From Simon Haykin's Adaptive Filter Theory: consider the characteristic equation is $1+a_{1}z^{-1}+a_{2}z^{-2}=0$, then for the roots to be inside the unit circle (or in the unit disk), the coefficients of the quadratic must satisfy the following conditions: $-1\le a_{1}+a_{2}$, $-1\le a_{2}-a_{1}$, and $-1\le a_{2}\le 1$.
I could figure this out some of the way. For complex roots it follows that the coefficients lie in the region defined by the intersection of $4a_2>a_{1}^2$ and $a_2<1$. For real roots, I could prove that $|a_2|<1$, and obviously $4a_2\le a_{1}^2$. But I cannot figure out how the line boundaries of the region can be derived. Can someone please explain how?
From the quadratic equation
$$z= \dfrac{-a_1}{2} \pm \sqrt{\left(\dfrac{a_1}{2}\right)^2 - a_2} $$
For the greater real root and the upper boundary on the real line
$$\begin{align*} 1 &\ge \dfrac{-a_1}{2} + \sqrt{\left(\dfrac{a_1}{2}\right)^2 - a_2}\\ \\ 1 + \dfrac{a_1}{2} &\ge + \sqrt{\left(\dfrac{a_1}{2}\right)^2 - a_2}\\ \\ 1 + a_1 + \left(\dfrac{a_1}{2}\right)^2 &\ge \left(\dfrac{a_1}{2}\right)^2 - a_2\\ \\ a_1 + a_2 &\ge -1 \end{align*}$$
For the lesser real root and the lower boundary on the real line
$$\begin{align*} -1 &\le \dfrac{-a_1}{2} - \sqrt{\left(\dfrac{a_1}{2}\right)^2 - a_2}\\ \\ -1 + \dfrac{a_1}{2} &\le - \sqrt{\left(\dfrac{a_1}{2}\right)^2 - a_2}\\ \\ -1 + a_1 + \left(\dfrac{a_1}{2}\right)^2 &\le \left(\dfrac{a_1}{2}\right)^2 - a_2\\ \\ -1 &\le a_2 -a_1 \\ \end{align*}$$
For the complex roots being inside the unit circle
$$\begin{align*} 1 &\ge \left|\dfrac{-a_1}{2} \pm i\sqrt{a_2 - \left(\dfrac{a_1}{2}\right)^2}\right|\\ \\ 1 &\ge \sqrt{\left(\dfrac{a_1}{2}\right)^2 + a_2 - \left(\dfrac{a_1}{2}\right)^2} \\ \\ 1 &\ge \sqrt{a_2}\\ \\ 1^2 &\ge \left(\sqrt{a_2}\right)^2\\ \\ 1 &\ge |a_2|\\ \\ -1 &\le a_2 \le 1 \\ \end{align*}$$