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October 3rd, 2017, 10:10 AM   #21
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Quote:
Originally Posted by Maschke View Post
No that's not true. Your previous post claimed to not use analysis, but you are relying on the completeness of the complex plane.

Your "algebraic" proof is no proof at all, for the same reason.

This is the state of failed proofs of FTA from the 1700's. Nobody had the vocabulary to express completeness.

How do you know that as your circle expands it must necessarily pass through zero? That's the key point.

And by the way didn't I already outline this exact argument in post #4?
And did not Gauss give a similar proof over 200 years ago?
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October 3rd, 2017, 10:18 AM   #22
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And did not Gauss give a similar proof over 200 years ago?
As I understand it, Gauss gets credit for the first complete (no pun intended) proof, but by modern standards of rigor he didn't have a proof either. In fact the last hole in Gauss's proof was not resolved till as recently as 1920.

https://en.wikipedia.org/wiki/Fundam...rem_of_algebra

Or are you saying that I shouldn't brag about posting Zylo's proof sketch earlier in this thread? Well you're right I guess.

Last edited by Maschke; October 3rd, 2017 at 10:21 AM.
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October 3rd, 2017, 11:00 AM   #23
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Quote:
Originally Posted by Maschke View Post
As I understand it, Gauss gets credit for the first complete (no pun intended) proof, but by modern standards of rigor he didn't have a proof either. In fact the last hole in Gauss's proof was not resolved till as recently as 1920.

https://en.wikipedia.org/wiki/Fundam...rem_of_algebra

Or are you saying that I shouldn't brag about posting Zylo's proof sketch earlier in this thread? Well you're right I guess.
No. I was not trying to snark at you. I just remembered that Gauss had constructed a proof using a spiral on the complex plane. I only vaguely remembered that his proof did not meet modern standards of rigor: how could it when real and complex analysis had not yet been developed.

Perhaps I was being a bit snarky with Zylo for proposing a proof that may not have met the standards of rigor of 200 years ago.
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October 4th, 2017, 06:01 AM   #24
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Post #19, the "algebraic proof," is wrong.

Consider the following algebraic proof:

General third degree complex polynomial:
$\displaystyle z^{3}+pz^{2}+qz=c$
$\displaystyle z^{3}$ maps to all points in the complex plane
$\displaystyle pz^{2}$ maps to all points in the complex plane
1) $\displaystyle qz$ maps to all points in the complex plane.
Therefore $\displaystyle z^{3}+pz^{2}+qz$ maps to all points in the complex plane.
Therefore $\displaystyle z^{3}+pz^{2}+qz=c$ has a solution for any c.

Same argument for n-degree polynomial.

EDIT: 1) does not follow- has to be proven. Back to proof I was working on when above suddenly occurred to me. I note none of the replies to previous proof saw what was wrong with it, so they just contained irrelevant vaque generalities.

Last edited by zylo; October 4th, 2017 at 06:20 AM.
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October 4th, 2017, 06:19 AM   #25
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Quote:
Originally Posted by zylo View Post
Therefore $\displaystyle z^{3}+pz^{2}+qz$ maps to all points in the complex plane.
This is where your argument breaks down. If two functions $\mathbb{C} \to \mathbb{C}$ each map to all points in the complex plane, their sum need not. For example, $z$ and $-z$ both map to all points in the complex plane, but their sum clearly doesn't.

Quote:
Originally Posted by zylo View Post
1) $\displaystyle qz$ maps to all points in the complex plane.

EDIT: 1) does not follow- has to be proven
1) here is fine so long as $q \neq 0$ (and if $q = 0$, then you just have fewer terms to worry about). Indeed, for any $a \in \mathbb{C}$, $qz = a$ has a solution $z = \frac{a}{q}$. The issue with this argument is the one I've pointed out above.

Last edited by cjem; October 4th, 2017 at 06:31 AM.
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October 5th, 2017, 07:21 AM   #26
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Quote from post # 19
"Previous post #18 gave topological proof of FTA. An algebraic proof follows:

Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form
$\displaystyle f(z)=z(z^{2}+pz+q)=C$
Assuming induction for the second degree polynomial,
$\displaystyle f(z)=z(z-a)(z-b)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$

$\displaystyle z=\rho e^{i\theta} \quad z-a=\rho _{a}e^{i\theta _{a}}, \quad z-b=\rho _{b}e^{i\theta _{b}}$
$\displaystyle a=|a|e^{i\alpha},\quad b=|b|e^{i\alpha}$
-------------------
Argument exactly same for nth degree polynomial."

The above was incomplete but contained core concept of proof:

Objective: Show that $\displaystyle z(z-a)(z-b)$ maps to entire complex plane, ie, $\displaystyle z=\rho e^{i\theta}$ exists for any value of $\displaystyle z(z-a)(z-b)=c.$

The argument is easier to follow for $\displaystyle z(z-a)=\rho \rho_{a}e^{i(\theta+\theta_{a})}$.

