October 12th, 2017, 11:03 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,640 Thanks: 119  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 
October 12th, 2017, 11:25 AM  #2 
Senior Member Joined: Aug 2012 Posts: 2,135 Thanks: 621  
October 12th, 2017, 12:03 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics 
Again with this nonsense? It isn't even true for polynomials since $f(z) = c$ is analytic for any constant $c$.

October 13th, 2017, 06:38 AM  #4  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,640 Thanks: 119  Quote:
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 06:47 AM.  
October 13th, 2017, 08:21 AM  #5  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895  Quote:
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.  
October 13th, 2017, 08:25 AM  #6  
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics  It is a constant polynomial. I don't know what you mean by an equation or identity. Quote:
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Stop writing and start reading.  
October 13th, 2017, 10:01 AM  #7  
Senior Member Joined: Oct 2009 Posts: 695 Thanks: 232  Quote:
 
October 16th, 2017, 09:42 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,640 Thanks: 119 
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. 
October 16th, 2017, 07:22 PM  #9  
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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.  
October 17th, 2017, 06:21 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,640 Thanks: 119 
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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