October 17th, 2017, 02:04 PM  #11  
Senior Member Joined: Aug 2012 Posts: 1,620 Thanks: 411  Quote:
Only because of the lax immigration policies of those commie liberals. Last edited by Maschke; October 17th, 2017 at 02:08 PM.  
October 17th, 2017, 04:24 PM  #12  
Senior Member Joined: Aug 2012 Posts: 1,620 Thanks: 411  Quote:
$f(z + \Delta z)$ with $f(z) + f'(z) \Delta z$ that is an approximation and not an equality. We may only write $f(z + \Delta z) \approx f(z) + f'(z) \Delta z$. You can get them within $\epsilon$ of each other by suitably small choice of $\Delta z$; but there is never equality except in the limit as $\Delta z \to 0$. To see why this is so let's review the derivation. The derivative of a function of a complex variable is defined by the same formula as that of a function of a real variable: (*) $\displaystyle f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z)  f(z)}{\Delta z}$ [Aside: In the complex case the limit is only defined if $\Delta z$ can approach $0$ from any direction in the plane. This turns out to be a profound restriction, making complexdifferentiable functions behave much more nicely than realdifferentiable ones. For example complexdifferentiable functions are infinitely differentiable and always represented by their power series.] Now we'd like to multiply both sides of (*) by $\Delta z$, but $\Delta z$ appears on both sides of the equation bound by the limit expression. Therefore we must write $\displaystyle \lim_{\Delta z \to 0} f'(z) \Delta z = \lim_{\Delta z \to 0} (f(z + \Delta z)  f(z)) = \lim_{\Delta z \to 0} f(z + \Delta z)  f(z)$ and then $\displaystyle \lim_{\Delta z \to 0} f(z + \Delta z) = f(z) + \displaystyle \lim_{\Delta z \to 0} f'(z) \Delta z $ There is only equality in the limit. For $\Delta z \neq 0$ at best we have only approximate equality, I don't know if this affects the rest of your argument, which is only an outline of a much more rigorous argument that you have not given. Also as noted the same argument proves a falsehood in the real case, which tells us that an error lurks within the vagueness. 2) I have another quibble. You write Quote:
$F(z+\Delta z) = (z+ \Delta z)f(z+ \Delta z) = (z+ \Delta z)(f(z)+f'(z) \Delta z)$ and using the distributive law the last term is $(z+ \Delta z)f(z) + (z+ \Delta z)f'(z) \Delta z)$ Then the second line isn't right, it should be $F(z + \Delta z) = F(z) + (z+ \Delta z) f'(z) \Delta z$ and then changing the $=$ to $\approx$ per the first quibble. 3) I don't follow your border crossing argument but if $F$ misses a point then there's a little region around $F(z)$ for any $z$ that also misses that point. That's because a point is a closed set and its complement is open, so $F(z)$ is always an interior point of the complement of a hole in the plane. I don't know if that's relevant, I really can't understand what you're saying. $e^z$ misses $0$ but hits everything arbitrarily close to $0$. For all you know, $F$ does the same thing. Last edited by Maschke; October 17th, 2017 at 05:20 PM.  
October 18th, 2017, 10:06 AM  #13  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  Quote:
Quote:
 
October 18th, 2017, 11:53 AM  #14  
Senior Member Joined: Aug 2012 Posts: 1,620 Thanks: 411  Quote:
Also I pointed out two specific errors in your chain of equalities that you did not address.  
October 18th, 2017, 12:51 PM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra  
October 18th, 2017, 07:36 PM  #16  
Senior Member Joined: Sep 2016 From: USA Posts: 197 Thanks: 105 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
This doesn't matter however because he is trying to apply these to prove surjectivity of the map which is absolutely unrelated. After trying to decipher it, the nearest approximation I have to his misunderstanding is that he thinks he can do induction on $\mathbb{C}$ and is trying to begin with the assumption that his function maps onto a punctured neighborhood of a fixed $c \in \mathbb{C}$ and use this to prove it to $c$ as well. Not only is this assumption complete nonsense, but it has already been pointed out that $f(z) = e^z$ maps onto a punctured neighborhood of 0 yet 0 is not in the image. This has only caused him to double down on his nonsense and I feel compelled to reply here again only so that unfortunate students who don't know better don't mistake it for math.  
