How many zeroes ? -A

[Camillo]

How many roots has the polynomial : $z^7-2z^5+6z^3-z +1 $ in $ |z| <1 $ ?

[raffamaiden]

Dovresti specificare a quale insieme appartiene z (anche perchè tra reale e complesso il modulo viene interpretato differentemente)

[dissonance]

Dovresti specificare a quale insieme appartiene z (anche perchè tra reale e complesso il modulo viene interpretato differentemente)

I guess that \(z \in \mathbb{C}\).

[Camillo]

Your guess is right .

[Camillo]

For finding the solution the Rouchè theorem is useful.

[Zero87]

For finding the solution the Rouchè theorem is useful.

I try to describe my solution in english (but i advertise the forumers that my english is awful). I like that someone sent a comment about my english because, in a not far future, i have to write my thesys in english (uff!).

Testo nascosto, fai click qui per vederlo

The polynomial function is $P(z)=z^7-2z^5+6z^3-z+1$: i have to study the zeroes in the domain $|z|<1$
So, i decided to use the Rouche's theorem which says that if i have a cycle $\gamma$ ~ $0$ (mod $\Omega$) like that $Ind_\gamma (z)=0,1$ $\forall z\in \Omega$ and $f,g$ holomorphics in $\Omega$ t.c. $|f(z)-g(z)|<|f(z)|$ $\forall z\in \gamma$ so, $f$ and $g$ have the same number of zeroes in the domain $\Omega_1={z\in \Omega : Ind_\gamma (z)=1}$.

So, i would choose $f(z)=6z^3$ and $g(z)=P(z)-f(z)$ and i have
- $|f(z)|_{|z|=1}$ that is equal to 6;
- $|g(z)|=|P(z)-f(z)|$ extimated (or calculated? i don't know which word i have to use in this case) on $|z|=1$ is equal to $1$.

So, this will respect the hipothesys of Rouché's Theorem and I will conclude that the function has 3 zeroes in the considered domain.

[EDIT] I didn't know that i had to use the spoiler...

[Rigel]

Since nobody tries...

Testo nascosto, fai click qui per vederlo

Let $f(z) = z^7-2z^5+6z^3$ and $g(z) = -z+1$. If $|z| = 1$ we have that
\[ |f(z)| = |z^3(z^4-2z^2+6)| = |z^4-2z^2+6| \geq 6 - 1 - 2 = 3,\qquad
|g(z)| = |-z+1| \leq 2,\]
hence $|g(z)| < |f(z)|$ for every $|z| = 1$.
By Rouché's theorem we deduce that $f$ and $f+g$ have the same number of zeroes in the unit disk (counted with their multiplicity).
Now it is enough to observe that $f$ has $7$ zeroes: $z_0 = 0$ with multiplicity $3$, and the solutions to the biquadratic equation $z^4-2z^2+6 = 0$, that are easy to compute and lie outside the unit disk.

[dissonance]

my thesys in english (uff)

I'm not 100% sure but I believe the anglo-saxon people use the word dissertation. And you should br glad they let you write in English. I proposed to do that myself but university staff won't let me do so.

So, i decided to use the Rouche's theorem which says that if i have a cycle $\gamma$ ~ $0$ (mod $\Omega$) like that $Ind_\gamma (z)=0,1$ $\forall z\in \Omega$ and $f,g$ holomorphics

Such that, not "like that". Also, "holomorphic" is an adjective and so it comes before the noun and without the final "s", even if it is plural: holomorphic \(f, g\).

extimated (or calculated?

I'd say evaluated.

So, this will respect the hipothesys of Rouché's Theorem and I will conclude that[...]

I don't like this sentence. I would have said something like "so the hypotheses of Rouche's Theorem are met, from which we conclude that[...]".

Any corrections and/or suggestions are welcome. This is what this forum stands for.

[Zero87]

Thanks for these suggestions!

Testo nascosto, fai click qui per vederlo

Come minimo ho sbagliato a scrivere anche i ringraziamenti...
PS: ho scritto "like that" perché non sapevo proprio come si scrivesse e ho sparato a caso!
@Rigel: non sapevo si dicesse "unit disk", thank you!

[Camillo]

@Zero87
Why do you write that for $|z| =1 $ , it is $|g(z)| =|z^7-2z^5-z+1| = 1$ , while it is $|g(z)| <= 5 $ ?.