October 12th, 2017, 10:03 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90  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, 10:25 AM  #2 
Senior Member Joined: Aug 2012 Posts: 1,564 Thanks: 377  
October 12th, 2017, 11:03 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 165 Thanks: 72 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, 05:38 AM  #4  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90  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 05:47 AM.  
October 13th, 2017, 07:21 AM  #5  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699  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, 07:25 AM  #6  
Senior Member Joined: Sep 2016 From: USA Posts: 165 Thanks: 72 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:
Quote:
Quote:
Quote:
Stop writing and start reading.  
October 13th, 2017, 09:01 AM  #7  
Senior Member Joined: Oct 2009 Posts: 142 Thanks: 60  Quote:
 
October 16th, 2017, 08:42 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 
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, 06:22 PM  #9  
Senior Member Joined: Sep 2016 From: USA Posts: 165 Thanks: 72 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, 05:21 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 
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. 

Tags 
analytic, function, range 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Show function is not analytic  WWRtelescoping  Complex Analysis  1  February 25th, 2014 07:53 PM 
What is an analytic function  aaronmath  Complex Analysis  4  October 14th, 2013 04:44 PM 
Strange analytic function  bach71  Calculus  3  April 17th, 2013 12:26 PM 
Existance of analytic function at C  {0}  vitalik.t  Complex Analysis  1  January 9th, 2011 10:08 AM 
analytic function  krackwacker  Complex Analysis  2  November 6th, 2008 10:19 PM 