--- title: "The Argument Principle and Rouche's Theorem" description: "Count zeros and poles inside a contour by watching how the argument turns, and transfer a count to a nearby function." canonical: https://lightmysky.com/learn/mathematics/the-argument-principle-and-rouches-theorem-mt_J1dC428M84 source: https://lightmysky.com/learn/mathematics/the-argument-principle-and-rouches-theorem-mt_J1dC428M84.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`. # The Argument Principle and Rouche's Theorem Count zeros and poles inside a contour by watching how the argument turns, and transfer a count to a nearby function. Subject: Mathematics · Area: Complex Analysis · Ages 21 to 22 Page: https://lightmysky.com/learn/mathematics/the-argument-principle-and-rouches-theorem-mt_J1dC428M84 ## Ready when they can - State the argument principle and apply it to count zeros in a region - Use Rouche's theorem to locate the roots of a polynomial inside a disc - Deduce the open mapping theorem or the maximum modulus principle from the same idea ## Lesson: Count roots by watching winding Let f be meromorphic on a neighbourhood of a simple closed positively oriented piecewise C1 curve C and its filled interior, with no zeros or poles on C. Then one over 2 pi i times the integral of f prime over f along C equals N minus P, the number of zeros minus the number of poles inside C, each counted with multiplicity. Near a zero a of order m, f(z) = (z-a)^m u(z) with u holomorphic and nonzero near a. Then f prime over f = m over (z-a) plus u prime over u. The second term is analytic, so only m over (z-a) contributes residue m. Near a pole of order m, f(z) = (z-a)^(-m) v(z) with v holomorphic and nonzero near a, and f prime over f = minus m over (z-a) plus v prime over v, giving residue minus m. The normalized integral equals the winding number of the image f(C) around 0, from the continuous unwrapped argument change. The unnormalized integral is 2 pi i times this winding. Let f and g be holomorphic on a neighbourhood of a simple closed curve C and its filled interior. If |g(z)| < |f(z)| at every point z on C, then f and f+g have the same number of zeros inside C, counted with multiplicity. The dominant term f may be a group of several monomials, not just one. The homotopy f + t*g stays nonzero on C for 0 ≤ t ≤ 1, preserving the winding count. **Example.** Worked Rouche application. For p(z) = z^2 + 3z + 1 on |z| = 1, take f(z) = 3z and g(z) = z^2 + 1. On |z| = 1, |f(z)| = 3 and |g(z)| ≤ |z|^2 + 1 = 2 < 3. So p has the same number of zeros inside as 3z, which is 1. A nonconstant holomorphic function f on a connected open domain D maps open sets to open sets. Let a in D and set h = f - f(a). By local Taylor factorization h has an isolated zero at a, so choose a closed disk inside D with no other h-zero. On its boundary, |h| attains a minimum delta > 0. For any w with |w - f(a)| < delta, Rouche applied to h and f(a) - w shows f - w has the same positive zero count as h inside the disk, so w is attained. Openness does not imply injectivity: z^2 folds the plane. A nonconstant holomorphic function on a connected open domain cannot attain its maximum modulus at an interior point. If |f| peaked at a, the open image of a small neighbourhood of a would contain values of larger modulus, contradicting the peak. Constants may attain interior maxima. For a boundary maximum, extend f continuously to the compact closure of a bounded domain; the maximum is attained on the boundary. A failed rough upper bound is inconclusive: it does not prove the actual strict inequality fails, that counts differ, or that the radius must change. Regrouping terms or sharper bounds can work on the same contour. A tie on the boundary breaks the strict inequality Rouche needs. The argument principle counts zeros minus poles; a negative result means more poles than zeros. **Recap.** The argument principle counts zeros minus poles via the log derivative; Rouche transfers counts between nearby functions; openness forces maximum modulus. ## Practice 17 questions on this page, each with its working shown. ## Needs first - [The Residue Theorem](https://lightmysky.com/learn/mathematics/the-residue-theorem-mt_BN2biZ_zos) ## Optional connections - [Evaluating Real Integrals by Residues](https://lightmysky.com/learn/mathematics/evaluating-real-integrals-by-residues-mt_Lj_HXtjzAh) ## Opens up - [Conformal Maps and Mobius Transformations](https://lightmysky.com/learn/mathematics/conformal-maps-and-mobius-transformations-mt_WtHCbAT4GI)