[Math] Zeros of $f(z) = z^5+3z^4+9z^3+10$ in the unit disk

Question

Show that $f(z) = z^5+3z^4+9z^3+10$ has $2$ zeros in the unit disk

I'm trying to use Rouche's theorem.

So I tried to find a function $g$ that has 2 zeros in the unit disk and:

$$|f(z)- g(z)| < |f(z)|+|g(z)| \quad \forall z \in \mathbb{D} \quad \text{(1)}$$

However, I couldn't find such function.

I tried $g(z) = 3z^4+9z^3+10$. This function has $2$ zeros in the unit disk according to Wolfram Alpha. I wasn't able to prove $(1)$ and that $g$ has
$2$ zeros in the unit disk with an analytical method.

I did the same for the function $g(z) = z^5+9z^3+10$ that has two zeros in unit disk by Wolfram Alpha. It didn't work either.

Could somebody help out to prove that $f$ has $2$ zeros in the unit disk?

Answer 1

This is a technique to solve the more general problem of counting the number of zeros of a polynomial inside the unit circle. One could use it for other curves other than the circle. All is needed is to be able to map it to a line by a rational function.

The idea is to use the argument principle instead: The number of zeros of a polynomial lying inside a loop is the number of times the image of that loop by the polynomial winds around the origin. But the unit circle is hard on additions. That is why a pretty proof by Rouche's can be tricky sometimes.

Let's instead map the unit circle to the imaginary line.

You might know a rational function that does the map, but we can derive it step by step.

1. Translate the circle one unit to the right. $z= x-1$.
2. Then we do inversion. Inversion would be $x = 1/\overline{y}$. But since the coefficients are real the conjugate won't matter. So we do $x=1/y$. We get a rational function of which we only care about the numerator (a polynomial). If zero is a solution, then $-1$ was a solution of the original polynomial and that we should've tested before hand. After this the circle got mapped to the vertical line passing through (1/2,0).
3. Finally we translate to the left by 1/2. y = w+1/2.

So, we get some polynomial with real coefficients. Let's evaluate it at $w = ir$ with $r$ real.

Now, separate imaginary part and real part. Both a polynomials of smaller degree. For this particular problem I think we get $$(880r^4-392r^2+23)+r(96r^4-912r^2+54)i$$

Now, to determine the number of times this winds around the origin we just need to see how it jumps from quadrant to quadrant. The counting of roots (no need of precise determination) can be done with Sturm's theorem in general.

For this particular problem the work is much easier. For $r=0$ we are at the point (23,0). The polynomials $880r^4-392r^2+23$ and $96r^4-912r^2+54$ are just quadratics in disguise. One can compute the roots if so inclined.

But all it matters is their relative position, which I think it is $ABBAABBA$, where the $A$'s represent roots from the second polynomial and the $B$'s represent the roots of the first one. Take into account the factor $r$ in the imaginary part which also changes its sign when $r$ crosses zero.

That order of the roots tells you the sign of the imaginary part and real part on each of the intervals between the roots. This tells you to which quadrant the whole expression is moving. From the succession of quadrants you count the winding number and that is your number.