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 /' (z) = 0 always has a root 
within a certain circle. 
Let f(z) = a 0 z n + a x z n - 1 + a 2 z n ~ 2 . . . + a n , 
and suppose the relation in question is 
a 0 b 0 + afx + a 2 b 2 . . . + a n b n = 0. 
Now f(z) is apolar to the form 
0 (z) = C 0 Z n + C x Z n ~ x . . . + C n , 
when 
12 3! 
n n(n — 1) n (n — 1) (n — 2) 
f^3^n—S ... — 0 , 
so that in our case f{z) is apolar to the form given by 
c n — 6 0> 
Cn — l = Ylb u 
n (n — 1) . 
Cn- 2 = + — J b 2 , 
etc., 
i.e. to the form 
yjr (z) = b 0 - nb x z + 11 — b 2 z 2 . . . (- 1 ) n b n z n , 
and hence it always has a root within any circle enclosing all the 
roots of (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 
/(*)- 0 . 
(6) Now let /' (z) be given, namely 
f 0 ) = + lhz 1l ~ 2 • • • Pn-i , 
so that 
/(Z) = ^ Z n + -At 2**- ... + Pn-I Z + fn, 
J w n n — 1 
where p n is an arbitrary constant. 
Hence if f(z) has two given roots a and /3, we have 
p -° (a n - /3 n ) + % (a" -1 - /S’*- 1 ) + . . . p n -i (<*-£) = 0, 
a homogeneous linear relation between the coefficients of f ( z ). 
