polynomial such that $\sum |b_i| < 2$ and $(x^2-1)^m | P(x)$. Letting $P(x) = Q(x) (x^2-1)^m$, we have for $\lambda_i$ sufficiently close to $1$ that $$\begin{aligned}
P^*(x) &=& Q(x) (x^2-1/\lambda_1^2) (x^2-1/\lambda_2^2) \cdots
              (x^2-1/\lambda_m^2) \\
       &=& x^{m n + 2} + b^*_{nm+1} x^{nm+1} + \cdots + b_0^*\end{aligned}$$ will also have $\sum |b_i^*| < 2$.

It seems highly likely that the same would be true for the case where $M$ contains non-trivial Jordan blocks, although the analysis becomes much messier.

The mixed real case
-------------------

Here we apply Theorem \[thm:tool2\] with roots $-\lambda$ and $\mu$ and $k_1 = k_2 = 1$. The polynomial we use is ($N=n=2$): $$P(x) = x^2 + \left(\frac{1}{{\lambda}}-\frac{1}{\mu}\right) x - \frac{1}{\mu{\lambda}}.$$

Observe that $P(-1/{\lambda}) = P(1/\mu) = 0$. The matrix $B$ in this case is $$B =
\begin{pmatrix}
 B_0(-{\lambda}) & B_1(-{\lambda})  \\
 B_0(\mu)  & B_1(\mu)
\end{pmatrix}
=
\begin{pmatrix}
- \frac{1}{{\lambda}\mu} & \frac{1}{{\lambda}} \\
- \frac{1}{{\lambda}\mu} & -\frac{1}{\mu}
\end{pmatrix}.$$ We see that this has determinant $\frac{{\lambda}+\mu}{{\lambda}^2 \mu^2} \neq 0$, as we are assuming both ${\lambda}, \mu > 0$. Since $$\left|\frac1{{\lambda}}-\frac1{\mu}\right|+\frac1{|\mu{\lambda}|}\le2, \quad \frac1{\sqrt2}\le {\lambda},\mu\le1,$$ we infer

For all $\frac1{\sqrt2} \leq {\lambda}, \mu \leq 1$ we have that $(0,0)$ lies in the interior of $A_{-{\lambda}, \mu}$.

The above gives us a 