Question

1 Let $P ( z )= z ^3+ az z ^2+ bz + c$ where $a , b$ and $c$ are real. There exists a complex number $\omega$ such that the three roots of $P(z)$ are $\omega+3 i, \omega+9 i$ and $2 \omega-4$ where $i^2=-1$. Find the value of $\mid a+b+$ $c \mid$.

Answer 1

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, $\omega+3 i , \omega+9 i$ and $2 \omega-4$
let $ \omega+3 i$ is real, hence $\omega=\alpha-3 i$ where $\alpha \in R$
then $(\alpha-3 i+9 i)$ and $(2 \alpha-6 i-4)$
i.e. $ \alpha+6 i$ and $(2 \alpha-4)-6 i$ must be complex conjugate $\Rightarrow \alpha=2 \alpha-4$
$\therefore \alpha=4$
Hence the roots are
4,4+6i, 4-6i (other options not possible)
the equation is
$(z-4)\left(z^2-8 z+5 z\right) \Rightarrow z^3-12 z^2+84 z-208$
Hence $| a + b + c |=|-12+84-208|=136$.