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December 5th, 2017, 05:24 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, 06:38 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. Tags algebra, fundamental, proof, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zylo Complex Analysis 50 December 5th, 2017 09:10 AM zylo Complex Analysis 15 November 30th, 2017 11:39 AM AspiringPhysicist Abstract Algebra 3 October 2nd, 2014 01:19 AM king.oslo Algebra 2 September 19th, 2013 11:48 PM johnny Algebra 11 December 21st, 2007 10:51 PM

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