = 2πi f^(n)(a)/n!, with n a nonnegative integer and 0! = 1. Worked inverse: for f(z) = z³, a = 1 and n = 2, f''(1) = 6, so I_2 = 2πi × 6/2 = 6πi on any circle centred at 1 traversed once counterclockwise. Why differentiation is justified. Fix the contour γ and let d be the distance from a to γ, which is positive. For any w with |w - a| < d/2, every contour point z has |z - w| ≥ d/2, so each finite-order kernel derivative is uniformly bounded on γ. Repeated differentiation under the integral sign is therefore legitimate. This needs f holomorphic on an open neighbourhood of a. The function |z|² is complex differentiable only at 0, with no neighbourhood, and gets no such formula. The Cauchy estimate. If f is holomorphic on a neighbourhood of the closed disk |z - a| ≤ R and |f| ≤ M on its boundary circle, then |f^(n)(a)| ≤ n! M/R^n. The factorial comes from differentiating the kernel n times, and the radius power from the contour length 2πR against the denominator R^(n+1). Worked estimate: with M = 4, R = 2 and n = 1, the bound on |f'(a)| is 1 × 4/2 = 2. Liouville's theorem. A function is entire when holomorphic on the whole complex plane. If one bound M holds for |f(z)| at every z, then at any centre a the estimate gives |f'(a)| ≤ M/R. Letting R grow without bound forces f'(a) = 0 at every a, so f is constant. Per-circle bounds alone do not imply this: a bound like M(R) = R on each circle of radius R is consistent with f(z) = z, which is not constant. Fundamental theorem of algebra. Let p be a nonconstant polynomial with no zero. Then 1/p is entire. Outside a large disk the leading term dominates, so |1/p| is small there; inside the fixed disk, continuity on the compact set with no zeros gives a finite bound. One global bound makes 1/p constant by Liouville, contradicting nonconstancy. So p has a root. Factoring out each root and repeating, counting multiplicity, gives exactly deg p roots. **Recap.** One loop rebuilds inside values and every derivative, a fixed global bound forces constant, and a zero-free polynomial would freeze its reciprocal. ## Practice 17 questions on this page, each with its working shown. ## Needs first - [Cauchy's Theorem and Deforming a Contour](https://lightmysky.com/learn/mathematics/cauchys-theorem-and-deforming-a-contour-mt_aacYFVN-HH) ## Optional connections - [Roots of Polynomials and Roots of Unity](https://lightmysky.com/learn/mathematics/roots-of-polynomials-and-roots-of-unity-mt_XAISygqdQr) ## Opens up - [Power Series and Analyticity in the Complex Plane](https://lightmysky.com/learn/mathematics/power-series-and-analyticity-in-the-complex-plane-mt_Mwxapy_pU8)