On the limit set of eigenvalues of banded Toeplitz Hessenberg matrices Let $T_{n}(b)$ be the $n\times n$ Toeplitz matrix determined by the symbol
$$
 b(z)=\frac{1}{z}+\sum_{j=0}^{k}a_{j}z^{j}
$$
where $k\in\mathbb{N}$ and $a_{0},\dots,a_{k}\in\mathbb{R}$, $a_{k}\neq0$. That means $T_{n}(b)$ is the $n\times n$ left-top truncation of the banded Toeplitz Hessenberg matrix
$$    T(b)=\begin{pmatrix}
 a_{0} & 1 & 0 & 0 &  \\
 a_{1} & a_{0} & 1 & 0 & \ddots \\
 a_{2} & a_{1} & a_{0} & 1 & \ddots \\
    a_{3} & a_{2} & a_{1} & a_{0} & 1 & \ddots \\
  & \ddots & \ddots & \ddots & \ddots\\
 \vdots &  & \\
 a_{k} \\
  & \ddots \\
      \end{pmatrix}\!.$$
Further, for $\lambda\in\mathbb{C}$, let $z_{0}(\lambda),\dots,z_{k}(\lambda)$ denote the roots of the equation $b(z)=\lambda$ (repeated according to their multiplicity) arranged such that
$$|z_{0}(\lambda)|\leq|z_{1}(\lambda)|\leq\dots|z_{k}(\lambda)|.$$
Schmidt and Spitzer (1960) showed that the set
$$\Lambda(b):=\{\lambda\in\mathbb{C} \mid |z_{0}(\lambda)|=|z_{1}(\lambda)|\}$$
coincides with the set of limit points of eigenvalues of matrices $T_{n}(b)$, as $n\to\infty$.
There is a strong numerical evidence (I add 3 pictures approx. $\Lambda(b)$ below) that the set $\mathbb{C}\setminus\Lambda(b)$ is connected though all my attempts to prove this assertion failed. So my question is:
$$\textbf{Is the set } \mathbb{C}\setminus\Lambda(b) \textbf{ connected?}$$
Relevant remarks:


*

*If $b$ is a general Laurent polynomial in $z$ (the Schmidt and Spitzer Theorem holds as well), the set $\mathbb{C}\setminus\Lambda(b)$ need not be connected as shown, e.g., in Proposition 5.2. in Böttcher, Grudsky, LAA, 2002. So the particular form of $b$ (or the Hessenberg form of $T(b)$), which we assume, is essential.

*If $a_{1}=\dots=a_{k-1}=0$, the set $\mathbb{C}\setminus\Lambda(b)$ is know to be connected.

*The set $\Lambda(b)$ is always connected, Ullman, Bull. AMS, 1967.



 A: This is more of a longer comment.
Note that your question does not really have anything to do with Toepliz matrices, even thought it is a nice application. In fact, it can be rephrased as a question about boundedness of linear recurrences.
What you are asking is the following:
Let $\lambda \in \mathbb{C}$ be the set of numbers such that the the two smallest roots (magnitude) of $P(z)=\lambda z$ agree in magnitude, where $P$ is a polynomial. Is the complement of this set connected?
Can this question be generalized, perhaps?
Is the corresponding question for $P(z) = \lambda z^k$, $P(z) = \lambda Q(z)$ or perhaps even a polynomial in two variables, $P(z,\lambda)=0$.
The latter seems too general to be true though.
Some more computer experiments are welcome!
Also, there is a higher-dimensional analogue of Schmitd and Spitzers statement, and perhaps the corresponding statement holds there?
A: This is not a complete answer to the question. However, the following example indicates that the curve $\Lambda(b)$ actually can separate the plane $\mathbb{C}$. However, this is just a numerically computed plot and I have no rigorous proof showing this to be a counterexample, indeed.
For the symbol
$$b(z)=1/z+z+2 z^2+3 z^3+4 z^4+5 z^5+6 z^6+7 z^7+8 z^8+9 z^9+10 z^{10}$$
the set $\Lambda(b)$ is as follows:

