( 219 ) 



equation is found most easily by regarding the rays of the complex 

 normal to XOZ. From x = x\ z = z' ensues p^ = 0, i)^ = 0, 

 2)^ = zp^, p,^ = 0, j^g = — ivp^. By substitution in (i) we lind 



{A -\- Da' -{- Ez' — 2 Fz)p^' = 0, 



and from this for the indicated surface 



J) (.^.^ j^ ,f) ^ Ez' ~ 2 Fz -\- A = . . . . (9) 



For the symmetrical complex this parallel surface is according to 

 (7) the second sheet of the singular surface. 



The planes of the pencils of rays of the bisingular points B^, B^ 

 form the lacking part of the axial surface of l<x,. We can show this 

 by determining the equation of the axial surface of the right line 

 z' ^0, y' =zb, and by putting in it b = co. We then find 



{Ez' -2Fz -}- A) {I) {w' J^ tf)-\- Ez' — 2Fz ^ A\ = . (10) 



The meridian surface, the parallel surface, and the two parts 

 of the singular surface belong to a selfsame pencil, having the skew 

 quadilateral B^I^BJ^ as basis. 



If in the equation of the cone of the complex the sum of the 

 coefficients of x"^, y^ and z'^ is equal to zero, then the edges form x^ 

 triplets of mutually perpendicular rays. The vertices of the triortho- 

 gonal (equilateral) cones of the complex belonging to ^ form the 

 surface of revolution 



{D + E) {x^ + 2/') + '^Ez' — 4i^^ + (2^ + ^) = . . (11) 



It has two circles in common with each of the parts of ^. These 

 contain the vertices of the cones of the complex which break up 

 into two perpendicular planes. 



^ 4. The distance 4 from a right line to OZ is determined by 



l^-^^-^l— (12) 



Pi + P, 



the angle X between a ray and XOY by 



tg^X= — — (13) 



So the condition l^ tang X = a furnishes the complex 



P8P9 = «(i>l' +Pï') (14) 



Here we have a simple example of a symmetrical complex of 

 revolution. 

 The equation 



V = «(Pa^+P,') (15) 



