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October 12th, 2017, 10:03 AM   #1
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Range of Analytic Function

Theorem: If F'(z) exists for all z (analytic), w=F(z) maps to entire complex plane.

Proof Outline: Assume F(z) doesn't exist at a point w$\displaystyle _{0}$ but does at any point in a neighborhood of w$\displaystyle _{0}$. Then, $\displaystyle \Delta$w =F'(z)$\displaystyle \Delta$z and F(z) exists at w$\displaystyle _{0}$.
Same reasoning applies at a boundary point.

Example: Polynomials
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October 12th, 2017, 10:25 AM   #2
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Theorem: If F'(z) exists for all z (analytic), w=F(z) maps to entire complex plane.
$e^z$ satisfies your hypotheses but never hits $0$.
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October 12th, 2017, 11:03 AM   #3
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Again with this nonsense? It isn't even true for polynomials since $f(z) = c$ is analytic for any constant $c$.
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October 13th, 2017, 05:38 AM   #4
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Again with this nonsense? It isn't even true for polynomials since $f(z) = c$ is analytic for any constant $c$.
f(z)=c is either an equation or an identity. As an equation, the theorem applies to, say, f(z)=z^3+az^2+bz, which does map everywhere if you accept FTA as independently proven.

f'(z)=0 everywhere is exceptional case, which doesn't happen for, say, f(z)=z.

EDIT
e^z = c has a solution for all c, including 0. $\displaystyle \infty$ is a legitimate complex number.

Last edited by zylo; October 13th, 2017 at 05:47 AM.
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October 13th, 2017, 07:21 AM   #5
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Quote:
Originally Posted by zylo View Post
f(z)=c is either an equation or an identity. As an equation, the theorem applies to, say, f(z)=z^3+az^2+bz, which does map everywhere if you accept FTA as independently proven.

f'(z)=0 everywhere is exceptional case, which doesn't happen for, say, f(z)=z.

EDIT
e^z = c has a solution for all c, including 0. $\displaystyle \infty$ is a legitimate complex number.
No, $\displaystyle \infty$ is NOT a "legitimate complex number".

Both Wolfram (Complex Number -- from Wolfram MathWorld) and
Wikipedia (https://en.wikipedia.org/wiki/Complex_number) define "complex number" as "a number of the form x+ iy where x and y are real numbers".

And $\displaystyle \infty$ is NOT a real number either:
https://en.wikipedia.org/wiki/Real_number

We can add $\displaystyle \infty$ to the real numbers, getting the "extended real number system". Note that it is possible to do that in different ways. We can add both $\displaystyle \infty$ and $\displaystyle -\infty$ to the real numbers, getting a system that is topologically equivalent to a closed, bounded interval, or add only [math]\infty[/b], getting a system that is topologically equivalent to a circle.

Similarly, we can add a single "infinity" to the set of complex numbers, getting a system that is topologically equivalent to a sphere, or we can add "infinitely many" "infinities" getting a system that is topologically equivalent to a disk.
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October 13th, 2017, 07:25 AM   #6
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Originally Posted by zylo View Post
f(z)=c is either an equation or an identity.
It is a constant polynomial. I don't know what you mean by an equation or identity.

Quote:
As an equation, the theorem applies to, say, f(z)=z^3+az^2+bz, which does map everywhere if you accept FTA as independently proven.
I have no clue what this means. You claim every analytic function maps onto $\mathbb{C}$. I claim that constant polynomials are analytic and do not.

Quote:
f'(z)=0 everywhere is exceptional case, which doesn't happen for, say, f(z)=z.
So your argument when presented with a counterexample to your "theorem" is "Hey look at this other function which isn't a counterexample". Do you have brain damage?

Quote:
e^z = c has a solution for all c, including 0.
No it doesn't.

