33So sum of roots must be zero.
a1=0 but it given they are not zero [7]
if the equation z4+a1z3+a2z2+a4=0,wherea1,a2,a3,a4 r real coeffiients diff. from zero has pure imaginary root then the expression a3/a1a2 + a1a4/a2a3 ha d value
1. 0
2. 1
3. -2
4. 2
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10 Answers
33
62two roots are ik, -ik
k4-a1ik3-a2k2+ia3k+a4 = 0
comparing real and imaginary part,
k4-a2k2+a4 = 0
a1k3-a3k = 0
k2 = a3/a1
(a3/a1)4-a2(a3/a1)2+a4 = 0
let the roots be x,y , ik, -ik
sum of roots is x+y = - a1
product of roots is xyk2 = a4
sum of roots taken two together is xik-xik+yik-yik+xy-k2 = xy-k2 = a2
now i think this one is solved :)
can you get the answer ?
62priyam, i think it says that a couple of roots are "purely" imaginary..
not that all four roots are purely imaginary..
62not just that.. two roots can still be complex but not purely imaginary... :)