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November 13th, 2017, 10:01 AM   #21
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Quote:
 Originally Posted by zylo Particularly,from definition of derivative: $\displaystyle |\Delta z||F'-\epsilon|< |\Delta F| < |\Delta z||F'+\epsilon|$ if $\displaystyle |\Delta z| < \delta$ ie, if $\displaystyle |\Delta F|$ sufficiently small for F to be defined. Same argument applies for zf(z) where f(z) has a derivative for all z, and f(z) maps to entire complex plane. Because you don't know that (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z) maps to w$\displaystyle _{p}$.
Your post is unclear. If you're agreeing you haven't got a proof (as in your other thread) you're right.

Last edited by Maschke; November 13th, 2017 at 10:06 AM. November 14th, 2017, 04:21 AM   #22
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Quote:
 Originally Posted by Maschke Your post is unclear. If you're agreeing you haven't got a proof (as in your other thread) you're right.

Quote:
 Originally Posted by zylo Let w=F(z) have a derivative for all of z. Assume there is a point w$\displaystyle _{p}$ which F(z) doesn't map to. I can define F'{z) at any point w$\displaystyle _{0}$ in a neighborhood of w$\displaystyle _{p}$, and w$\displaystyle _{0}$ has a neighborhood which doesn't include w$\displaystyle _{p}$. Visually, draw a circle around w$\displaystyle _{p}$. Draw w$\displaystyle _{0}$ inside the circle. There is a circle around w$\displaystyle _{0}$ which doesn't include w$\displaystyle _{p}$. ie, the initial assumption doesn't lead to a contradiction, and F'(z) exists for all z doesn't imply F maps z to all of w. Particularly,from definition of derivative: $\displaystyle |\Delta z||F'-\epsilon|< |\Delta F| < |\Delta z||F'+\epsilon|$ if $\displaystyle |\Delta z| < \delta$ ie, if $\displaystyle |\Delta F|$ sufficiently small for F to be defined. Same argument applies for zf(z) where f(z) has a derivative for all z, and f(z) maps to entire complex plane. Because you don't know that (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z) maps to w$\displaystyle _{p}$.
The visual description should be clear.

The inequality, from the definition of derivative, confirms this. I can choose $\displaystyle \Delta$z small enough so that F is confined to a neighborhood which doesn't include w$\displaystyle _{p}$ so that it is defined at all points in the neighborhood, so that the existence of the derivative doesn't provide a contradiction.

There is still the interesting question, show when a complex function maps to the entire complex plane. Then FTA is just a trivial corollary:

Any complex polynomial has a root.
Proof: Any complex polynomial maps to entire complex plane.

EDIT:
Of course there is always the possibility that F'(z) exists for all z implies F(z) maps to entire complex plane (cosz for ex) and I wasn't able to prove it.

Last edited by zylo; November 14th, 2017 at 04:31 AM. November 14th, 2017, 09:00 AM   #23
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Quote:
 Originally Posted by zylo Of course there is always the possibility that F'(z) exists for all z implies F(z) maps to entire complex plane (cosz for ex) and I wasn't able to prove it.
What do you think about $F(z) = e^z$? November 14th, 2017, 07:22 PM #24 Senior Member   Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics His nonsensical "proof" works just as well for a constant polynomial. I am amazed he is still spouting this. There is no way he actually believes this. Even if a counterexample where difficult to find, his assumption that $z_0$ is not in the image of a map implies it maps onto a neighborhood of $z_0$ is so flawed I can't even begin. November 15th, 2017, 05:50 AM #25 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 e$\displaystyle ^{-\infty}$ =0 as was pointed out before. If F(z) maps the entire complex plane, you obviously have to deal with infinity. F(z)=c is an identity or an equation, as was pointed out before. As an identity it is a special case. F'(z) defined for all z doesn't imply F(z) maps to entire complex plane. It held in a few cases which raised the question if it was generally true. It was shown in post #20 that it wasn't. Showing F(z) maps to entire complex plane is still interesting, and any positive contributions are welcome. November 15th, 2017, 09:38 AM   #26
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Quote:
 Originally Posted by zylo e$\displaystyle ^{-\infty}$ =0 as was pointed out before.
FTA applies to the usual complex plane. There's no point at infinity. Secondly, in the complex numbers there is no such thing as $-\infty$. The occasional poster who confused you on this point frankly should know better.

Last edited by Maschke; November 15th, 2017 at 09:46 AM. November 15th, 2017, 10:07 AM #27 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 e$\displaystyle ^{-\infty}$=1/e$\displaystyle ^{\infty}$ Look up extended complex plane. But since my proof didn't work, it's academic. November 15th, 2017, 10:34 AM   #28
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Quote:
 Originally Posted by zylo e$\displaystyle ^{-\infty}$=1/e$\displaystyle ^{\infty}$ Look up extended complex plane. But since my proof didn't work, it's academic.
FTA refers to the complex plane, not the extended complex plane. On the complex plane, $e^z$ has no zero.

It may not be relevant to your proof, since you haven't got a proof.

But it IS highly relevant to your understanding of what FTA says, and for that reason you should look up FTA and verify that it refers to the complex plane and not the Riemann sphere, which does have a point at infinity.

$e^z$ also falsifies your belief that a function differentiable on the entire plane must have a zero. That is simply false. Studying this will move your mathematical understanding forward.

Remember the point is to make the effort to learn more math. If your proof failed, understanding why is a step toward understanding the actual proof of FTA.

Last edited by Maschke; November 15th, 2017 at 10:43 AM. November 15th, 2017, 03:00 PM #29 Senior Member   Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics 1. The question of whether or not you can make your argument in the extended complex plane is already a red herring since the fundamental problem here is your argument is complete nonsense. It isn't wrong, its "not even wrong". You use so many terms which either don't have any meaning, or don't mean what you seem to think they do that it is impossible to point out any single flaw in your argument. Its like if I were to claim that every matrix whose entries are made up of fish is invertible. There is nothing here to refute, its just nonsense. 2. In the interest of playing devil's advocate however, please tell me what the root is of the following polynomial $f(z) = z^2 + \infty$ Tags analytic, function, range Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Complex Analysis 1 February 25th, 2014 07:53 PM aaron-math Complex Analysis 4 October 14th, 2013 04:44 PM bach71 Calculus 3 April 17th, 2013 12:26 PM vitalik.t Complex Analysis 1 January 9th, 2011 10:08 AM krackwacker Complex Analysis 2 November 6th, 2008 10:19 PM

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