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November 18th, 2017, 09:21 AM  #11  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,217 Thanks: 93  Quote:
FTA proof fails. The real problem with the proof is not that it was wrong, I make mistakes all the time, but that it was apparently transparent when it really wasn't. I can see where that would be very irritating. It would infuriate me. Thanks to the responders. But I was thinking ahead. Let's see now, 02? Maybe I'll stay away from mapping, infinite sets are treacherous.  
November 29th, 2017, 11:22 AM  #12 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,217 Thanks: 93 
Let w=f(z) and f'(z) be defined for all z and f(z) unbounded in any direction in the wplane. Assume there is a point w$\displaystyle _{0}$ which f(z) doesn't map to. Draw a curve in wplane ending at w$\displaystyle _{0}$. The corresponding curve in zplane is open at z$\displaystyle _{0}$. But $\displaystyle \lim_{z \rightarrow z_{0}} = f(z_{0}) = w_{0}$. Contradiction. If w$\displaystyle _{0}$ is a border point, existence of derivative leads to existence of w in a neighborhood of w$\displaystyle _{0}$. Contradiction. Therefore f(z) maps to every point w. Therefore any polynomial P$\displaystyle _{n}(z)$ = 0 has a solution: $\displaystyle z^{n}+a_{n1}z^{n1}+..+a_{1}z$ maps to entire complex plane. FTA Last edited by zylo; November 29th, 2017 at 11:53 AM. 
November 29th, 2017, 03:42 PM  #13  
Senior Member Joined: Aug 2012 Posts: 1,700 Thanks: 448  Quote:
Now let $w_0 = 1$. Certainly $w_0$ is not hit by $f$. So take some curve that terminates at $1$. The limit of $f$ applied to any continuous curve terminating at $1$ is going to be $1$, which isn't anywhere near $w_0$ and doesn't prove anything at all. Can you explain what is different about the complex case that makes your limit expression valid? And what is $z_0$? You haven't defined it. You think it's some magic point that maps to $w_0$ but how do you know there is any such point? In the example I showed, the curve converges to $1$ and hits $1$. There's no hole there. But $f$ still never hits $w_0 = 1$. There's no hole at the end of the image of the curve. There is a hole, but it's somewhere else. Now it's on you to justify the limit in the complex case, if it's true. I don't happen to know offhand. Your requirement that $f$ is unbounded in every direction is a little murky. By the way, what do you mean by that? That the function's image is unbounded in every direction? Or that the function is unbounded when the inputs are constrained to a particular direction? Example: $f(z) = e^z$. That hits everything in the entire plane except for $0$, so the image is unbounded. On the other hand if you constrain the input to the imaginary axis and go in the northern direction, the output is the unit circle. It's bounded. So can you please say exactly what you mean by "unbounded in every direction?" And when did you use that hypothesis? Last edited by Maschke; November 29th, 2017 at 04:32 PM.  
November 29th, 2017, 06:11 PM  #14 
Senior Member Joined: Sep 2016 From: USA Posts: 272 Thanks: 139 Math Focus: Dynamical systems, analytic function theory, numerics 
To beat a dead horse There is no reason for $f$ to map onto the curve you have drawn at all. You don't get to assume $w_0$ is an isolated point which is not in the image. You also haven't fixed any of your previous mistakes. Its the same argument as before. Spend more time reading imo. 
November 30th, 2017, 08:28 AM  #15  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,217 Thanks: 93  Quote:
z$\displaystyle {_0}$ is the limit point of corresponding curve in z plane, and limit f(z) along the curve as you approach end point exists by definition of continuity. Unbounded isn't necessary. I was thinking of f(z)=c but later realized that case was covered. All that is required is derivative exists for all z, and the conditions of the proof. e$\displaystyle ^{z}$ is tricky for e^{z} = 0. However, it satisfies the condition of the theorem. If a ray in the wplane goes to 0, the corresponding curve* in the z plane is a horizontal line and the limit of e^$\displaystyle {z}$ as you go to the end of the line is 0. * e$\displaystyle ^{x+iy}$ = Re$\displaystyle ^{i\Theta}$, $\displaystyle \Theta$ = constant. e$\displaystyle ^{x}$ = R, y = $\displaystyle \Theta$. e$\displaystyle ^{z}$ is interesting because it's an infinite polynomial, but doesn't effect result (FTA} for finite polynomials.  
November 30th, 2017, 11:39 AM  #16  
Senior Member Joined: Aug 2012 Posts: 1,700 Thanks: 448  But your proof never uses the fact that $z$ is complex. So the real case is a counterexample to your proof until you justify otherwise. Quote:
Last edited by Maschke; November 30th, 2017 at 11:44 AM.  

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