Determine if an equation's roots are inside the unit circle

Question

I have a polynomial such as,

$z^2 - 1.5z + 0.9 = 0$

I need to know if the roots are inside the unit circle and if the transfer function is stable if the equation is the denominator. But I am not exactly sure of how to proceed and I found a negative $\Delta$ which would mean the equation has no solution. So I am a bit lost here.

I would appreciate any help, thanks.

Answer 1

Let you are given a quadratic equation that may have complex roots

$$\begin{align*} az^2+bz+c = 0 \end{align*}$$

$$\implies z = \frac{-b±\sqrt{b^2-4ac}}{2a}$$

Let $\sqrt{b^2-4ac} = \Delta$

Thus $$z_1 = \frac{-b+\Delta}{2a}$$, $$z_2 = \frac{-b-\Delta}{2a}$$

Mod(Z): $||Z = a+ ib|| = \sqrt{a^2+b^2}$

Now,

If $||z||< 1$: 
    Then your roots are inside of the unit circle
else: 
    Outside or on the unit circle.

For the above problem: $z^2-1.5z-0.9 = 0$

$z_1 = 0.75-0.5809i$ which is complex $\implies z_2 = 0.75+0.5809i$ will be the conjugate

$||z_1|| = \sqrt{0.75^2+0.5809475^2} = \sqrt{0.9}<1$ Same for the other root

However, If you find this for quadratic equation with real coefficients if either of one roots is complex other will be its conjugate. Thus if either of the roots is within the unit circle other will also be inside of the unit circle and vice-versa

Answer 2

This gets you part of the answers you are looking for. Multiple people have commented on finding the roots and if they are in the unit circle, so I didn't go into that any further.

I'm pretty sure this only applies to a linear system. <- don't quote me on that!

Your formula has a constant. If you did not, you would need to factor it to get a constant. (For example, $z^3 + z^2 + z$ versus $z(z^2 + z + 1)$ - the latter has a constant of one. You would need to use the equation inside the parentheses in the latter equation for this to work.)

How many roots are there?

What's the value of your polynomial exponent?

• This tells you how many roots, positive, negative, real, or imaginary, that exist.

How many positive roots are there?

Is the largest polynomial's exponent odd or even?

• This tells you that you either have two or zero or positive roots.

How many times does the sign change in the quadratic equation when placed in order by highest order polynomial?

• Positive -> negative -> positive = 2 changes
• There are 2 positive roots, if they are not complex roots.

How many negative roots are there?

Flip the sign of z, $-z^2 + 1.5z + .9$ -- keep in mind that if the exponent is even, the result is even, regardless of the sign, and fix the equation knowing that a negative times a negative is always positive - so now your equation looks like this: $z^2 + 1.5z + .9$. Now count the sign changes.

• There are no sign changes. Therefore there are no negative roots. If there were any sign changes, the quantity would equate to the number of negative roots, if they are not complex roots.

What does all this have to do with the transfer function? What's the point?

At this point, you know there are no negative poles. The system is unstable. If the poles are all positive, that infers that the polynomials are negative.

For imaginary roots, the sign of the imaginary coefficient doesn't matter because there is always a conjugate.

An answer like this should have references. I did look a little, but it seems like everything on state-space modeling reads like stereo instructions. Since everything animate uses the concepts of state-space all the time, shouldn't it be simpler to explain?? I digress... I will keep looking for something substantial in terms of a reference and edit my answer if I find something.