356 Mr Grace, On the Zeros of a Polynomial. 



and hence all the roots 1 of 



(a - z) n -(8- z) n = 

 lie within a circle whose centre is and whose radius is 



OA cot-. 



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 A B 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 -(/3- z) n = 0. 



(8) The results just arrived at can be illustrated geometrically 

 when the equation f{z) = is a cubic. 



In fact suppose that A, B, G represent the roots of 



/<*) = <>, 



and that P 1 and P 2 represent those of 



then Pi, P» are the foci of the maximum ellipse inscribed in the 

 triangle ABG. 



Hence if be the middle point of 

 AB we have 



OPi . OP 2 = square of semi-diameter 

 parallel to A B 



so that if A and B are given we have 



OPi.OP 2 = 30A*; 



1 Except two which lie on the circle. If n > 3 some roots certainly lie inside the 

 circle, but if n = B all the roots lie on the circle and we can construct an apolar form 

 having its two roots on this circle. 



