Consider polynomial below:
$P(s) = a_4s^4 + a_{3}s^{3} + a_2s^2 + a_1s + a_0 $
The question is: Under which conditions on coefficient of above polynomial, we claim that the zeros of polynomial constitute such a special distribution like this :
$ z_1 = α $
$z_2= -β $
$z_{3,4} = -ζ∓jγ$
where $ |z_1 |<|z_2 |$ and $ (α,β,γ,ζ>0)$ .
Any Ideas would be appreciated.
First note that $a_4 \ne 0$, otherwise there couldn't be four zeros. To simplify, divide the polynomial by $a_4$, so we can assume $a_4 = 1$, i.e. it's a monic polynomial.
The set of roots is symmetric under complex conjugation, so your monic polynomial is as well: the $a_i$ are real. Since $0$ is not a root, $a_0 \ne 0$.
Now you want one root on the positive real axis, one on the negative real axis, and the other two in the left half plane off the real axis. Let $p_0 = P$, $p_1 = p'$, $p_2$, $p_3$, $p_4$ be the Sturm sequence for $P$ (in degenerate cases there might be fewer than these). The number of roots in the positive real axis and in the negative real axis can be obtained by Sturm's theorem in terms of the number of of sign changes in the Sturm sequence at $x = +\infty$, at $x=0$ and as $x = -\infty$: there should be one more sign change at $-\infty$ than at $0$, and one more at $0$ than at $+\infty$.
For there to be three roots in the left half plane and one in the right half plane, take the generalized Sturm sequence using the Euclidean algorithm starting with $R_0(z) = -a_2 z^2 + a_0$ and $R_1(z) = -z^3 + a_1 z$: there should be two more sign changes at $+\infty$ than at $-\infty$.