Mr Grace , On the Zeros of a Polynomial. 
355 
Consequently by the result just proved f'(z) is apolar to the 
form 
" ■ V" - Z n —\ («-> - / 3 "-) + ( !L - 1 . K^- 2 ) 1 (aK - 2 _ ^ 
n — z 
n 
or to 
n - 1 v ^ ' 2 
(n — 1) (n — 2) (n — 3) 1 
+ Z 3 
3! 
n — 3 
- (a n ~ 3 — ft n ~ 3 ) . . . = 0, 
(a n - ft 11 ) - nz (a"" 1 - /3 n ~ J ) + U ^ ^ (a n ~ 2 - ft n ~ 2 ) ...=0, 
A 
that is to the form of degree n — 1, 
(a — z) n — (j3 — z) n . 
Thus if f(z) — 0 has two given roots a and ft, then f'(z) is 
apolar to the form (a — z) n — ({3 — z) n . 
(7) Now the roots of the equation 
(a - z) n ~(ft~z) n = 0 
are easily constructed, for they are given by 
(a — z) = co ((3 — z), 
where w is any nth. root of unity which is not unity itself. 
To construct the points representing these roots take A, B for 
the points a, ft and let 0 be the middle point of A, B, then 
we have 
a 
2 rir 
z-ft 
a + ft 
2 
, r = 1 , 2, ... to - — 1; 
, r 7r 
_ = fc cot — 
a — ft n 
a + ft , rir a — ft 
.*. z = — + i cot — , 
2 n 2 
which shews that to arrive at z we have to travel a distance 
rir 
0 A cot 
n 
from 0 along a line at right angles to AB. 
The greatest distance of one of these (n — 1) points from 0 is 
OA cot — , 
YOL. XI. pt. v. 
26 
