Suppose a fixed $n^{\text{th}}$ degree monic polynomial is given
$$ p(x) = x^{n} + a_{n-1} x^{n-1} + \dots + a_0,$$
with coefficients vector $a = (a_{n-1}, \dots , a_0) \in \mathbb R^n$. Now we consider a parameterized family
\begin{align*}
p(x, r) = x^{n} + ra_{n-1} x^{n-1} + \dots + ra_0,
\end{align*}
with $r \in \mathbb R$.
Suppose for some $r_0 \in \mathbb R$, we have $p(i \alpha, r_0) = 0$, i.e., $p(x, r_0)$ has a zero on the imaginary axis. Let us further assume for some $\delta > 0$, we have $p(x, r)$ has all its zeros on the right half plane of $\mathbb C$ for each $r \in (r_0, r_0 + \delta)$.
My question is: is it possible that for all $r \in (r_0 – \delta, r_0)$ we also have $p(x, r)$ has all its zeros on the right half plane (we may include the imaginary axis). That is, is there a monic polynomial, that allows us to parametrize as above, such that the parametrized family has zeros touching the imaginary axis but all the zeros are confined in the right half plane when we vary $r$ continuously?
As commented by @saulspatz, $x^2 + r$ will have all its zeros on the imaginary axis if $r < 0$. I was in mind asking the case that at least there exists some $r \in (r_0 – \delta, r_0)$ such that some of the roots of $p(x, r)$ will move off the imaginary axis.
Best Answer
I see it is essentially the same question as another one of yours: Can we have a simple root of real polynomials with coefficients as linear functions such that the curve of the root is tangent to imaginary axis?. For somebody who is curious about the result, I leave the link.