Quote:
$\displaystyle \infty$ is a legitimate complex number.
No it isn't. But even if it were, are you claiming that $e^{\infty} = 0$? If so, then $\lim_{z \to \infty} e^z = 0$ by analyticity but taking $z \to \infty$ along the positive real line clearly violates this.

Stop writing and start reading.
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October 13th, 2017, 09:01 AM   #7
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Quote:
Originally Posted by zylo View Post
f(z)=c is either an equation or an identity. As an equation, the theorem applies to, say, f(z)=z^3+az^2+bz, which does map everywhere if you accept FTA as independently proven.

f'(z)=0 everywhere is exceptional case, which doesn't happen for, say, f(z)=z.

EDIT
e^z = c has a solution for all c, including 0. $\displaystyle \infty$ is a legitimate complex number.
Then you probably want to talk about entire meromorphic functions. But then the only relevant functions are the rational functions (= quotients of polynomials).
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October 16th, 2017, 08:42 AM   #8
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f(z) = c is an identity if z is given and an equation if c is given- but that's irrelevant. All posts irrelevant except last one which I don't understand because it has big words. If it says rational functions map to entire complex plane I would be interested if it referred to an intelligible theory.

There is a problem:

Assume f(z) has a derivative for all z and maps to w plane with a hole in it (area where f(z) doesn't map to). The hole is closed because f'(z) is undefined on it's border. In this case, f'(z) is defined and f'(z)$\displaystyle \Delta$z exists in f(z) for $\displaystyle \Delta$z sufficiently small no matter how close you get to the border.

For example, f(z) could map to the inside of a circle for all z and f'(z) exist everywhere inside the circle.


The only way out would be to show that at a point near the border f'(z)$\displaystyle \Delta$z is greater than the distance to the border.
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October 16th, 2017, 06:22 PM   #9
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Quote:
Originally Posted by zylo View Post
f(z) = c is an identity if z is given and an equation if c is given- but that's irrelevant. All posts irrelevant except last one which I don't understand because it has big words. If it says rational functions map to entire complex plane I would be interested if it referred to an intelligible theory.

There is a problem:

Assume f(z) has a derivative for all z and maps to w plane with a hole in it (area where f(z) doesn't map to). The hole is closed because f'(z) is undefined on it's border. In this case, f'(z) is defined and f'(z)$\displaystyle \Delta$z exists in f(z) for $\displaystyle \Delta$z sufficiently small no matter how close you get to the border.

For example, f(z) could map to the inside of a circle for all z and f'(z) exist everywhere inside the circle.


The only way out would be to show that at a point near the border f'(z)$\displaystyle \Delta$z is greater than the distance to the border.
This is all nonsense. There is no relationship between the image of a function and its smoothness properties.

Researching meromorphic functions won't help either, as analytic functions are a strict subset of meromorphic functions and you don't have any understanding of analytic functions. Certainly, your theorem isn't true for analytic functions so it won't be true for meromorphic functions.

I think micrm@ass was just exercising his/her vocabulary. In this case, your vocabulary has vastly outgrown your understanding which is your problem.

Or you are trolling which is what I still suspect.
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October 17th, 2017, 05:21 AM   #10
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If F(z) = zf(z) where f(z) exists, has a derivative everywhere, and maps to entire complex plane, then so does F(z).

Again, assume a closed island in w where F(z) undefined. Let F(z) be arbitrarily close to the border.

F(z+$\displaystyle \Delta$z) = (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z)= (z+$\displaystyle \Delta$z}(f(z)+f'(z)$\displaystyle \Delta$z)

F(z+$\displaystyle \Delta$z)=F(z)+(f(z)+f'(z)$\displaystyle \Delta$z)

where f(z) and f'(z) exist and $\displaystyle \Delta$z is associated with f'(z) so not restricted by the border.

It follows F(z) can cross the border, a contradiction, so the island doesn't exist and F(z) maps to entire complex plane.
Example: Proof of FTA by induction where f(z) and zf(z) are polynomials.
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