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December 5th, 2017, 05:24 AM  #1  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91  Fundamental Theorem of Algebra Proof 02 Quote:
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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra  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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