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November 18th, 2017, 09:21 AM   #11
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Quote:
Originally Posted by zylo View Post

{z} $\displaystyle \subset$ {f(z)} T
z $\displaystyle \rightarrow$ f(z) T
$\displaystyle z^{2} \rightarrow zf(z)$ T
{$\displaystyle z^{2}$} $\displaystyle \subset$ {zf(z)} F
For last step to be true, you have to show $\displaystyle z^{2}$ $\displaystyle \epsilon$ {zf(z)}, which entails showing zf(z) maps to entire complex plane, which is what you are trying to prove.

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.
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November 29th, 2017, 11:22 AM   #12
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Let w=f(z) and f'(z) be defined for all z and f(z) unbounded in any direction in the w-plane.

Assume there is a point w$\displaystyle _{0}$ which f(z) doesn't map to.

Draw a curve in w-plane ending at w$\displaystyle _{0}$. The corresponding curve in z-plane 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:
$\displaystyle z^{n}+a_{n-1}z^{n-1}+..+a_{1}z$ maps to entire complex plane. FTA

Last edited by zylo; November 29th, 2017 at 11:53 AM.
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November 29th, 2017, 03:42 PM   #13
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Quote:
Originally Posted by zylo View Post
Let w=f(z) and f'(z) be defined for all z and f(z) unbounded in any direction in the w-plane.

Assume there is a point w$\displaystyle _{0}$ which f(z) doesn't map to.

Draw a curve in w-plane ending at w$\displaystyle _{0}$. The corresponding curve in z-plane is open at z$\displaystyle _{0}$.

But $\displaystyle \lim_{z \rightarrow z_{0}} = f(z_{0}) = w_{0}$. Contradiction.
This doesn't make a lot of sense without further explanation. Consider the analogous case in the reals. Let $f(x) = x^2$. Then $f$ is defined and differentiable for all reals.

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.
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November 29th, 2017, 06:11 PM   #14
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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.
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November 30th, 2017, 08:28 AM   #15
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Quote:
Originally Posted by Maschke View Post
Consider the analogous case in the reals. Let $f(x) = x^2$.

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.

$f(z) = e^z$. That hits everything in the entire plane except for $0$, so the image is unbounded.
f(z)=z$\displaystyle {^2}$ hits -1.

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 w-plane 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.
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November 30th, 2017, 11:39 AM   #16
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Quote:
Originally Posted by zylo View Post
f(z)=z$\displaystyle {^2}$ hits -1.
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:
Originally Posted by zylo View Post

e$\displaystyle ^{z}$ is tricky for e^{z} = 0. However, it satisfies the condition of the theorem. If a ray in the w-plane 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.
Yet $e^z$ never hits zero. It's a counterexample to your flawed proof. You've had this pointed out to you about 20 times already.

Last edited by Maschke; November 30th, 2017 at 11:44 AM.
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