If the roots of $z^4+az^3+bz^2+z$ are distinct and concyclic in the complex plane, does $ab\in\mathbb R$ imply $1<ab<9$?

Question

HMMT February 2022, Team Round, Problem 6 (proposed by Akash Das) is:

Let $\operatorname{\it P\!}{\left(x\right)}=x^4+ax^3+bx^2+x$ be a polynomial with four distinct roots that lie on a circle in the complex plane. Prove that $ab\neq9$.

Its solution may be found here.
Recently, certain netizens claimed that there exists a stronger result when $ab$ is real:

Let $P(z)=z^4+az^3+bz^2+z$ be a polynomial with four distinct roots that are concyclic in the complex plane; then $ab\in\mathbb R\implies1<ab<9$.

Nevertheless, I could find neither any proof nor any counterexample of it.
Does the above proposition hold? (And what if $ab\in{\mathbb C\setminus\mathbb R}$?)

Edit. It appears that when $ab\in{\mathbb C\setminus\mathbb R}$, $ab$ can take any values in the complex plane other than the real axis. Here is a simulation:

Answer 1

As said in the comments, the problem is equivalent to showing that if $ a b \in \mathbb{R}$ and the polynomial

\begin{equation}P \left(z\right) = {z}^{3}+b {z}^{2}+a z+1\end{equation}

has three colinear distinct roots on a line that does not contain zero, then $1 < a b < 9$. We prove here that this result is true in the general case, when $ a$ and $ b$ are complex numbers. Let us define

\begin{equation}a b = {m}^{3} , \qquad b = m c , \quad a = \frac{{m}^{2}}{c} , \quad m \in {\mathbb{R}}^{\ast } , \quad c \in {\mathbb{C}}^{\ast }\end{equation}

hence

\begin{equation}P \left(z\right) = {z}^{3}+m c {z}^{2}+\frac{{m}^{2}}{c} z+1\end{equation}

Let $ \frac{c}{w} \in {\mathbb{C}}^{\ast }$ a direction vector of the line $ L$ containing the roots. This line contains the average of the three roots, which is $-\frac{m c}{3}$, hence if $ z$ is a root,

\begin{equation}z = m c \left(\frac{t}{w}-\frac{1}{3}\right) , \qquad t \in \mathbb{R}\end{equation}

Substituting into $ P$ gives

\begin{equation}P \left(z\right) = {t}^{3} \frac{{m}^{3} {c}^{3}}{{w}^{3}}+t \left(1-\frac{{c}^{3}}{3}\right) \frac{{m}^{3}}{w}+\frac{2 {m}^{3} {c}^{3}}{27}-\frac{{m}^{3}}{3}+1\end{equation}

By multiplying by $ \frac{{w}^{3}}{{m}^{3} {c}^{3}}$, one finds a new depressed cubic polynomial which must now have three real roots

\begin{equation}Q \left(t\right) = {t}^{3}+p t+q = {t}^{3}+t \left(\frac{1}{{c}^{3}}-\frac{1}{3}\right) {w}^{2}+{w}^{3} \left(\frac{2}{27}+\frac{1}{{c}^{3}} \left(\frac{1}{a b}-\frac{1}{3}\right)\right)\end{equation}

Obviously the conditions are now $ \left(p , q\right) \in {\mathbb{R}}^{2}$ and $ 27 {q}^{2}+4 {p}^{3} < 0$. In particular, it is necessary that $ p < 0$. As $ w$ is determined up to an arbitrary non-zero factor, it can be scaled so that $ p =-1$. It follows that

\begin{equation}{c}^{3} = \frac{1}{\frac{1}{3}-\frac{1}{{w}^{2}}}\end{equation}

then

\begin{equation}q = \left(\frac{2}{27 \left(\frac{1}{3}-\frac{1}{{w}^{2}}\right)}-\frac{1}{3}+\frac{1}{a b}\right) \left(\frac{1}{3}-\frac{1}{{w}^{2}}\right) {w}^{3} \in \mathbb{R}\end{equation}

Let us now define the complex numbers

\begin{equation}{\mu} = {\xi}+i {\zeta} = \frac{2}{27 \left(\frac{1}{3}-\frac{1}{{w}^{2}}\right)}-\frac{1}{3} , \qquad {\nu} = {{\xi}'}+i {{\zeta}'} = \left(\frac{1}{3}-\frac{1}{{w}^{2}}\right)^{-1}{w}^{-3}\end{equation}

One has

\begin{equation}{\xi}+\frac{1}{a b}+i {\zeta} = q {{\xi}'}+i q {{\zeta}'} \qquad \text{hence} \qquad q = \frac{{\zeta}}{{{\zeta}'}} \quad \text{and} \quad \boxed{\frac{1}{a b} = {{\xi}'} \frac{{\zeta}}{{{\zeta}'}}-{\xi}}\end{equation}

