353 
Mr Grace , On the Zeros of a Polynomial. 
the roots of f(z) = 0 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 0 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 f (z) he 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 £ by 
x x and make the expression homogeneous by means of x 2 . 
Then let 
f(x x x 2 ) = a x n , 
and let </> ( x x x 2 ) = b x n be the apolar form so that we have ( ah) n = 0, 
or if the zeros of f be (y x , y 2 )(z x , z 2 ) ... (w x , w 2 ) we have 
h y h z . . . h w — 0 . 
Thus of the n ratios (y x , y 2 )...(w x , w 2 ) all but one may be 
chosen arbitrarily and the wth is then given by the mixed polar of 
the preceding values with respect to 
h x n = 0. 
Now returning to the Argand diagram suppose that 'the roots 
of the fixed apolar form 
<#> O) = o 
are all inside a given circle S and suppose that P x , P 2 . ... P n are 
the roots of 
/(*)- 0 . 
so that P n is determined by equating to zero the mixed polar of 
P x , P 2 , ... P n -i with respect to f(z). 
If P x be outside S then all the roots of its first polar f x (z) = 0 
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 
fn — i ( z ) = 
which determines P n , hence if P x , P 2 , ... P n - X are all outside the 
circle S then P n is inside the circle, i.e. the equation f(z) — 0 has 
at least one root within a circle enclosing all the roots of the fixed 
equation 
4 > 0 ) = o. 
If the circle S contains all the roots of </> (z), then when 
P x> P 2 , ... P n _ x are all on the circle so also is P n . 
