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.


1 Answer 1


Your solution is mostly correct, but there are some errors.

First, the rational root theorem does not tell you that there are no real roots. It tells you that there are no rational roots. The fact that the polynomial has no real roots in $\Re(z) > 0$ is a consequence of the fact that $x^4+3x^2+x+1 > 0$ when $x > 0$.

You're correct that the coefficient of $z^3$ being zero tells us immediately that all of the roots may not lie in $\Re(z) > 0$. This means that there are either $0$ or $2$ roots in $\Re(z) > 0$.

For the same reason, they may not all lie in $\Re(z) < 0$. So, if there are $0$ roots in $\Re(z) > 0$, all $4$ roots must lie on the line $\Re(z) = 0$. Since the roots will come in conjugate pairs, the polynomial would then have the form

$$ (z^2+a^2)(z^2+b^2) $$

for some real $a$ and $b$. In this case the substitution $z = ix$ with $x$ real (i.e. a restriction to the imaginary axis) will yield another polynomial with real coefficients. Your polynomial does not have this property, and thus cannot have this factorization. We must conclude that there are exactly $2$ roots in $\Re(z) > 0$.

  • $\begingroup$ Antonio, fantastic, thanks. For the last part of the answer, can we also note that if the polynomial was of the form $(z^2+a^2)(z^2+b^2)$, then the coefficient of $z$ should be zero, contradiction? $\endgroup$
    – Conan Wong
    Jan 2, 2013 at 3:28
  • $\begingroup$ @ConanWong You're very welcome. And yes, that definitely works too. $\endgroup$ Jan 2, 2013 at 4:21

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