November 10th, 2017, 08:41 AM  #17 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 
Theorem: If F'(z) exists for all z, F(z) maps to entire complex plane w. Proof: Assume map of F(z) has a hole in w. Let F(z$\displaystyle _{0}$) be near border of the hole. F(z) defined in a neighborhood of z$\displaystyle _{0}$. F(z$\displaystyle _{0}$+z)$\displaystyle \approx$ F(z$\displaystyle _{0}$)+F'(z$\displaystyle _{0}$)$\displaystyle \Delta$z F(z) crosses the border. Contradiction. F(z) maps to entire complex plane. Which is also Fundamental Theorem of Algebra if F(z) is a complex polynomial. Example: y=cosx has a derivative for all x but does not map to all of y. No solution for y=2. w=cosz has a derivative for all z and maps to all of w. A solution for any w, including 2. REF: Fundamental Theorem of Algebra Proof Post #46 Last edited by zylo; November 10th, 2017 at 08:44 AM. Reason: Add REF 
November 10th, 2017, 10:31 AM  #18 
Senior Member Joined: Aug 2012 Posts: 1,620 Thanks: 411  $e^z$ is a counterexample. How many times has that already been mentioned?
Last edited by Maschke; November 10th, 2017 at 10:33 AM. 
November 10th, 2017, 06:22 PM  #19  
Senior Member Joined: Aug 2012 Posts: 1,620 Thanks: 411  Quote:
With $F(z) = e^z$, we have that $F$ is defined and complexdifferentiable on the entire plane, and fails to hit $0$. Is that what you mean by a "hole?" You haven't really said, but I'll assume this fits your definition. If not, say what a hole is. Ok. Let's take some number really really close to $0$ in the $w$plane, say $e^{10^6}$. This is a number on the positive real axis, very close to $0$. In this case $z_0 = 10^6$, or negative one million. A number way out there to the left on the real axis. In fact the only way to get $e^z$ close to zero is for $z$ to approach $\infty$. I hope you can see that numbers very far to the left on the negative real axis get mapped to a small interval to the right of $0$ on the positive real axis. To visualize how neighborhoods work, it's helpful to think in terms of rectangles. A rectangle about a point gets mapped to a wedge shape truncated by concentric circles. It never crosses any border, never hits $0$. It's helpful to play around with this visualization app. https://www.geogebra.org/m/Ac43DNNP Here's the image of a rectangle around the origin. As you move the rectangle to the left, it gets smaller. As you move it up or down, it rotates around a circle. If you moved the rectangle leftward so that it was centered on $10^6$, the wedge would be very small and its left end would be very close to $0$. But $0$ would still be outside the wedge. Because $e^z$ never hits $0$. Yes that's fine. You can take any neighborhood of $10^6$ and map it to an interval to the right of $0$ on the real axis. But you can see that no "border crossing," whatever the heck that means, is occurring. Rectangular neighborhoods of points way out to the left get mapped to small wedges to the right of $0$. [Circular neighborhoods are very tricky, I'm still trying to visualize those!] Quote:
$F(z  10^6) \approx e^{10^6} + e^{10^6} \Delta z$. Which tells us what, exactly? What does that mean? What border has been crossed? There's nothing in this example or in the visualization of the image of a neighborhood of $z_0$ that fits a picture of "crossing a border." On the contrary, the wedges get close to zero and rotate around zero but no wedge ever contains zero. What is the contradiction? Since we already know that $F$ misses $0$, your conclusion is false. Can you shed some light on your argument? It really makes no sense and has a very obvious counterexample. Last edited by Maschke; November 10th, 2017 at 06:33 PM.  
November 13th, 2017, 08:13 AM  #20 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 
Let w=F(z) have a derivative for all of z. Assume there is a point w$\displaystyle _{p}$ which F(z) doesn't map to. I can define F'{z) at any point w$\displaystyle _{0}$ in a neighborhood of w$\displaystyle _{p}$, and w$\displaystyle _{0}$ has a neighborhood which doesn't include w$\displaystyle _{p}$. Visually, draw a circle around w$\displaystyle _{p}$. Draw w$\displaystyle _{0}$ inside the circle. There is a circle around w$\displaystyle _{0}$ which doesn't include w$\displaystyle _{p}$. ie, the initial assumption doesn't lead to a contradiction, and F'(z) exists for all z doesn't imply F maps z to all of w. Particularly,from definition of derivative: $\displaystyle \Delta zF'\epsilon< \Delta F < \Delta zF'+\epsilon$ if $\displaystyle \Delta z < \delta$ ie, if $\displaystyle \Delta F$ sufficiently small for F to be defined. Same argument applies for zf(z) where f(z) has a derivative for all z, and f(z) maps to entire complex plane. Because you don't know that (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z) maps to w$\displaystyle _{p}$. 

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