proof of Rouché’s theorem

Consider the integral


where 0λ1. By the hypotheses, the function f+λg is non-singular on C or on the interior of C and has no zeros on C. Hence, by the argument principle, N(λ) equals the number of zeros (counted with multiplicity) of f+λg contained inside C. Note that this means that N(λ) must be an integer.

Since C is compactPlanetmathPlanetmath, both |f| and |g| attain minima and maxima on C. Hence there exist positive real constants a and b such that


for all z on C. By the triangle inequalityMathworldMathworldPlanetmathPlanetmath, this implies that |f(z)+λg(z)|>a-b on C. Hence 1/(f+λg) is a continuous functionMathworldPlanetmathPlanetmath of λ when 0λ1 and zC. Therefore, the integrand is a continuous function of C and λ. Since C is compact, it follows that N(λ) is a continuous function of λ.

Now there is only one way for a continuous function of a real variable to assume only integer values – that function must be constant. In particular, this means that the number of zeros of f+λg inside C is the same for all λ. Taking the extreme cases λ=0 and λ=1, this means that f and f+g have the same number of zeros inside C.

Title proof of Rouché’s theoremMathworldPlanetmath
Canonical name ProofOfRouchesTheorem
Date of creation 2013-03-22 14:34:26
Last modified on 2013-03-22 14:34:26
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 6
Author rspuzio (6075)
Entry type Proof
Classification msc 30E20