Region of the coefficients of a quadratic equation that cause the roots of it to be in the unit disk
It may be helpful to look at how the roots and the coefficients are connected to each other. Let's start with a complex root. We can write this as
$$ H(z) = (1-r/z) \cdot (1-r'/z) = 1 - (r+r')/z +r\cdot r'/z^2$$ $$ = 1 -2 \cdot Re\{r\}\cdot z^{-1}+ |r|^2 \cdot z^{-2}$$
We can see that $a1$ is simply twice the negative real part of the root and that $a2$ is the magnitude squared of the root. In this case the condition $|a_2| &lt; 1$ is sufficient and $a_1 is automatically taken care off by the constraint that the root is complex.
Let's do the same for a pair of real roots, $r_1$ and $r_2$
$$H(z) = (1-r_1/z) \cdot (1-r_2/z) = 1-(r_1+r_2) \cdot z{^-1} + r_1 \cdot r_2 \cdot z^{-2} $$
So here $a1$ is the negative sum of both roots and $a_2$ is the product of the roots. Again, we get $|a_2| &lt; 1$ since the product of two numbers smaller than one cannot be larger than one.  If both roots have the same sign, we can write the constraint as
$$ 0&lt; (r_1-1) \cdot (r_2 -1) = r_1 \cdot r_2 - r_1 - r_2 +1 = a_2+a_1 + 1$$
which gives you
$$-1 &lt; a_1 + a_2$$
Doing the same exercise for mixed signs gives you the other other inequality.