Q.
1 Let P(z)=z3+azz2+bz+c where a,b and c are real. There exists a complex number ω such that the three roots of P(z) are ω+3i,ω+9i and 2ω−4 where i2=−1. Find the value of ∣a+b+c∣.
331
110
Complex Numbers and Quadratic Equations
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Answer: 136
Solution:
It is to be noted that two of the numbers need to be conjugates and one number must be real, as the coefficients of the cubic are all real.
Three roots are, ω+3i,ω+9i and 2ω−4
let ω+3i is real, hence ω=α−3i where α∈R
then (α−3i+9i) and (2α−6i−4)
i.e. α+6i and (2α−4)−6i must be complex conjugate ⇒α=2α−4 ∴α=4
Hence the roots are
4,4+6i, 4-6i (other options not possible)
the equation is (z−4)(z2−8z+5z)⇒z3−12z2+84z−208
Hence ∣a+b+c∣=∣−12+84−208∣=136.