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November 14th, 2017, 10:34 AM   #1
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Fundamental Theorem of Algebra Proof 01

1) Let w = z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz = z(z$\displaystyle ^{2}$+az+b)
2) Is z$\displaystyle ^{3}$ a subset of z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz? Yes, if z$\displaystyle ^{2}$ is a subset of z$\displaystyle ^{2}$+az+b, which it is by induction.*
3) Since z$\displaystyle ^{3}$ maps to all of complex plane, so does w = z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz.
4) The extension should be obvious: P$\displaystyle _{n-1}$ $\displaystyle \rightarrow$ P$\displaystyle _{n}$. FTA

5) * z$\displaystyle ^{2}$ is a subset of z$\displaystyle ^{2}$+az+b if z_{0}^{2} = z$\displaystyle ^{2}$+az+b has a solution for all z$\displaystyle _{0}$.
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November 14th, 2017, 10:50 AM   #2
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Quote:
Originally Posted by zylo View Post
1) Let w = z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz = z(z$\displaystyle ^{2}$+az+b)
That has an obvious zero, namely $z = 0$. Can you walk through your proof with, say, $f(z) = z^3 + a z^2 + b z + 5$?

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2) Is z$\displaystyle ^{3}$ a subset of z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz?
No, not by the common meaning of subset. It's not helpful for you to redefine standard terminology to mean something that only you understand.

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3) Since z$\displaystyle ^{3}$ maps to all of complex plane, so does w = z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz.
Can you prove it hits 12?

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4) The extension should be obvious:
No. Not obvious. Not even apparently related to FTA.

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5) * z$\displaystyle ^{2}$ is a subset of z$\displaystyle ^{2}$+az+b if z_{0}^{2} = z$\displaystyle ^{2}$+az+b has a solution for all z$\displaystyle _{0}$.
Unclear what you mean.
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November 15th, 2017, 07:46 AM   #3
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The point of the proof is to show z$\displaystyle ^{n}$+a$\displaystyle _{n-1}$z$\displaystyle ^{n-1}$+..+a$\displaystyle _{1}z$ maps to entire complex plane: FTA

It is shown by induction that {z$\displaystyle ^{n}$} is a subset of {z$\displaystyle ^{n}$+a$\displaystyle _{n-1}$z$\displaystyle ^{n-1}$+..+a$\displaystyle _{1}z$} and z$\displaystyle ^{n}$ maps to entire complex plane.

z$\displaystyle ^{n}$+a$\displaystyle _{n-1}$z$\displaystyle ^{n-1}$+..+a$\displaystyle _{1}z$ maps to entire complex plane iff z$\displaystyle ^{n}$+a$\displaystyle _{n-1}$z$\displaystyle ^{n-1}$+..+a$\displaystyle _{1}z$+a$\displaystyle _{0}$ does. It is convenient to use both.

Last edited by zylo; November 15th, 2017 at 08:02 AM.
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November 15th, 2017, 10:02 AM   #4
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OP Clarified

1) Let $\displaystyle w=z^{3}+az^{2}+bz=z(z^{2}+az+b) $

2) {$\displaystyle z^{2}$} $\displaystyle \equiv$ {$\displaystyle z^{2}+az+b$} by induction##, ie, with each $\displaystyle z^{2}$ I can associate $\displaystyle z^{2}+az+b$, ie
$\displaystyle z^{2} \leftrightarrow z^{2}+az+b \rightarrow z^{3} \leftrightarrow z^{3}+az^{2}+bz \rightarrow$ {$\displaystyle z^{3}$} $\displaystyle \equiv$ {$\displaystyle z^{3}+az^{2}+bz$} and $\displaystyle z^{3}$ maps to all of z. Therefore

3) $\displaystyle w=z^{3}+az^{2}+bz$ maps to all of z.

