Fudametal Theorem of Algebra Yvoe Lai March 010 We prove the Fudametal Theorem of Algebra: Fudametal Theorem of Algebra. Let f be a o-costat polyomial with real coefficiets. The f has at least oe complex root. The first half of the proof aalyses the images of sufficietly large loops (Sectio 1) ad sufficietly small loops (Sectio ). We will quatify i these sectios what sufficietly large ad sufficietly small mea. The secod half of the proof aalyses the images of loops close together (Sectio 3), ad applies the properties of images to show the existece of a z C so that f (z) = 0 (Sectio 4). The existece proof relies o a examiatio of widig umber. 0 Prelimiaries Let f be defied with the coefficiets f (z) = a 0 + a 1 z + a z +... + a 1 z 1 + a z, where a = 0 (to guaratee that f (z) is ocostat). Note that sice f is ocostat, 1. Notatio. Throughout the proof, we will use the otatio: a = max{ a 0, a 1,..., a 1, 1} coefficiets of f. C R = (R, t), t [0, π], where (R, t) is the parametrizatio, writte i polar form, of the couterclockwise path i C traversig oce couterclockwise aroud the circle with radius R ad ceter 0. Simplifyig Assumptios. Note that if a 0 = 0, the f (0) = 0, so 0 is a root. So from ow o, we work with the case a 0 = 0. It is sufficiet to prove the theorem for a = 1. For suppose that we have show that all ocostat polyomials with leadig coefficiet 1 must have at least oe complex root. The the polyomial g(z) = f (z) = a 0 + a 1 z +... + a 1 z 1 + z a a a has at least oe root; suppose the root is z 0. It follows that a f (z 0 ) = a g(z 0 ) = a 0 = 0, so z 0 is a root of f. Thus provig the Fudametal Theorem of Algebra for the case a = 1 implies the theorem for the case whe a = 1. So, i what follows, we assume a = 1. 1
1 Large Iput Loops Our goal i this sectio is to prove Propositio 1 below. The key idea of its statemet is If R is big eough, the f seds C R to a path cotaied i a aulus aroud the circle with radius R, cetered aroud the origi. A schematic of the statemet ca be foud at the ed of this sectio. Propositio 1. If R a, the { f (C R ) z C Lemma 1. If R a, the for 0 k 1. 1 R z 3 } R. ar k R Proof of Lemma 1. Note that a > 1, so if R a, the R m R a for m 1, icludig m = k, for 0 k 1. It follows that as desired. R k a R ar k R ar k Private math: For this lemma, we seek a R so that ar k R, which ca be rewritte ar k R a R k k a R ( ) Oe R that satisfies ( ) is a, because a is at least as large as k a. Proof of Propositio 1. To prove the propositio, we eed to show that R a implies the radius of f (C R (t)) is betwee 1 R ad 3 R, for all t. The radius of f (C R (t)) is (R, t) + a 1 (R, t) 1 +... + a 1 (R, t) + a 0, which is bouded below by R ( a 1 R 1 +... + a 1 R + a 0 ), ad bouded above by ( ) R + a 1 R 1 +... + a 1 R + a 0. ( ) Suppose R a. We have just show that this coditio implies ar k R for 0 k 1. Cosequetly, a 1 R 1 +... + a 1 R + a 0 R +... + R + R }{{ } times = R.
Pluggig i our calculatios ito ( ) ad ( ), we coclude that R a implies the radius of f (C R (t)) satisfies R R f (C R(t)) R + R for all times t, as claimed. 1 R f (C R (t)) 3 R We have show that whe R is large eough, meaig its radius is at least a, the f seds C R to a loop that is cotaied i a eighbourhood of a circle with radius R. See Figure 1 for a visual schematic of the meaig of Propositio 1. C R a R R R iput output Figure 1. Visual schematic for Propositio 1. If C R is cotaied i the shaded iput regio, the its image is cotaied i the shaded output regio. 3
Small Iput Loops Propositio. Whe R a 0 a, the f (C R) is cotaied i a disk of radius a 0 cetered at a 0. Lemma. Suppose R a 0 a. The ark a 0 for 1 k. Proof of Lemma. Suppose R a 0 a, ad let 1 k. Note that a 0 a < 1, as a 0 is ( ) a0 k at most a. This implies a a 0 a. We have as desired. ar k a ( ) a0 k a a 0 a a = a 0 Proof of Propositio. To prove the propositio, we eed to show that R a 0 a implies that a 0 f (C R (t)) a 0 for all t. Note that ( a 0 f (C R (t)) = a 0 a 0 + a 1 (R, t) +... + a 1 (R, t) 1 + (R, t) ) Private math: For this lemma, we seek a R so that ar k a 0, which ca be rewritte ar k a 0 R k a 0 a R k a0 a ( ) Oe R that satisfies ( ) is a 0 a, because a 0 a is at most k a0 a if a 0 a < 1. a 1 R + a R +... + a 1 R 1 + R. Suppose R a 0 a. We have just show this coditio implies ark a 0 1 k. Cosequetly, for a 0 f (C R (t)) a 1 R + a R +... + a 1 R 1 + R for all t, as claimed. a 0 + a 0 +... + a 0 }{{ } times = a 0 We have show that whe C R is small eough, meaig its radius is at most a 0 a, the f seds C R to a loop that is cotaied i a eighbourhood of a 0, the costat term. 4
