354 Mr Grace, On the Zeros of a Polynomial. 



(5) If the coefficients of the form f(z) satisfy any linear homo- 

 geneous relation, then the equation f (z) = always has a root 

 within a certain circle. 



Let f(z) = a z n + a x z n ~ l + a 2 z n ~ 2 ... + a n , 



and suppose the relation in question is 



a b + aj?! + a 2 b 2 ... + a n b n = 0. 



Now f(z) is apolar to the form 



<f> (z) = c Q z n + c 1 z n ~ 1 ...+c n , 

 when 



1 2 3! 



a c n n a lCn ^ + ^ _ X) a 2 c n _ 2 ^ ^ _ ^ ^ _ ^ a 3 c n . 3 . . . - U, 



so that in our case f{z) is apolar to the form given by 



C n = Co > 



c n —i = nOi , 



n {n — 1 ) . 



etc., 

 i.e. to the form 



^ (/) = 6 - nb Y z + n {n ~^ b 2 z> ...(-l) n b n z n , 



and hence it always has a root within any circle enclosing all the 

 roots of -ty (z) = 0. If we choose the least of these circles we 

 have a definite circle within which there is always a root of the 

 equation 



f(z) = 0. 



(6) Now let /' (z) be given, namely 



/' 0) = P oz 11 ' 1 + PiZ n ~ 2 ■ • ■ Pn-1 , 

 so that 



f(£) =h z n + J^ 2 n-i _ +p n _ lZ +p n , 



J x n Ti — 1 



where p n is an arbitrary constant. 



Hence if f(z) has two given roots a and /?, we have 



Pj , a n _ '@n\ + Pi ( a «-i _ ygn-i) + . . . p n _ x (a-j3)= 0, 



n n — 1 



a homogeneous linear relation between the coefficients of /' (z). 



