Florian Enescu, Fall 2010 Polynomials: Lecture notes Week 9.

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1 Floria Eescu, Fall 200 Polyomials: Lecture otes Wee 9 Gauss-Lucas Theorem; Jese diss, Jese Theorem The followig theorem by Gauss ad rediscovered by Lucas describes a beautiful relatio ship betwee the roots of a polyomial ad the roots of its derivative For a polyomial P, we call the roots of its derivative critical poits First we eed to state a result that is iterestig i its ow right ad will be used i the proof of the Gauss-Lucas theorem We will refer to a half-plae as ay of the two regios determied by a lie i the plae Lemma The sum of a fiite umber of complex umbers that are situated i a ope half-plae is situated i that ope half-plae Proof We ca traslate the lie ad the umber such that the lie that determies the half-plae passes through the origi It is eough to show the statemet for two complex umbers, sice we ca the iterate our sum We ca rotate our lie so that the lie is the x-axis ad the complex umbers are above it It is clear the that is the pricipal argumets of the two complex umbers are θ ad θ 2, the their sum will have the pricipal argumet betwee θ ad θ 2, sice the sum of two complex umber z,z 2 is the vertex of the paralellogram with the other vertices equal to z,z 2, 0 I coclusio θ will be aywhere betwee 0 ad π ad hece i the upper half-plae We ca ow state the Gauss-Lucas Theorem Theorem 2 Let P be a complex polyomial The the roots of P belog to the covex hull determied by the roots of P Proof Write P (z =(z z (z z, where = deg(p The P /P = z z + + z z Suppose that P (w = 0 ad P (w 0 ad that w does ot belog to the covex hull determied by z,,z We ca draw a lie through w that does ot itersect the covex hull of z,, z The all umbers w z i will all be situated i oe of the ope half-plaes determied by the traslatio of the lie through the origi The iverses of w z i will have the pricipal argumets give by subtractig the origial pricipal argumets from 2π respectively We will ow move to aother statemet about the positio of the critical poits of a polyomial with respect to the roots of the polyomial itself We have oted before that for a real polyomial P,ifP (z = 0, the P (z =0 Let us cosider z 0 a root of P such that z 0 C \ R The z 0 is also a root The circle with ceter at Re(z 0 ad radius Im(z 0 is called a Jese dis of P

2 Theorem 3 (Jese Let P be a polyomial with real coefficiets The ay o-real critical poit of P lie iside or o the boudary of a Jese dis of P Proof Let = deg(p ad let z,, z be its complex roots, possible o distict The as i proof of the Gauss-Lucas Theorem, P (z P (z = i= z z i Assume that w is o-real critical poit of P,soP (w = 0 If P (w = 0, the w = z i for some i ad w lies o the boudary of the Jese dis determied by w = z i So, let us assume that P (w 0 Let w = u + iv, v 0 For a oreal root z i = a + bi, a, b R, its cojugate is also a root ad equals a bi The we have But (w a 2 + b 2 =(w a 2 + b 2 So, w a bi + w a + bi = 2w 2a (w a 2 + b 2 2w 2a (w a 2 + b 2 = (2w 2a(w a2 + b 2 (w a 2 + b 2 2 We would lie to compute the sig of the imagiary part of this expressio Sice the deomiator is real ad positive the it is eough to loo at the sig of the imagiary part of the umerator But (2w 2a(w a 2 + b 2 =2( w a 2 (u iv a+b 2 (u + iv a, ad the imagiary part of this umber is 2( v w a 2 +b 2 v=2v( w a 2 +b 2 Note that this expressio has the sig of v, because w outside the Jese dis of z = a + ib meas exactly that b 2 < w a 2 Let us loo at the sig of the imagiary part of the expressio, w z i i the case z i = a real umber The w a w a w a = u a iv w a 2, ad the imagiary of this expressio is exactly the sig of v Now i the sum

