(319) 



If we write : 



a = a x -f 2 (a IS — o x ) a -f (a, -f a, — 2a M ) as* 



and 



6- = V + aft i:r «H- - 



ax V ''•'' 



the latter equation way also be written in the form 



0< 



a,-!- a, — 2rt 1; 





+ « 



" 2K,-at) 



a i + « 3 — 2 «1! 



2 A 



d6 



+•'...(» 



If we demand that the locus be imaginary all over the width of 

 the figure, so for all the values of x between and 1, there are 

 three conditions : 



Ik 



b 



1st . ' l _v> L_ 2 nd L____\__l__ 



and a third condition which is still to be derived. 



Let us write for this purpose a x -\-a t — 2a ls = c and b i = nb l , 



o, 1 a, n 3 



then 1st— > __ 2 nd ->- — - and when we introduce the 



c (/* — 1) c (n — 1) 



a, l+*i . <?, >< s (1+6, ) 



auxiharv quantities f, and £,, so that — = — — and — — — : ti 



* J l c (w-1)' c (»-l) s 



(gj andf s positive), we find from 2a la = a x -\-a % — c or — — = — -f — — 1 



2 c c 



2 <h* _ *+*! W' + M * F, _ j 



or 



(n— 1)' (ra— l) 5 

 2w4-«i+w J f a 



2*= = 



(n-l) s 



So the condition that this locus be imaginary is thai for all the 

 values of x : 



— x i + - 1 + # a > o 



("-If I («— 1)* i 



tf') 



or 



«,>0 and e,>0 and l+^-^< 2 J^I 



The last condition may be written more symmetrically in the form: 



n -1 (n — l)*^(n— l)* 



or 



