The Fundamental Theorem of Algebra and the Minimum Modulus Principle

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1 FudametalThmAlgAdMiModulusPriciple.tex November 8, The Fudametal Theorem of Algebra ad the Miimum Modulus Priciple Joh A. Guber Departmet of Electrical ad Computer Egieerig Uiversity of Wiscosi Madiso Abstract A direct proof of the fudametal theorem of algebra is give. I other words, we show that every polyomial of degree greater tha or equal to oe has at least oe root i the complex plae. The reader is assumed to be familiar with the followig facts. The triagle iequality [2, pp , Theorem 1.13 ad p. 23, Problem 13]: For complex a ad b, a b a + b a + b. The biomial theorem: For complex a ad b, (a + b) ( ) a k b k,,1,... k k Closed ad bouded subsets of the plae are compact (the Heie Borel Theorem) [2, p. 4, Theorem 2.41]. A cotiuous fuctio o a compact set is bouded ad achieves its miimum ad maximum values o the set [2, pp. 89 9, Theorem 4.16]. A slight modificatio of the proof yields the miimum modulus priciple. If you fid this writeup useful, or if you fid typos or mistakes, please let me kow at guber@egr.wisc.edu

2 FudametalThmAlgAdMiModulusPriciple.tex November 8, Prelimiary Lemmas Lemma 1. If p is a polyomial of degree 1, the p(z) as z. Proof. Suppose p(z) k c kz k, where c. Write [ p(z) z c + c c ] z z. Let z be so large that c 1 /z + + c /z < c /2. The ( p(z) z c 1 c + z z [ c c 1 z + + c ) z + + c ], by the triagle iequality, z [ c c /2 ] z c /2. Lemma 2 (Taylor s Theorem for Polyomials). If p is a polyomial of degree 1, the for ay z C, p(z) p(z ) + z l1 Proof. Suppose p(z) k c kz k. The Now write l p (l) (z) p (l) (z ) (z z ) l l! p (l) (z ) (z z ) l. l! k(k 1) (k [l 1])c k z k l l 1 l! (z z ) l k c k k l k k ( k l ) k! (k l)! c kz k l (z z ) l z k l c k [(z z ) + z ] k, c k z k p(z). k! (k l)! c kz k l. by the biomial theorem,

3 FudametalThmAlgAdMiModulusPriciple.tex November 8, Lemma 3. Suppose f (z) ae jα + be jβ (z z ) l + (z z ) l+1 ϕ(z), where a ad b are positive, ad l 1. If ϕ is cotiuous o some closed disk D cetered at z, the there is some z D with f (z) < f (z ). Proof. Sice ϕ is cotiuous o the compact set D, ϕ is bouded there, say by M, ad we may write ϕ(z) M for all z D. Cosider z of the form z z + re jθ, where r is small eough that z D. The use the triagle iequality to write f (z) ae jα + be jβ (z z ) l + z z l+1 M ae jα + be jβ r l e jθl + r l+1 M e jα( a + br l e j(β+θl α)) + r l+1 M a + br l e j(β+θl α) + r l+1 M. Choose θ so that e j(β+θl α) 1; i.e., θ : (π + α β)/l. The f (z) a br l + r l+1 M. Now further reduce r so that br l < a ad rm < b. We ow have f (z) a br l + r l+1 M a r l [b rm] < a f (z ). 2. The Fudametal Theorem of Algebra Theorem 4 (Fudametal Theorem of Algebra). If p is a polyomial of degree greater tha or equal to oe, the p has a root i the complex plae. Proof. (Based o [3].) By Lemma 1, we ca choose R so large that for all z R, we have p(z) > p(). Sice D : {z C : z R} is closed ad bouded, ad sice p(z) is cotiuous, p(z) achieves its miimum value o D at some z D. We claim that p(z ) ; i.e., z is the required root. To obtai a cotradictio, suppose otherwise that p(z ). Sice z D, z R. I fact, z < R. To see this, otice that poits o the boudary of D, i.e., poits with z R, have p(z) > p() ad so caot miimize p(z) o D. With miimizer z, Taylor s Theorem for Polyomials implies p(z) ae jα + k1 c k (z z ) k,

4 FudametalThmAlgAdMiModulusPriciple.tex November 8, where c k : p (k) (z )/k!, ad writig p(z ) ae jα for some a > is justified because we assumed p(z ). Next, we kow that c sice deg p 1. However, some of the other coefficiets may be zero. Suppose c k for k 1,...,l 1, but c l. The writig c l be jβ, we have p(z) ae jα + be jβ (z z ) l + (z z ) l+1 +1 c k (z z ) k (l+1). } {{ } :ϕ(z) Sice z < R, there is a closed disk D cetered at z ad satisfyig D D. Sice ϕ is cotiuous o D, by Lemma 3, there is some z D D with p(z) < p(z ), which cotradicts z miimizig p o D. 3. The Miimum Modulus Priciple Theorem 5. Let f be a cotiuous fuctio o a closed disk D, ad suppose f is aalytic o the iterior of D. If f is ozero o D, the the miimum of f o D is achieved o the boudary of D. Proof. Sice f is cotiuous o the closed ad bouded set D, the miimum of f o D is achieved by some z D. We must show that z lies o the boudary of D. Suppose otherwise that z is a iterior poit of D. Sice f is aalytic o the iterior of D, we ca expad f i a Taylor series about z, f (z) f (z ) + k1 c k (z z ) k, where c k : f (k) (z )/k!. If all the c k, the f is costat o D, ad the theorem is trivially true. Otherwise, let l : mi{k 1 : c k }. Writig c l be jβ ad f (z ) ae jα, we have f (z) ae jα + be jβ (z z ) l + (z z ) l+1 +1 c k (z z ) k (l+1). } {{ } :ϕ(z) This expasio is oly valid i a eighborhood of z, but we may restrict z to a closed disk D cetered at z such that D is a subset of the iterior of D. The ϕ is cotiuous o D D, ad by Lemma 3, there is a z D D with f (z) < a f (z ). This cotradicts the fact that z miimizes f o D.

5 FudametalThmAlgAdMiModulusPriciple.tex November 8, Refereces [1] R. V. Churchill, J. W. Brow, ad R. F. Verhey, Complex Variables ad Applicatios, 3rd ed. New York: McGraw-Hill, [2] W. Rudi, Priciples of Mathematical Aalysis, 3rd ed. New York: McGraw-Hill, [3] Wikipedia, Fudametal theorem of algebra Wikipedia, The Free Ecyclopedia, [Olie]. Available: of_algebra&oldid , accessed Oct. 1, 214.

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