1) Hold $\displaystyle \theta$ constant. Then $\displaystyle |z(z-a)| = \rho \rho_{a}$ varies from 0 to $\displaystyle \infty$ as $\displaystyle \rho$ goes from to 0 to $\displaystyle \infty$, variation depending on $\displaystyle \theta$.

2) Hold $\displaystyle \rho$ constant. Then $\displaystyle arg [z(z-a)]$ varies from $\displaystyle 0+\theta_{a}(0)$ to $\displaystyle 2\pi +\theta_{a}(0)$, variation depending on $\displaystyle \rho$.

3) Think of the process as creating a polar map consisting of rays $\displaystyle \theta$ and circles $\displaystyle \rho$.
On each ray label points $\displaystyle |z(z-a)|=\rho \rho_{a}$.
On each circle label points $\displaystyle arg [z(z-a)]$.

4) Then, given $\displaystyle c=Re^{i\Theta}$, locate intersection of R and $\displaystyle \Theta$ on the map to determine $\displaystyle \rho$ and $\displaystyle \theta$.

Visualize by sketching $\displaystyle z-a$: draw $\displaystyle a$ and a circle of radius $\displaystyle \rho$ at the end of $\displaystyle a$. Line from origin to point on the circle is $\displaystyle z-a$.

The proof is geometric/algebraic and requires nothing beyond elementary complex algebra.

Last edited by zylo; October 5th, 2017 at 07:48 AM. Reason: Change quote to direct copy
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October 5th, 2017, 08:13 AM   #27
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Quote:
Originally Posted by zylo View Post

The argument is easier to follow for $\displaystyle z(z-a)=\rho \rho_{a}e^{i(\theta+\theta_{a})}$.

1) Hold $\displaystyle \theta$ constant. Then $\displaystyle |z(z-a)| = \rho \rho_{a}$ varies from 0 to $\displaystyle \infty$ as $\displaystyle \rho$ goes from to 0 to $\displaystyle \infty$, variation depending on $\displaystyle \theta$.

2) Hold $\displaystyle \rho$ constant. Then $\displaystyle arg [z(z-a)]$ varies from $\displaystyle 0+\theta_{a}(0)$ to $\displaystyle 2\pi +\theta_{a}(0)$, variation depending on $\displaystyle \rho$.
1. Your earlier assertions without proof that $f$ maps onto $\mathbb{C}$ is true. However, it is a consequence of the FTA. You certainly can't state this without justification in your "proof" of the FTA.

2. Above you are attempting to prove this fact. However, your argument is far from rigorous. One problem is that $\theta_a$ varies as $\rho$ varies even for fixed $\theta$. You have yet to give a reasonable argument for why a polynomial can't "skip" values. Furthermore, you seem to think it maps circles to circles and rays to rays. Neither of these are true and counterexamples have already been given. The most you can say is that a simple closed curve will map to a closed curve. Even this image need not be simple and a counterexample of this has also been given. Even if it were, this is far from sufficient to show what you are trying to show.

3. Your induction argument still runs into the problems pointed out earlier in this thread if you try to proceed for arbitrary $n$. If you can't address this, then the best you can hope for is a proof of FTA for degrees no greater than 3.

I suggest looking at Rouche's theorem if you are really interested in proving FTA via complex analysis.
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October 5th, 2017, 10:48 AM   #28
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FTA Quickie

SDK, I apply FTA by induction to the lower order polynomial f(z), regardless of degree. I map on to circles and lines, not to circles and lines.

Sorry you didn't understand my post #26. You may prefer the following:

FTA Quickie:

1) Given 3rd degree complex polynomial
$\displaystyle z(z^{2}+pz+q)=zf(z)=c$
By induction applied to the second degree polynomial, $\displaystyle f(z)$ maps z to entire complex plane.

2) $\displaystyle zf(z)$ maps continuously to entire complex plane (from 0 to infinity).

3) $\displaystyle zf(z)=c$ has a solution (root) z=d for any c.

4) Same argument for any degree.


I have googled FTA. The complexity and abstraction are not my cup of tea. I prefer a simple, transparent, comprehensible, easy to remember, proof, where possible. If it bothers you, be consoled by fact that it is virtually unnoticed and writ in water, and the others will be around a long time. Anyhow, thanks for your interest.
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October 5th, 2017, 01:08 PM   #29
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Even simpler:

Given any n degree complex polynomial written as:
$\displaystyle F_{n}(z)=-a_{0}$
$\displaystyle F_{n}(z)$ is continuous for all $\displaystyle z$ and varies continuously from complex $\displaystyle 0$ to complex $\displaystyle \infty$ (eventually $\displaystyle z^{n}$ predominates).
It follows $\displaystyle F_{n}(z)=-a_{0}$ has a root for any $\displaystyle a_{0}$.
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October 5th, 2017, 01:24 PM   #30
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Quote:
Originally Posted by zylo View Post
$\displaystyle F_{n}(z)$ is continuous ...
What does continuous mean in a purely algebraic setting as you are claiming?
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