November 30th, 2017, 08:19 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  Range of e^z
Range of e$\displaystyle ^z$ Let F$\displaystyle _{n}$(z) be n terms of the expansion of e$\displaystyle ^{z}$1. For any n, F$\displaystyle _{n}$(z) (nth degree polynomial without the constant term) maps to all of complex plane. Therefore, for ANY n, F$\displaystyle _{n}(z)$+1 maps to entire complex plane. Therefore, e$\displaystyle ^{z}$ = F$\displaystyle _{\infty}$(z) + 1 maps to entire complex plane 
November 30th, 2017, 09:22 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1065 
Please solve $e^z1=1$ $1 \in \mathbb{C}$ so by your reasoning $\exists z_0 \in \mathbb{C} \ni e^{z_0}1=1$ 
November 30th, 2017, 09:24 AM  #3  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,346 Thanks: 2466 Math Focus: Mainly analysis and algebra  Quote:
 
November 30th, 2017, 01:34 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 417 Thanks: 231 Math Focus: Dynamical systems, analytic function theory, numerics 
This is an impressive level of narcissism and ego even for the internet. After numerous people have pointed out your nonsense "proof" in the other thread(s) are extremely flawed and offered $e^z$ as a counterexample, you have decided to claim $e^z$ is not a counterexample, rather than just admit your mistake and spend some time learning. Breathtaking.

December 1st, 2017, 01:31 PM  #5  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
$\displaystyle e^{z} = e^{x+iy} = e^{x}(\cos y+i\sin y) = 0$ $\displaystyle e^{x}\cos y=0, e^{x}\sin y=0$ $\displaystyle z=\infty+iy$ In R, if $\displaystyle \lim_{x \rightarrow \infty} f(x) = \infty$ or L, then $\displaystyle f(\infty) = \infty$ or f$\displaystyle (\infty)$ = L is clear,transparent, and conventional. The meaning is explicitly implied by the fact that $\displaystyle \infty$ is not a number. As an example in complex variables, $\displaystyle f(z)=f(x,y)= \sin z =\frac{e^{iz}e^{iz}}{2i}=\frac{e^{y}(\cos x +i\sin x)e^{y}(\cos xi\sin x)}{2i}$ $\displaystyle f(\infty,0)= (\sin\infty,0)\\ f(\infty,0)= (\sin\infty,0)\\ f(0,\infty)=(\infty,0)\\ f(0,\infty)=(\infty,0)\\ f(\infty,\infty)=\infty(\sin\infty + i\cos\infty)\\ f(\infty,\infty)=\infty(\sin\infty + i\cos\infty)\\ f(\infty,\infty)=\infty(\sin\infty  i\cos\infty)\\ f(\infty,\infty)=\infty(\sin\infty  i\cos\infty)\\ $ In doing this, one unhesitatingly uses $\displaystyle e^{\infty}$ = 0. Quote:
A polynomial in z maps to entire complex plane for all n. Convergence is a different matter. The RANGE of infinite polynomial $\displaystyle e^{z}$ is all of complex plane, regardless of whether or not it converges. The subject is RANGE, RANGE. EDIT: There is an infinite variety of infinite degree complex polynomials (depending on their constants); some converge and some don't. But they all map to entire complex plane by OP, i.e., their RANGE is all of complex plane. Last edited by skipjack; December 2nd, 2017 at 08:27 PM.  
December 1st, 2017, 02:14 PM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,346 Thanks: 2466 Math Focus: Mainly analysis and algebra  Quote:
Quote:
Last edited by skipjack; December 2nd, 2017 at 08:30 PM.  
December 2nd, 2017, 07:27 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
1) The complex polynomial P$\displaystyle _{n}$(z) maps all of z to the entire complex plane, for any n (FTA: P$\displaystyle _{n}(z)$ = 0 always has a solution). 2) It may happen that P$\displaystyle _{n}$(z$\displaystyle _{0}$) approaches a limit as more terms are added in a prescribed fashion (n gets larger). That leads to the subject of convergence, not addressed here. The OP topic is the range of e$\displaystyle ^{z}$, not its convergence. (y = cos x, x,y real, converges for all x, but its range is [1,1], i.e., not all of y. One could, for example, show the series for cos x converges for all x without knowing what it converges to (its range)). As a matter of interest, I was just showing what sin z was for various large values of z, "z=$\displaystyle \infty$" is way out on any ray extending from the origin. I just picked a couple specific examples. Last edited by skipjack; December 2nd, 2017 at 08:33 PM. 
December 2nd, 2017, 09:35 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,346 Thanks: 2466 Math Focus: Mainly analysis and algebra 
You're still writing nonsense and demonstrating a complete lack of understanding of the subjects at hand. The equation $z=\infty$ is a dead giveaway. $\sin(\infty)$ and $\cos(\infty)$ are even worse, it's not even obvious what (wrongheaded) idea you have in mind for those.

December 4th, 2017, 07:25 AM  #9  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  Quote:
 
December 4th, 2017, 10:04 AM  #10  
Senior Member Joined: Aug 2012 Posts: 1,973 Thanks: 551  Quote:
It's true that $\forall n \in \mathbb N, \frac{1}{n} > 0$. But the limit of the sequence is not greater than zero. The limit is equal to zero. The behavior at the limit may differ from the behavior at each point. Something can be true for each $n$ yet false in the limit. That's exactly what's happening in this case. Each partial sum of the power series for $e^z$ satisfies FTA, yet $e^z$ does not. And by the way, a power series is not a polynomial. Polynomials by definition have a finite number of terms.  

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