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.
What a learner can do afterwards
- 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
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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.
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.
The argument principle counts zeros minus poles via the log derivative; Rouche transfers counts between nearby functions; openness forces maximum modulus.
All nine extra tasks and three readings are original authorship created for this topic.
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