The cubic equation has three real roots if and only if $ 27 {q}^{2} < 4$ that is to say

\begin{equation}\boxed{27 {{\zeta}}^{2} < 4 {{{\zeta}'}}^{2}}\end{equation}

We now use Sympy to perform symbolic calculations. Let

\begin{equation}w = x+i y\end{equation}

According to Sympy, the above three real roots condition reduces to

\begin{equation}27 {{\zeta}}^{2} - 4 {{{\zeta}'}}^{2} = \frac{12 y^2 (4 {x}^{2}-3)}{|w|^4 |w^2 -3|^2} < 0\end{equation}

which means that $\Im(w)\not = 0$ and $ \Re{\left(w\right)} \in \left]{-\frac{\sqrt{3}}{2}} , \frac{\sqrt{3}}{2}\right[$. The computation gives

\begin{equation}a b = 9 \frac{3+{y}^{2}-3 {x}^{2}}{9+{y}^{2}-3 {x}^{2}}\end{equation}

As $ {y}^{2}-3 {x}^{2}$ can take any value in $ \left]{-\frac{9}{4}} , \infty \right[$, it follows that

\begin{equation}1 = 9 \frac{3-\frac{9}{4}}{9-\frac{9}{4}} < a b < 9 \frac{\frac{3}{\infty }+1}{\frac{9}{\infty }+1} = 9\end{equation}

All the values in this interval can be reached.

Let us conclude this with the Python code that performs the symbolic calculations

from sympy import re, im, I, E, symbols, ratsimp, factor, expand_complex

x, y = symbols("x y", real=True)
U = -I * I


def abs2(z):
    return expand_complex(z * z.conjugate()).simplify()


w = x * U + y * I

d = U / 3 - U / w**2

W = abs2(w)
T = abs2(w**2 - 3)
print(f"W = |w|**2 = {W}")
print(f"T = |w**2 - 3|**2 = {T}")

mu = expand_complex(2 / (27 * d) - U / 3).simplify()
nu = expand_complex(1 / (d * w**3)).simplify()
print(f"mu: {mu}")
print(f"nu: {nu}")

xi, zeta = mu.as_real_imag()
xip, zetap = nu.as_real_imag()
print(f"T * xi = {factor((xi*T).simplify())}")
print(f"T * zeta = {factor((zeta*T).simplify())}")
print(f"T * W * xip = {factor((xip*T*W).simplify())}")
print(f"T * W * zetap = {factor((zetap*T*W).simplify())}")

crit = factor(((27 * zeta**2 - 4 * zetap**2) * T * W**2).simplify())
print(f"criterion: {crit} < 0")

ab = 1 / ratsimp(xip * zeta / zetap - xi).simplify()
print(f"ab: {ab}")

The output is

W = |w|**2 = x**2 + y**2
T = |w**2 - 3|**2 = x**4 + 2*x**2*y**2 - 6*x**2 + y**4 + 6*y**2 + 9
mu: (-x**2 - 2*I*x*y + y**2 + 9)/(9*(x**2 + 2*I*x*y - y**2 - 3))
nu: 3/(x**3 + 3*I*x**2*y - 3*x*y**2 - 3*x - I*y**3 - 3*I*y)
T * xi = -(x**4 + 2*x**2*y**2 - 12*x**2 + y**4 + 12*y**2 + 27)/9
T * zeta = -4*x*y/3
T * W * xip = 3*x*(x**2 - 3*y**2 - 3)
T * W * zetap = -3*y*(3*x**2 - y**2 - 3)
criterion: 12*y**2*(4*x**2 - 3) < 0
ab: 9*(-3*x**2 + y**2 + 3)/(-3*x**2 + y**2 + 9)

Answer 2

Suppose $a,b\in\mathbb{R}$. Let $$ f(x)=x^3+ax^2+bx+1 $$ have three distinct roots $x_1,x_2,x_3$. Since $f(x)$ is a real polynomial, it has at least a real root $x_1$. Since $0,x_1,x_2,x_3$ are concyclic, $x_2,x_3$ must be a pair of purely imaginary roots and hence the circle has the center $\frac12x_1$ with radius $\frac12|x_1|$. Let $x_{2,3}=r\pm si$ ($r,s\in\mathbb{R},s>0$). By Vieta's Theorem $$ x_1+x_2+x_3=-a,x_1x_2+x_2x_3+x_1x_3=b,x_1x_2x_3=-1$$ which give $$ x_1+2r=-a,2rx_1+r^2+s^2=b,,x_1(r^2+s^2)=-1. \tag{1}$$ Also $|x_2-\frac12x_1|=|\frac12x_1|$ gives $$ r^2+s^2=rx_1.\tag{2}$$ From (1) and (2), one has $$ x_1=-\frac{1}{\sqrt {-r}}, a=\frac1{\sqrt{-r}}-2r,b=3\sqrt{-r},s=\sqrt{\sqrt{-r}-r^2}. $$ From this we can see $x_1<0,r<0$. Since $s>0$, one must have $r\in(-1,0)$. So $$ ab=3+6(-r)^{3/2}.$$ Thus $$ 3<ab<9. $$