2

Let us consider $$\omega_4 z^4+\omega_3 z^3+\omega_1 z+\omega_0=0\quad (1)$$ where $\omega_0,\omega_4>0$, $\omega_1,\omega_3\in\mathbb R$ ($\omega_1,\omega_3$ both negative or both positive). Let us consider the following situations:

  1. All the roots of (1) are not-real.
  2. All the roots of (1) have negative real part.
  3. All the real roots of (1) have modulus $> 1$. For any complex number $z=x+iy$, where $x$ and $y$ are real numbers, the absolute value or modulus of $z$ is denoted $|z|$ and is defined by $|z|=\sqrt {x^{2}+y^{2}}$.

Is it possible to find conditions on $\omega_i$ ($i=0,1,3,4$) that satisfy one of the above points?

Mark
  • 7,841
  • 6
  • 38
  • 72

1 Answers1

3

For 1.

According to here, all the roots of $(1)$ are non-real iff

  • $\Delta\gt 0$ and $D\gt 0$

  • $\Delta\gt 0$ and $P\gt 0$

  • $\Delta=0$ and $D=0$ and $P\gt 0$ and $R=0$

where $$\begin{align}\Delta&=256\omega_4^{3}\omega_0^{3}-192\omega_4^{2}\omega_3\omega_1\omega_0^{2}-27\omega_4^{2}\omega_1^{4}-6\omega_4\omega_3^{2}\omega_1^{2}\omega_0-27\omega_3^{4}\omega_0^{2}-4\omega_3^{3}\omega_1^{3} \\\\D&=64\omega_4^{3}\omega_0-16\omega_4^{2}\omega_3\omega_1-3\omega_3^{4} \\\\P&=-3\omega_3^2 \\\\R&=\omega_3^3+8\omega_1\omega_4^2 \end{align}$$

Since $P=-3\omega_3^2\lt 0$, we see that all the roots of $(1)$ are non-real iff $$\Delta\gt 0\qquad\text{and}\qquad D\gt 0,$$ i.e. $$\color{red}{256\omega_4^{3}\omega_0^{3}-192\omega_4^{2}\omega_3\omega_1\omega_0^{2}-27\omega_4^{2}\omega_1^{4}-6\omega_4\omega_3^{2}\omega_1^{2}\omega_0-27\omega_3^{4}\omega_0^{2}-4\omega_3^{3}\omega_1^{3}\gt 0}$$ $$\color{red}{\text{and}\qquad 64\omega_4^{3}\omega_0-16\omega_4^{2}\omega_3\omega_1-3\omega_3^{4}\gt 0}$$

For 2.

Let $f(z)$ be LHS of $(1)$. Then, we have $$f'(z)=4\omega_4z^3+3\omega_3z^2+\omega_1$$ Since the discriminant $\Delta_3$ is negative$$\Delta_3=-108\omega_3^3\omega_1-432\omega_4^2\omega_1^2\lt 0$$we see that $f'(z)=0$ has only one real root.

It follows that the number of the real roots of the quartic equation $(1)$ is at most two.

Suppose that all the roots of $(1)$ have negative real part.

Let us separate it into cases :

Case 1 : The roots are $a,b,c+di,c-di$ where $a,b,c,d\in\mathbb R$ with $a\lt 0,b\lt 0,c\lt 0,d\gt 0$. This case includes the case where $a=b$.

Since the coefficient of $z^2$ is $0$, we get, by Vieta's formulas, $$\frac{0}{\omega_4}=ab+2ac+2bc+c^2+d^2$$ The LHS equals $0$ while the RHS is positive. This is impossible.

Case 2 : The roots are $a+bi,a-bi,c+di,c-di$ where $a,b,c,d\in\mathbb R$ with $a\lt 0,b\gt 0,c\lt 0,d\gt 0$.

Since the coefficient of $z^2$ is $0$, we get, by Vieta's formulas, $$\frac{0}{\omega_4}=a^2+b^2+4ac+c^2+d^2$$ The LHS equals $0$ while the RHS is positive. This is impossible.

Therefore, it is impossible that all the roots of $(1)$ have negative real part.

For 3.

From 2., we already know that $f'(z)=0$ has only one real root.

Also, we have$$f''(z)=0\iff z=-\frac{\omega_3}{2\omega_4},\ 0$$

  • Case 1 : $\omega_1\gt 0$ and $\omega_3\gt 0$.
    Since $f'(0)=\omega_1\gt 0$, we see that $f'(z)=0$ has only one real root $z=\alpha$ where $\alpha\lt 0$.
    It is necessary that $$f(-1)=\omega_4-\omega_3-\omega_1+\omega_0\gt 0$$ and that $f(z)=0$ has at least one real root, i.e. $\Delta\le 0$ or $D\le 0$ from 1..
    Since we have $$-\frac{\omega_3}{2\omega_4}\lt 0\qquad\text{and}\qquad f'(0)=\omega_1\gt 0$$ considering the graphs, we see that it is necessary that $f'(-1)=-4\omega_4+3\omega_3+\omega_1\gt 0$.

    $\qquad$enter image description here

    On the other hand, if $$f(-1)\gt 0,\quad f'(-1)\gt 0\quad\text{and}\quad (\Delta\le 0\ \ \text{or}\ \ D\le 0)$$then, we see that all the real roots of $(1)$ have modulus $> 1$.

  • Case 2 : $\omega_1\lt 0$ and $\omega_3\lt 0$
    Since $f'(0)=\omega_1\lt 0$, we see that $f'(z)=0$ has only one real root $z=\beta$ where $\beta\gt 0$.
    It is necessary that $$f(1)=\omega_4+\omega_3+\omega_1+\omega_0\gt 0$$ and that $f(z)=0$ has at least one real root, i.e. $\Delta\le 0$ or $D\le 0$ from 1..
    Since we have $$0\lt -\frac{\omega_3}{2\omega_4}\qquad\text{and}\qquad f'(0)=\omega_1\lt 0$$ considering the graphs, we see that it is necessary that $f'(1)=4\omega_4+3\omega_3+\omega_1\lt 0$.

    $\qquad$enter image description here

    On the other hand, if $$f(1)\gt 0,\quad f'(1)\lt 0\quad\text{and}\quad (\Delta\le 0\ \ \text{or}\ \ D\le 0)$$then, we see that all the real roots of $(1)$ have modulus $> 1$.

From the two cases, all the real roots of $(1)$ have modulus $> 1$ iff

$$\small\color{red}{(\omega_1\gt 0,\quad\omega_3\gt 0,\quad\omega_4-\omega_3-\omega_1+\omega_0\gt 0,\quad -4\omega_4+3\omega_3+\omega_1\gt 0,\quad (\Delta\le 0\ \ \text{or}\ \ D\le 0))}$$ $$\color{red}{\text{or}}$$ $$\small\color{red}{(\omega_1\lt 0,\quad\omega_3\lt 0,\quad\omega_4+\omega_3+\omega_1+\omega_0\gt 0,\quad 4\omega_4+3\omega_3+\omega_1\lt 0,\quad (\Delta\le 0\ \ \text{or}\ \ D\le 0))}$$

mathlove
  • 139,939