3 0= P (w P (w = i= w z i, appear two types of roots: complex o-real roots which ca be paired together ad the sum correspodig to them has the sig of v, ad terms correspodig to real roots which have also sig equal to the sig of v I coclusio, the above sum has sig exactly the sig of v, ad we ow that v 0, so the sum caot be zero Cotradictio! Therefore our w must belog iside or o the boudary of oe of the Jese diss I what follows we show some more applicatios of Laguerre s theory developed i lecture 6 Let f be a polyomial of degree, ad ξ a complex umber differet from The (apolar derivative of f with respect to ξ is defied as A ξ f(z =(ξ zf (z+f(z, for ξ, or A ξ f(z =f (z, for ξ = Oe should ote that this defiitio is such that for a polyomial of degree, the ceter of mass of z with respect to the roots of f equals ξ if ad oly if A ξ f(z =0 To faciliate the computatios we will write our polyomials i a differet form Namely, every polyomial f(z = =0 d z, ca be rewritte as f(z = =0 a z Remember that ( =!!(! So for example ( 3 ( =3, 3 3 =, ad f(z =z 3 z = The beefits of this writig will be clear later Util the, let us ote that ad so f (z = = f (z = (! (!(! a z = = With this computatio ow oe ca see that But, A ξ f(z =0 ( 3 z 3 3 ( a z, = ( 3 3 z ( a z = ( =(ξ z a + z + =0 =0 ( a z a + z

4 sice Therefore ξ a + z ( a z + A ξ f(z ( ( = = =0 ( a z = (a + a + ξz This way we obtaied a ew polyomial of degree If we repeat our computatio ad compute we will get A η (A ξ f(z ( 2 ( 2 ((a + a + ξ+(a + + a +2 ξηz =0 Note that a + a + ξ +(a + + a +2 ξη = a + a + (ξ + η+a +2 ξη So, if we iterate the procedure we ca see that (a + a + ξz, where! A ξ A ξ2 A ξ f(z =a 0 + a σ + + a σ, σ = ξ + + ξ σ 2 = ξ ξ 2 + ξ ξ ξ ξ σ = ξ ξ Let cosider aother polyomial g(z= =0 b z Deote z,, z the roots of f(z ad ξ,, ξ be the roots of g Therefore Viète relatios for the roots of g tell us that ( b σ =, b ( b 2 σ 2 =, 2 b

5 Therefore we ca otice ow that σ =( 0 b0 b is i fact equivalet to a 0 b ( a b + A ξ A ξ2 A ξ f(z =0 ( a 2 b 2 + +( a b 0 =0 2 a z ad g(z = Defiitio 4 Two polyomials f(z = =0 =0 b z, are called apolar if ad oly if their coefficiets satisfy ( ( a 0 b a b + a 2 b 2 + +( a b 0 =0 2 Let us costruct a apolar polyomial for f(z =2z 2 +4z +=+2 ( 2 ( z z 2 So, a 0 =,a =2,a 2 =2 The we are looig for a polyomial g of degree 2 such that g(z=b 0 +2b z + b 2 z 2, ad ( 2 b 2 2b +2b 0 =0 There are ifiitely may possibilities for b 0,b,b 2 therefore ad we ca choose for example b 0 =,b =,b 2 =2, or i other words g(z=+2z +4z 2 The followig result is a cosequece of Laguerre s wor ad it is what is eeded i our applicatios We remid you that a circular domai (ope or closed is the iterior or the exterior of a circle or a half-plae Theorem 5 (J H Grace, 902 Let f, g be two apolar polyomials If all the roots of g belog to a circular domai K, the at least oe of the roots of f also belogs to K We postpoe the proof of this theorem so that we ca illustrat how it ca be applied Let f(z = z + cz, with c C, 2 We will show that f has at least oe root iside the dis z To this purpose, we loo for a polyomial g(z= =0 b z, such that b ( b + cb 0 =0 Let us tae b =, so we eed to choose b,b 0 such that + b + cb 0 = 0 Note that there are NO coditios imposed o b,, b 2 A possibility would hece be ay polyomial with b 0 =0,b =

6 Let ξ = ɛ, where ɛ are th roots of uity, =,, Note that ξ are o the circle z = Costruct the polyomial g(z=π =(z ξ The sum of the roots of g is ad the product of the roots is 0 (sice oe of the roots is actually 0 Hece g(z =z z + +0, which is a polyomial with b =,b 0 = 0 This meas that g(z is apolar with f(z Moreover all its roots belog to the closed circular domai z, ad hece by Grace s Theorem we ow that at least a root of f must be iside K Oe ca agree that this is quite a beautiful applicatio of Laguerre s wor Our ext lecture will deal with a proof of Grace s Theorem

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