My Math Forum Fundamental Theorem of Algebra Proof 02

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December 5th, 2017, 04:24 AM   #1
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Fundamental Theorem of Algebra Proof 02

Quote:
 Originally Posted by zylo Let w=f(z), and f'(z) be defined for all z 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

w=e$\displaystyle ^{z}$ was offered as contradiction because assertedly w=0 doesn't exist. w=0 is perfect example of isolated point of the Theorem above.

Consider the rays in w heading toward the origin: $\displaystyle w=Re^{i\Theta}, \Theta = constant, R \rightarrow 0$. The corresponding "curve" in z is $\displaystyle e^{z}= Re^{i\Theta}$, or
$\displaystyle e^{x}=R$ and $\displaystyle y=\Theta$ =constant.
$\displaystyle R \rightarrow 0$, if $\displaystyle e^{x} \rightarrow 0$, which it does if $\displaystyle x \rightarrow -\infty$, and the limit exists,
ie, limit point w=0 of the line approaching origin exists, conditions of the theorem above are satisfied, and e$\displaystyle ^{z}$ maps to all of w.

If you are talking about maps from all of z to all of w, you have to deal with infinity, ie, possibility of existence of limits as something gets unboundedly large.

December 5th, 2017, 05:38 AM   #2
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Quote:
 Originally Posted by zylo If you are talking about maps from all of z to all of w, you have to deal with infinity, ie, possibility of existence of limits as something gets unboundedly large.
No, because $\infty \not \in \mathbb C$. If something gets unboundedly large, it has no limit by definition. Moreover, the limit is not part of the range of a function. The real function $f(x)=\frac1x$ with domain $D=\{x \in \mathbb R: x \gt 1\}$ has, as it's range, the open interval $(0,1)$. It does not include either $0$ or $1$ even though they are limits because the function never attains those values.

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