Given the polynomial $x^5 + a^2 x^4 + 1 = 0$, find the set of values of $a$ such that $2$ of the roots have negative imaginary part.

Question

I am given the equation:

$$x^5 + a^2 x^4 + 1 = 0$$

with $a \in \mathbb{R}$. I have to find the set of values of $a$ such that $2$ of the roots of the equation have negative imaginary part.

If we use the notation:

$$f(x) = x^5 + a^2 x^4 + 1$$

I know that since $a \in \mathbb{R}$ we have that $f \in \mathbb{R}[X]$ and so, if a complex number $z \in \mathbb{C}$ is a root of the equation, then its conjugate, $\overline{z}$ is also a root of the equation. So the statement of the problem is equivalent to asking for the equation to have $4$ complex roots in total, since we would have $2$ with negative imaginary part and another $2$ complex roots (the conjugates of the first $2$) with positive imaginary part. I think my reasoning is correct, however I don't know how to use this knowledge to find the set of values of $a$ such that the requirements of the problem are satisfied.

Answer 1

You have made a good start. Now you can note that having four complex roots means there is only one real root and the fact that the polynomial is of odd degree guarantees that there is at least one. You can use Decartes' rule of signs to show there is one negative root and no positive root for any value of $a$.

Answer 2

As you said, we want the equation to have only $1$ real root. When $a=0$, we obtain $x^5+1=0$ which clearly satisfies the requirements. When $a\ne 0$, the derivative of $f$ is $$f'(x)=5x^4+4a^2x^3=x^3\left(5x+4a^2\right) $$ Examining its sign, we see that $f$ is strictly increasing in the intervals $\left(-\infty,-4a^2/5\right]$ and $[0,\infty)$, and strictly decreasing in $\left[-4a^2/5,0\right]$. Thus $$f\left(-\frac{4a^2}{5}\right)>f(0)=1>0 $$ Also, $\lim_{x\to-\infty}f(x)=-\infty$. It follows that $f$ has exactly $1$ real zero in the interval $(-\infty,-4a^2/5)$ and no real zeroes in $[-4a^2/5,\infty)$ where $f$ is positive. Hence, all $a\ne 0$ are solutions as well.

Answer 3

At a point where the number of real roots changes, the discriminant would be $0$. The discriminant of your polynomial is $256 a^{10}+3125$, which is always strictly positive for real $a$, so the number of real roots is constant. And when $a=0$, this number is obviously $1$.

Answer 4

$x^5+a^2·x^4+1=0$

$x^4·(x+a^2)+1=0$

$f(x)=x^4$

$g(x)=\frac{-1}{x+a^2}$

the graph of the function g(x) has the vertical asymptote:

$x=-a^2$

looking at the graph, we can see that the graphs f (x) and g (x) intersect at point P to the left of the vertical asymptote and near to the asymptote.

...so the real root is of the form:

$x≈-(a^2+ε)$