356 Mr Grace , On the Zeros of a Polynomial. 
and hence all the roots 1 of 
(a — z) n —(8— z) n — 0 
lie within a circle whose centre is 0 and whose radius is 
OA cot — . 
n 
But f (z) is apolar to the form 
(a-z) n -(/3- z) n , 
and hence f (z) must have at least one zero within this circle. 
Cor. The zeros of f (z) cannot all lie on the same side of the 
line bisecting AB at right angles. 
This follows from the above reasoning by regarding the line in 
question as a circle of infinite radius which for the purpose of our 
argument practically encloses all the roots of 
(a - z) n - (8 - z) n = 0. 
(8) The results just arrived at can be illustrated geometrically 
when the equation f (z) = 0 is a cubic. 
In fact suppose that A, B, C represent the roots of 
/O) = o, 
and that P 1 and P 2 represent those of 
/'(*)- 0 , 
then P u P 2 are the foci of the maximum ellipse inscribed in the 
triangle ABC. 
Hence if 0 be the middle point of 
AB we have 
0P X . 0P 2 = square of semi-diameter 
parallel to AB 
= A *-l 
so that if A and B are given we have 
0P X . 0P 2 = 30.4 2 ; 
1 Except two which lie on the circle. If n > 3 some roots certainly lie inside the 
circle, but if n = S all the roots lie on the circle and we can construct an apolar form 
having its two roots on this circle. 
