GIVE THE IMAGINARY ROOTS OF AN ALGEBRAIC EQUATION. 479 
totes and Q has three, which all pass through the origin, and are arranged at 
equal angles around it as already explained. 
Next take an equation with all its roots imaginary, say 1+/—1, 
—1+,/—2, so that f(«)=(e°—2v+2)(v’+2xv+3) and the equation is 
+a? —20+6=0. 
Then the equations to the two curves are 
y—6ey +a*—-y +a°—224+6=0 . . . . (P) 
2y°x —20? —x+1=0 9 shai, hie inGQ) 
and the curves intersect in four points, as shown in fig. 21, 
When the four roots are all imaginary, the curves may be arranged?in a 
widely different way from that here shown, Thus, if the roots are 1+4,/—1, 
—1+8,/—1, so that 
a+ 78x? — 96x02 +1105=0 
the equations to the two curves may be put in the form 
y? =a? + 89+ /8x'+ 1562+ 9674416 . . . . (P) 
24 
y =e +39—— vase $a2 (Q) 
and the curves will lie as shown in fig. 22. They still intersect in four points, 
but the branches of the P curve now touch the asymptotes (1, 4), (2, 3), instead 
of (1, 2), (3, 4), as in fig. 21. It follows that there will be a transition position, 
in which the branches will touch the asymptotes (1, 3), (2, 4), and will cross, so 
that the curve P will have two double points. 
Lastly, take an equation with two real and two imaginary roots, say 
1+,/—1,2,—4, so that «*—1027+20%2—16=0. Then the equations to the 
two curves become 
yf =38e? —5 + J/8a*—2027—20a+41.. . . (P) 
5 
a) 
y =x? —5+— Sts» (OY, 
In this case the curves intersect in two points, as shown in fig. 23. Here the 
real part of the imaginary roots lies between the two real roots; let us there- 
fore take another instance in which this is not the case. Thus let the roots be 
2+,/—1, —1, —3, so that «*—8v’+8x%+15=0. Then the equations to the 
two curves are 
y= 30? —4& /8a'—160"—8e+1 . . (P) 
ypod—44— Fee Mees) 
VOL. XXX. PART II. 45 
