Mr Grace, On the Zeros of a Polynomial. 353 



the roots of f(z) = lie within a circle, then all the roots of the 

 first polar of any external point lie within that circle. The result 

 is still true when all the zeros of f(z) lie on the circle, but if be 

 also on the circle, then all the zeros of the first polar are also. 



Similarly if all the zeros lie outside the circle, then all the 

 zeros of the first polar of an internal point lie outside the circle. 



(4) If the form j (z) be apolar to a given form </>, then it 

 has a zero lying within any circle enclosing all the roots of <£ = 0. 



For a moment let the forms be homogeneous, i.e. replace z by 

 x 1 and make the expression homogeneous by means of x 2 . 



Then let 



J [X 1 X 2 ) = Q'x , 



and let $ {x x x 2 ) = b x n be the apolar form so that we have (ab) n = 0, 

 or if the zeros of/ be (y 1 , y 2 )(zi, z 2 ) ... (w 1} w 2 ) we have 



byb z ...b w = 0. 



Thus of the n ratios (y 1} y 2 )...(w 1} w 2 ) all but one may be 

 chosen arbitrarily and the nth. is then given by the mixed polar of 

 the preceding values with respect to 



Now returning to the Argand diagram suppose that the roots 

 of the fixed apolar form 



<£(*) = o 

 are all inside a given circle S and suppose that P lt P 2 , ... P n are 

 the roots of 



/(*) = <>, 



so that P n is determined by equating to zero the mixed polar of 

 P 1} P 2 , ... P n -i with respect to f(z). 



If P 1 be outside 8 then all the roots of its first polar f (z) = 

 are inside S, if P 2 be outside S then all the roots of its first polar 

 with respect to f (z), say f 2 (z) = 0, are inside S, and so on. 



But by continued polarization we come to a linear form 



A-i(*) = o, 



which determines P n , hence if P x , P 2 , ... P n -\ are a U outside the 

 circle S then P n is inside the circle, i.e. the equation f(z) = has 

 at least one root within a circle enclosing all the roots of the fixed 

 equation 



If the circle $ contains all the roots of <f> (z), then when 

 P 1} P 2 , ... P n -i are all on the circle so also is P n . 



