Region of the coefficients of a quadratic equation that cause the roots of it to be in the unit disk

Question

From Simon Haykin's Adaptive Filter Theory: consider the characteristic equation is $1+𝑎_1𝑧^{−1}+𝑎_2𝑧^{−2}=0$, then for the roots to be inside the unit circle (i.e. 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 other two line boundaries of the region can be derived. Can someone please explain how?

An approach for the case with complex roots is shown here: https://math.stackexchange.com/questions/1839654/show-the-roots-of-the-quadratic-equation-z2-bz-c-0-lie-in-or-on-the-unit

I've also asked this question here: https://math.stackexchange.com/questions/1839654/show-the-roots-of-the-quadratic-equation-z2-bz-c-0-lie-in-or-on-the-unit, although there is a typo in that posting in the third condition specifying the boundary of the region of the coefficients, and I don't have enough reputation there to fix it.

Answer 1

The roots of the quadratic equation are

$$z_{1,2}=-\frac{a_1}{2}\pm\sqrt{\frac{a_1^2}{4}-a_2}\tag{1}$$

For $a_1^2/4\ge a_2$, the roots are real-valued. In that case we require

$$-1<z_{1,2}<1\tag{2}$$

Let's start with the first inequality (from the left) in $(2)$:

$$-1+\frac{a_1}{2}<-\sqrt{\frac{a_1^2}{4}-a_2}\tag{3}$$

Note that the minus sign before the square root represents the more restrictive constraint.

Squaring $(3)$ (note that $<$ becomes $>$) and rearranging gives

$$a_1-a_2<1\tag{4}$$

In the same way you can obtain the other inequality by considering the second inequality in $(3)$ (with a positive sign before the square root):

$$a_1+a_2>-1\tag{5}$$

Answer 2

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 $|a_2| < 1$ 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 $|a_2| < 1$ 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 $$ 0< (r_1-1) \cdot (r_2 -1) = r_1 \cdot r_2 - r_1 - r_2 +1 = a_2+a_1 + 1$$ which gives you $$-1 < a_1 + a_2$$

Doing the same exercise for mixed signs gives you the other other inequality.