C R a 0 a a 0 a 0 iput output Figure. Visual schematic for Propositio. If C R is cotaied i the shaded iput regio, the its image is cotaied i the shaded output regio. 3 Neighbourhoods of Iput Loops The goal of this sectio is to prove Propositio 3. Propositio 3. Let (R, t) C. For every ɛ > 0, there exists a δ 0 > 0 such that δ < δ 0 implies f (R + δ, t) f (R, t) < ɛ. Recall that the degree of f is. I the proof of Propositio 3, we let C deote the largest coefficiet over all terms of the multiomials {(x + δ) k 0 < k }. Lemma 3. Give 0 < δ < R, for all itegers k 0. (R + δ) k R k δ kcr k 1 Proof of Lemma 3. After expasio, the expressio (R + δ) k R k cotais oly terms of the form R k 1 δ, R k δ,..., Rδ k 1, δ k, as the term R k cacels. To prove the lemma, first cosider positive δ. The (R + δ) k R k = (R + δ) k R k CR k 1 δ + CR k δ +... + CRδ k 1 + Cδ k δc(r k 1 + R k δ +... + Rδ k + δ k 1 ) δc(r k 1 + R k 1 +... + R }{{ k 1 ) } k times δckr k 1. 5
Now suppose δ is egative, so (R + δ) k R k = R k (R δ ) k. Because power fuctios are cocave up, it follows that We have show the lemma. R k (R δ ) k (R + δ ) k R k δ CkR k 1. Corollary 1. Let K = a C max{r, 1}. If 0 δ R, the Proof of Corollary 1. Note that f (R + δ, t) f (R, t) = f (R + δ, t) f (R, t) < δk. a It follows from Lemma 3 that ( a (R + δ) k R k) a k ( (R + δ, t) k (R, t) k) a ((R + δ, t) k (R, t) k) ( (R + δ) k R k). Hece f (R + δ, t) f (R, t) < δk as claimed. We ow prove the propositio. aδkcr k 1 < aδc max{r, 1} = aδc max{r, 1} = δk Proof of Propositio 3. Fix ɛ > 0. We seek δ 0 > 0 that guaratees that if δ < δ 0, the f (R + δ, t) f (R, t) < ɛ. We claim that δ 0 = ɛ K works. Ideed, suppose δ < δ 0 = ɛ K. By Corollary 1, as we sought to prove. f (R + δ, t) f (R, t) < δ K < δ 0 K = ɛ K K = ɛ 6
4 Proof of the Fudametal Theorem of Algebra Propositio 4. Let R > 0, ad suppose f (C R ) does ot cross the origi. Let ɛ 0 be mi{ f (C R (t)) t [0, π]}. Let δ 0 = ɛ 0 /K, where K is defied as i Corollary 1. The δ < δ 0 implies that the widig umbers of f (C R ) ad f (C R+δ ) are the same. Proof. We show that there is a deformatio of f (C R ) to f (C R+δ ) that does ot cross the origi, ad hece the widig umbers of the two paths must be equal. Defie p α (t), t [0, π] as the path f (C R (t)) + α( f (C R+δ (t)) f (C R (t))) for α [0, 1]. Note that p 0 = f (C R ) ad p 1 = f (C R+δ ), ad as α sweeps from 0 to 1, the path p α deforms from f (C R ) to f (C R+δ ). By Propositio 3, δ < δ 0 guaratees that f (C R+δ (t)) f (C R (t)) is strictly less tha ɛ 0, the miimum radius of ay poit i f (C R (t)). It follows that p α caot pass through the origi for ay α. Hece we ca deform f (C R ) to f (C R+δ ) without passig through the origi, so δ < δ 0 guaratees that the widig umbers of f (C R ) ad f (C R+δ ) are equal. By usig similar techiques, we obtai the corollary of Propositio 1: Corollary. Suppose R a. The the widig umber of f (C R ) is. Proof. We show that the widig umber of f (C R ) is by deformig a path that has widig umber to f (C R ). This path is (C R ) = (R, t). Defie the deformatio q α (t) = (C R ) + α( f (C R (t)) (C R (t)) ), t [0, π] from (C R ) to f (C R ). By Propositio 1, f (C R (t)) is at most R /. It follows that f (C R (t)) (C R (t)) R / R = R /, so q α does ot pass through the origi for ay α. Thus the widig umbers of (C R ) ad f (C R ) must be equal. 7
We ow complete the proof of the Fudametal Theorem of Algebra. I the followig proof, we use circle to deote a path of the form (r, t), t [0, π], where r > 0 is costat. Fudametal Theorem of Algebra. Let f be a ocostat polyomial. The f has at least oe complex root. Proof of Fudametal Theorem of Algebra. I this argumet, we aalyse widig umbers. Recall w(p) deotes the widig umber of p. Suppose by way of cotradictio that f has o roots. The, for ay R > 0, the path f (C R ) does ot cross the origi. I particular, w( f (C R )) is well-defied. By Propositio 4, there is a aulus N δ (C R ) such that if C is a circle i N δ (C R ), the w( f (C)) = w( f (C R )). For each R, pick such a aulus ad deote it N(C R ). Cosider the regio D = {z C 0 < z a}. This regio is covered by the collectio {N(C R ) 0 < R a} of auli. Circles from overlappig auli must produce images with the equal widig umbers. Because the regio D is coected ad covered by auli, each of which overlaps at least oe other aulus, we ca coclude that the widig umber of the images of all circles i D must be the same. The widig umber of f (C a ) is, by Corollary. Sice C a D, it follows that the widig umber of images of all circles i D must be. But by Propositio, the widig umber of f (C a0 /a) is 0, ad 0 =. Hece we have foud our cotradictio. There must exist a root of f i the complex plae. 8