How many roots does the polynomial $z^4 + 3z^2 + z + 1$ have in the right-half complex plane (i.e. $Re(z) \gt 0$)?
I honestly can't think of how to approach the problem as it seems different from the regular Rouche's Theorem problems.
I can only say that the answer is either 0, 2 or 4 as all the roots come in complex conjugate pairs. (By the Rational Roots Theorem tested on +1 and -1, the polynomial has no real roots.)
Attempt at Solution [1 hour after posting question]
After pondering on this question a bit, I wonder if the following argument will work:
[The Rational Roots Theorem bit above shows that the number of roots in the right-half plane is either 0, 2 or 4.]
The coefficient of $z^3$ in the polynomial is 0, indicating that the sum of the roots of the polynomial is 0.
If all 4 roots were in the right-half complex plane, or left-half complex plane, then this coefficient would not be 0.
Thus, the polynomial has 2 roots in the right-half complex plane.
Could someone comment/help to verify this please? Thanks.