Proof of the Fundamental Theorem of Algebra Author(s): S. Wolfenstein Source: The American Mathematical Monthly, Vol. 74, No. 7 (Aug. - Sep., 1967), pp. 853-854 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2315822 . Accessed: 02/09/2011 20:38 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The American Mathematical Monthly. http://www.jstor.org 1967] 853 CLASSROOM NOTES PROOF OF THE FUNDAMENTAL THEOREM OF ALGEBRA de Paris S. WOLFENSTEIN, Universite The so-called Fundamental Theorem of Algebra states that every nonconstant polynomial with complex coefficientshas at least one complex zero; or alternatively,that if p is any nonconstant polynomial functionof a complex variable, then the range of p contains every complex number. The purpose of the present note is to sketch a proofsimpler than appears currentlyavailable and in any case employing no mathematical tools more sophisticated than are already required for elementarycalculus. The proof is based upon the convergenceof an iterative procedure,a notion which should be familiar to beginningstudents (Newton's method, Picard's method). Prerequisitesare as follows: (A) A nonzeropolynomialhas only a finitenumberof zeros. (B) If f: Rm >Rnis a continuous functionsuch that If(x)I goes to infinity with lxI (in particular iff is a nonconstantpolynomial functionof a complex variable), then the range off is closed. (This followsat once fromWeierstrass's Theorem that everybounded sequence has a convergentsubsequence.) (C) For n> 1, the complement of a finitesubset of Rn is connected. In addition only the most elementarypropertiesof complex numbersare needed. LEMMA.Let p be a nonconstantpolynomialfunctionof a complexvariable,S therangeofp and T= {p(z): p'(z) =01. Then S-T is open. small the equation Proof. We firstshow that if Ic| is sufficiently (1) W+ n bmWmc m=2 has a root that can be approximatedby iteration.Let wo= 0, n >, Wk+1 = c - m bmWkX m=2 clearly if the sequence wk convergesat all it convergesto a root of (1). Suppose now that |c is small; specifically,let a be any positive number less than 1, p-=min(Oa/[E =2 mIbmj], 1) and I|c <P- En=2 |bin pm(this is a positive inductivecalculation shows that, forall k, Iwkf <p. number). A straightforward Further, for k> O, Wk+1 - Wk - WkI | < | Wk-1I n Zbm E rn=2 n - < at I Wk - Wk_l IE m=2 Wk-1 m Em8W J-I 8=1 m I bmI pm-l, by theforegoing, 854 CLASSROOM [September NOTES so that, forc as stated, Wk does convergeto a root of (1). Now let p be any nonconstant polynomial, zo any complex number such thatp'(zo) O. With the substitutionsz=zo+w, a==p(zo)+p'(zo)c, the result just obtained shows that the equation p(z)=a has a root for all values of a close enough to p(zo). In otherwords,ifzo is not a zero of p', p(z0) is an interior point of S. This proves the Lemma. THEOREM. Withthenotationof thelemma,S is theset ofall complexnumbers. Proof. The complementof S is open (B), and S- T is open by the lemma. But as T is finite(A) its complementcannot consist of two disjoint nonempty open sets (C). Hence the complementof S must be empty. AN EASY PROOF OF THE FUNDAMENTAL THEOREM OF ALGEBRA CHARLES FEFFERMAN, University ofMaryland +an2Znbe a complex polynomial. Then P THEOREM. Let P(z) ==ao+alz+ has a zero. Proof.We shall show firstthat IP(z) I attains a minimumas z varies over the entire complex plane, and next that if IP(zo) is the minimum of |P(z) |, then P(zo) =0. Since JP(z)j so large that |zfnIa+an +ao/znJ l/z+ (1) (z20), wecanfindan I aolI IP(z)|I (IzI M>O >M). Now, the continuous function|P(z)j attains a minimumas z varies over the compact disc { z Izj I M}. Suppose, then,that (2) | P(zo) I Z I| _ I P(Z) I ?<M). In particular, P(zo) | P(0) = Iao I so that, by (1), P(zo) ?<-| P(z)| (I zI >M) Comparing with (2), we have | P(zo) (3) I ' IP(Z) I Since P(z) =P((z-zo)+zo), (all complexz). we can write P(z) so that forsome complex polynomial Q, P(z) (4) Q(z - as a sum of powers of z-zo, Zo). By (3) and (4), (5) | Q(O)| Q(z) (all complexz). We shall show that Q(O) =0. This will establish the theorem since, by (4), P(zo) ==Q(O). Let j be the smallest nonzero exponent for which zi has a nonzero coefficient in Q. Then we can write Q(z) =co+cjzi+ +cnzn (cj1-zO). Factoring
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