So I know about this question and I've seen papers like this and this. But the former isn't exactly what I want and the latter two papers are too deep and I'm lazy and I wanna quick-and-easy answer and I'm sure you legitimate mathematicians are happy to supply this lazy electrical engineer with a quick-and-easy answer, right?
So, in general, an $N$th-order polynomial with real coefficients is:
$$ f(z) = \sum\limits_{n=0}^{N} a_n \, z^n \qquad \qquad \text{where }N\in\mathbb{Z}\ge0,\ a_n\in\mathbb{R},\ z\in\mathbb{C}$$
Now, consider this simple quadratic:
$$\begin{align} w^2 + bw + c &= w^2 (1 + bw^{-1} + cw^{-2})\\ &= (w-r_1)(w-r_2) \\ \end{align}$$
again $b,c\in\mathbb{R}$ and $w,r_1,r_2\in\mathbb{C}$ . We know that
$$\begin{align} r_1 &= -\tfrac{b}{2} + \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \\ r_2 &= -\tfrac{b}{2} - \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \\ \end{align}$$
if $b^2 \ge 4c$ and
$$\begin{align} r_1 &= -\tfrac{b}{2} + i\sqrt{c - \left(\tfrac{b}{2}\right)^2} \\ r_2 &= -\tfrac{b}{2} - i\sqrt{c - \left(\tfrac{b}{2}\right)^2} \\ \end{align}$$
if $b^2 < 4c$
In the latter complex-conjugate case, we know that
$$ |r_1| = |r_2| = c $$
which is both simple and handy.
Now, whether the roots are real or complex-conjugate, suppose we have chosen $b$ and $c$ so that both
$$\begin{align} |r_1| &< 1 \\ |r_2| &< 1 \\ \end{align}$$
Fine, now let's return to the sparse polynomial $f(z)$. Suppose $N$ is even and all coefficients $a_n$ are zero except:
$$\begin{align} a_0 &= 1 \\ a_{N/2} &= b \\ a_N &= c \\ \end{align}$$
What are the roots of $f(z)$? Suppose $N$ is pretty big (and even), say $N\approx 1000$. If I can guarantee that $b$ and $c$ are chosen to insure that $|r_1|$ and $|r_2|$ are less than $1$, can I rely on all $N$ roots of $f(z)$ also being less than $1$?
I think that I can. I've been fiddling with the substitution of
$$ w = z^{N/2} $$
and I know about the Nth-roots-of-unity, but am I guaranteed that all of the roots of the $N$th-order polynomial have magnitude less than $1$?
So you have the equation: $$cz^{2n}+bz^n+1 = 0$$, where it's guaranteed that $z^n_{1,2}$ have modulus less than $1?$ If this is the case, then your intuition is correct because if: $$|z^n| = |z|^n = |\omega| <1,$$ it must necessarily follow that: $$|z|<1$$ since $|z|$ is a positive real number.