7

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?

user344460
  • 225
  • 3
    tough problem. Working on it using $g(z) = (z+3/2+8i/3)(z+3/2-8i/3)(z+9/8)(z-9/16+7i/9)(z-9/16+7i/9)$ which seems to work but its hard to show it. – Mark Fischler Jun 15 '17 at 19:06
  • 1
    Correct me if I'm wrong, but given $|z| = 1$, then $\bar{z} = \frac{1}{z}$, and $z \neq 0$. So if $z^5 + 3z^4 + 9z^3 + 10 = 0$, then $\frac{1}{z^5} + \frac{3}{z^4} + \frac{9}{z^3} + 10 = 0$, so $z$ is also a root of $1 + 3z + 9z^2 + 10z^5$. It is then a root of the gcd of those two polynomials, but wolfram says it's gcd is 1, so someone is wrong here. – Henrique Augusto Souza Jun 15 '17 at 19:21
  • Easy problem if you understand complex analysis. Easier than Rouche's is the argument principle. Take that unit circle, which is hard to do additions, and map it to, say the imaginary axis. – OR. Jun 15 '17 at 19:21
  • 1
    My proposed $g(z)$ fails the condition by a tiny amount at $z = \frac{29}{30}e^{\frac{3\pi}{10}}$. – Mark Fischler Jun 15 '17 at 19:38
  • Do what I said. After the change of variable you end up with the equivalent problem of: How many times does $(880z^4-392z^2+23)+z(96z^4-912z^2+54)i$ winds around the origin when $z$ moves along the reals? Notice that that is just baby work. Since all that matters is when the real part and imaginary part change signs, and they are quadratic polynomials on z^2. – OR. Jun 15 '17 at 19:54
  • Notice that this technique shows that this type of problems can be solved by an algorithm (in a finite number of steps). For higher degrees we would need to start using Sturm's theorem to count changes of sign. So, this kind of problems belongs in the category of taking derivatives of elementary functions, or finding GCD of polynomials, etc. – OR. Jun 15 '17 at 20:02
  • 3
    This is a near duplicate. See https://math.stackexchange.com/q/2038467/254075 – sharding4 Jun 15 '17 at 20:10
  • @sharding4 Pretty solution. But after using it one would still need to count out the root between the two circles to solve this problem. – OR. Jun 15 '17 at 20:16
  • @MlazhinkaShungGronzalezLeWy Could you post your suggestion as an answer with a little more application? – Sahiba Arora Jun 15 '17 at 20:17
  • @SahibaArora what does it mean "more application"? Do you mean the solution for arbitrary polynomials? – OR. Jun 15 '17 at 20:17
  • @MlazhinkaShungGronzalezLeWy Sorry, long day. I meant more "explanation". – Sahiba Arora Jun 15 '17 at 20:18
  • @SahibaArora Let me try. – OR. Jun 15 '17 at 20:27
  • @SahibaArora ok. Let me know if you need more details. – OR. Jun 15 '17 at 20:46
  • It is unfortunate that so many students get failed over the years due to not being able to solve essentially the same problem with different coefficients. I see that more as a failure of the instructors. – OR. Jun 15 '17 at 21:14
  • And it is amusing when people that are supposed to have experience also fail to solve them. But it is still a failure of their instructor. – OR. Jun 15 '17 at 21:15
  • @MlazhinkaShungGronzalezLeWy Comments are suppose to address the post (which you did some comments above). Please using the comment sections for ranting and similar practices that do not contribute to the site. – Pedro Jun 15 '17 at 21:37
  • 1
    @HenriqueAugustoSouza I find your argument convincing; but what you've shown is that there aren't any roots along the unit circle. – Jonathan Y. Jun 15 '17 at 22:42
  • @JonathanY. Yeah, I thought that could be the case.. but Wolfram found some roots pretty close to the unit circle by a numerical method ($|z| \approx 0.97$), so I'm in doubt here. I've done the gcd myself too, check if it was right, and found that they are coprime as well. – Henrique Augusto Souza Jun 15 '17 at 22:46
  • @JonathanY. maybe "in the unit disk" means "inside the unit disk"? Then it may be true, given that result from the numerical method. – Henrique Augusto Souza Jun 15 '17 at 22:48
  • 2
    @HenriqueAugustoSouza Yes, the unit disc is the open domain $|z|<1$; it's boundary is the unit circle $|z|=1$. That $f$ doesn't vanish along the unit circle is at the basis of the attempts to use Rouche's theorem, here. – Jonathan Y. Jun 15 '17 at 22:50
  • 1
    @JonathanY. Oh, I see now. I was thinking on the boundary all this time and misunderstood the problem! – Henrique Augusto Souza Jun 15 '17 at 22:52

1 Answers1

2

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.

OR.
  • 5,941
  • 1
    "For this particular problem I think we get..." -- why not show your process? – Jonathan Y. Jun 15 '17 at 22:14
  • @JonathanY. Because anyone having to deal with this problem in complex analysis I assume is able to add and multiply polynomials. Also, the process is steps 1 to 3. Do you need help with that? – OR. Jun 15 '17 at 22:17
  • @JonathanY. Also the most important part of the answer is not the computation or even solving this particular problem, but to show how to solve it for any polynomial. – OR. Jun 15 '17 at 22:19
  • You'll note (a) I haven't asked the question, and (b) even though the question is currently highly rated, the answer hasn't been voted on. I therefore make suggestions to improve it: show your work, don't tell us about it. That includes explicitly composing $f$ with the mapping you used (a Mobius transform? Why go through this process?), and working out the order of roots of both polynomials you derived. – Jonathan Y. Jun 15 '17 at 22:35
  • @JonathanY. I don't care about votes. And if you care so much, anyone can edit an answer. You can do it. Or post your own. And I will tell "us" whatever I decide I want to tell. – OR. Jun 15 '17 at 22:38