7/13/211 Rouché s Theorem Prelimiary Iformatio Complex Aalysis I complex aalysis, variables have a real part (x) ad a imagiary part (iy), where i = 1 z = x + iy Fuctios of complex variables also have a real part ad a imagiary part which ca be iterpreted as real-valued fuctios of the two real variables x ad y. f(z) = u(x,y) +iv(x,y) Example: f(z) = (z 2 ) = (x + iy) 2 = (x 2 -y 2 ) + i(2xy) Additio of complex variables follows the rule: (a + ib) + (c + id) = (a + c) + i(b + d) Multiplicatio of complex variables follows the rule: (a + ib) (c + id) = (ac bd) + i(ad + bc) Prelimiary Iformatio Derivatives Takig the derivative of a complex fuctio is similar to takig the derivative of a real fuctio, but the limit must have the same value for ay sequece of complex values for z that approach the poit i questio o the complex plae. f '(z ) lim f (z) f (z ) zz z z 1
7/13/211 Example: Fid the derivative of f(z) = z 2. So, f (z) f (z ) (z)2 2 z (z z )(z z ) z z z z z z z z f (z) f (z lim ) lim (z z z z z z ) z z 2z z z It s the same as takig the derivative of f(x) = x 2! Critical poits of complex fuctios exist where the derivative is zero. Sigularities are poits z i the domai of a fuctio f(z) where f(z ) fails to be differetiable. Poles are sigularities at a poit a such that f(z) approaches ifiity as z approaches a. Whe a complex fuctio is differetiable everywhere it is aalytic or holomorphic. Examples of aalytic complex fuctios: Polyomials Expoetials Example of a fuctio which is NOT aalytic: Absolute value fuctio z x 2 y 2 Not differetiable whe z=. 2
7/13/211 Example of a pole: f (z) 3 z 1 Prelimiary Iformatio Taylor Series The Taylor series we are familiar with holds whe the real variable x is replaced by the complex variable z, However, the iterval of covergece is ow replaced by the idea of the disk of covergece, sice the iequality describes the iterior of a disk of radius R, cetered at the poit z. z z R Radius of Covergece Aalytic Fuctios: Redefied. R lim 1 a where a f ( ) (z )! Theorem: A fuctio is aalytic if ad oly if it is equal to its Taylor Series i some eighborhood of every poit. Coverges for all z that satisfy z-z < R Diverges for all z such that z-z > R. If lim a =, the the series coverges for every z. 3
7/13/211 Rouche s Theorem Rouche s Theorem: Let two fuctios f(z) ad g(z) be aalytic iside ad o a simple closed curve C, ad suppose that f(z) > g(z) at each poit o C. The f(z) ad f(z) + g(z) have the same umber of zeros, iside C. Geometrically Speakig Picture walkig a dog aroud a tree o a leash with variable legth. O your walk, you let the leash go log eough so the dog ca walk ahead of you. However, provided you keep the leash short eough so the dog caot reach the tree, you ad the dog will still walk aroud the tree the same umber of times! Usage of Rouche s Theorem Rouche s Theorem is used to simplify the problem of fidig the zeros i a give aalytic fuctio. I order to do so, write the fuctio as a sum of two fuctios, oe of which is easier to work with ad domiates (grows more quickly tha) the other. The zeros are the foud by lookig at oly the domiatig part of the origial fuctio. 4
7/13/211 Take, for example, the polyomial z 7 2z 3 + 7 We wat to fid how may zeros it has i the disk z <2. Set f(z) = z 7 ad g(z) = 2z 3 + 7. Note that g(z) < f(z) o z <2. Sice f(z) has 7 roots i the disk, f(z) + g(z) also has 7 roots, sice it satisfies Rouche s Hypothesis. A Little History Lesso Eugee Roche, NOT the Eugee we eed, Fu fact: he was the origial Ajax ma. Our Rouché Bor i souther Frace, Taught at the Charlemage college ad the was professor at the Academy of Arts ad Trades i Paris, Examier at the Polytechic istitute, Name is kow through Rouché's theorem which he published i the Joural of the École Polytechique i 1862. Fudametal Theorem of Algebra Rouche s Theorem ca be used to help prove the Fudametal Theorem of Algebra The Fudametal Theorem states: A polyomial 1 2 P( z) az a 1 z a 2z... a of degree always has roots. 5
7/13/211 Proof P( z) a z a a 1 2 1 z a 2z... Let f (z) a z ad g(z) a 1 z 1... a O the circle z R, f (z) a R, Ad g(z) a 1 R 1... a 1 R a Now, make R large eough so that a 1... a 1 a a R The, f(z) < g(z) holds o the boudary of the circle cetered at the origi of radius R. Sice f(z) clearly has zeros, we have proved that P( z) f ( z) g( z) a z a a 1 1 z... also has zeros usig Rouché s Theorem. Questios? 6