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Trouble understanding the proof of fundamental theorem of algebra using

  1. Oct 26, 2011 #1
    *using complex analysis. (sorry couldn't post the entire thing in the thread title). It seems there is a common proof of the fundamental theorem of algebra using tools from complex analysis, namely Louisville's theorem and the max. modulus theorem. Here is an example proof from proof wiki: (by the way, is this considered a homework type problem? I apologize if it doesn't belong here)

    Let p:C→C be a complex polynomial with p(z)≠0 for all z∈C.
    Then p extends to a continuous transformation of the Riemann sphere Cˆ=C∪{∞} (and this extension also has no zeros).
    Since the Riemann sphere is compact, there is some ε>0 such that |p(z)|≥ε for all z∈C.

    Now consider the holomorphic function g:C→C defined by g(z):=1/p(z).
    We have |g(z)|≤1/ε for all z∈C.
    By Liouville's Theorem, g is constant. Hence p is also constant, as claimed.

    I follow everything until the very end. This is a proof by contradiction, how does it contradict things that p is constant? This is going to sound like a really stupid question but couldn't a complex polynomial be a constant function so that any variable you plug in for x gives the same result because of the choice of coefficients? I am having a lot of difficulties in forming thoughts about functions of complex variables, maybe because I am just not used to them yet and so have a hard time being able to visualize them immediately.


    Sorry if this sound incredibly dumb. It's just my brain is fried and for some reason I am just not getting this. Also, I am having a bit of a difficulties differentiating (no pun intended) between analytic functions and holomorphic functions.
     
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  3. Oct 26, 2011 #2

    HallsofIvy

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    The thing you may be missing, certainly you don't give it here, is the actual statement of the "fundamental theorem of calculus. Part of the hypothesis must be that p(x) is NOT constant. Obviously, the polynomial p(x)= 1 is never equal to 0.
     
  4. Oct 26, 2011 #3
    thank you for the reply. Yeah, I know that the degree of the poly has to be ≥ 1. But I thought a constant function would be any function that regardless of the input you get the same number. I knwo this sounds really stupid but, can that ever happen with a complex function, i mean where you actually have a polynomial of degree greater than or equal to 1 but that the result is the same no matter what you plug in because of the coefficients in the poly?


    I guess it's that, I keep getting messed up with complex numbers. There is no order among th complex numbers, so sometimes I feel like I dont know what assumptions I can make!
     
  5. Oct 26, 2011 #4
    You cannot. If a polynomial over any infinite field has degree at least 1, it must be nonconstant. This statement does require proof, however, and it is good of you to consider the possibility that it isn't true.

    If you just want to prove the statement for the complex numbers, the easiest thing to do is to take limits as z→∞ in any direction. Specifically, if f(z) is a polynomial of degree n>0, then [itex]\frac{f(z)}{z^n}[/itex] looks like [itex]c_n + \frac{c_{n-1}}{z} + \frac{c_{n-2}}{z^2} + \cdots + \frac{c_0}{z^n}[/itex]. Each of the fractions approach zero, so the limit is [itex]c_n[/itex], which is not zero (since by hypothesis, f has degree n). But if f is constant, then [itex]\lim_{z\rightarrow \infty} \frac{f(z)}{z^n} = 0 \neq c_n[/itex]. So f is not constant.

    If you want to prove the statement for all infinite fields, you can use the fact that if r is a root of a polynomial, then (z-r) is a factor of it. So factoring the polynomial into irreducible factors, this shown there are at most n roots of a polynomial of degree n≥0. But if f is a constant polynomial of degree n>0, then there is some [itex]c\in \mathbb{C}[/itex] such that f(z)=c everywhere. So then f(z) - c is a polynomial of degree n which is zero everywhere, and so has infinitely many roots -- contradicting the fact that a polynomial of positive degree only has finitely many roots. This is in many ways a simpler proof, but it requires knowing that polynomials factor uniquely into irreducibles, and I don't know whether you've proven that fact yet.
     
  6. Oct 26, 2011 #5
    ooooh thank you thank you thank you!! Thanks to both of you who replied. Yeah this clears up all my doubts! And I don't feel so stupid anymore lol I have a big problem of hesitating even for basic things I think (like the fact that a poly of degree 1 or greater can't be positive) I am so happy to have this resolved! :D
     
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