4) The extension should be obvious: P$\displaystyle _{n-1}$ $\displaystyle \rightarrow$ P$\displaystyle _{n}$. FTA

##
Let $\displaystyle z_{A}=z_{B}^{2}+az_{B}+b$, for each $\displaystyle z_{A}$ you can find z$\displaystyle _{B}$ INDUCTION premise, and for each $\displaystyle z_{B}$ you can find $\displaystyle z_{A}$, ie, {$\displaystyle z^{2}$} $\displaystyle \equiv$ {$\displaystyle z^{2}+az+b$}
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November 15th, 2017, 10:36 AM   #5
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Quote:
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1) Let $\displaystyle w=z^{3}+az^{2}+bz=z(z^{2}+az+b) $
Of course that has an obvious zero. How does your proof work if there's a nonzero constant term?
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November 15th, 2017, 04:01 PM   #6
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\popcorn
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November 16th, 2017, 08:28 AM   #7
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Ex for n=4

Illustration of OP for n=4. The objective is to show that $\displaystyle z^{4}+az^{3}+bz^{2}+cz$ maps to all of complex plane (FTA) if $\displaystyle z^{3}+az^{2}+bz$ does.

{$\displaystyle z^{3}$} $\displaystyle \subset$ {$\displaystyle z^{3}+az^{2}+bz+c$} **
$\displaystyle \therefore z^{3} \rightarrow z^{3}+az^{2}+bz+c$
$\displaystyle \therefore z^{4} \rightarrow z^{4}+az^{3}+bz^{2}+cz$
$\displaystyle \therefore$ {$\displaystyle z^{4}$} $\displaystyle \subset$ {$\displaystyle z^{4}+az^{3}+bz^{2}+cz$}.
z$\displaystyle ^{4}$ maps to all of complex plane,
$\displaystyle \therefore z^{4}+az^{3}+bz^{2}+cz$ maps to all of complex plane, ***
which is same as
$\displaystyle z^{4}+az^{3}+bz^{2}+cz+d$ maps to all of complex plane,
which is used to prove n=5.


**$\displaystyle z_{0}^{3}=z^{3}+az^{2}+bz+c$ has a solution for every z$\displaystyle _{0}$ because n=3 given, ($\displaystyle z^{3}+az^{2}+bz+c$ maps to entire complex plane).

*** If this step bothers you, see proof in post #4 which has an extra step.
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November 16th, 2017, 09:57 AM   #8
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The following is a generalization illustrating some steps in previous post.

Theorem
If f(z) maps to entire complex plane, so does zf(z).
Proof
$\displaystyle z_{0} = f(z)$ has a solution for all of $\displaystyle z_{0}$ because f(z) maps to entire complex plane.
$\displaystyle \therefore$ {z} $\displaystyle \subset$ {f(z)}
$\displaystyle \therefore$ z $\displaystyle \rightarrow$ f(z), maps to
$\displaystyle \therefore z^{2} \rightarrow zf(z)$
$\displaystyle \therefore$ {$\displaystyle z^{2}$} $\displaystyle \subset$ {zf(z)}
z$\displaystyle ^{2}$ maps to entire complex plane.
$\displaystyle \therefore$ zf(z) maps to entire complex plane.


Alternate proof: Ref Post #4
$\displaystyle z_{0}$ = f(z) has a solution for all of $\displaystyle z_{0}$ because f(z) maps to entire complex plane, and every z determines a $\displaystyle z_{0}$.
$\displaystyle \therefore$ {z} $\displaystyle \equiv$ {f(z)}
$\displaystyle \therefore z \leftrightarrow f(z)$
$\displaystyle \therefore z^{2} \leftrightarrow zf(z)$
$\displaystyle \therefore$ {$\displaystyle z^{2}$} $\displaystyle \equiv$ {$\displaystyle zf(z)$}
$\displaystyle \therefore$ zf(z) maps to entire complex plane.
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November 16th, 2017, 11:02 AM   #9
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The pattern for FTA utilizing above theorem is:

$\displaystyle P_{n-1}(z) \rightarrow zP_{n-1}(z) \rightarrow P_{n}(z)$

It is essential to note that $\displaystyle z^2 +az$ maps to entire complex plane iff $\displaystyle z^2 +az +b$ does, and the n degree generalization.
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November 16th, 2017, 11:54 AM   #10
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Great job! You just proved that since $f(z) = \bar{z}$ maps to the entire complex plane, then so does $g(z) = z\bar{z} = |z|^2$. Keep up the great work proving solid mathematical theorems!
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