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One important case of the Fundamental Theorem of A

 
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wantian3688



Joined: 24 Feb 2013
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PostPosted: Thu Mar 21, 2013 5:37 am    Post subject: One important case of the Fundamental Theorem of A Reply with quote

RKEY,RKEYOne important case of the Fundamental Theorem of Algebra is that every nonconstant polynomial with real coefficients must have at least one complex root. Since it is not true that every such polynomial has to have at least one real root (as the example demonstrates),[url=http://oakleypascher.over-blog.com]lunette oakley pas cher[/url], many mathematicians feel that the complex numbers form the most natural setting for working with polynomials.
In fact,[url=http://lunettes-de-soleil.webnode.fr]chanel pas cher[/url], a stronger version of the Fundamental Theorem of Algebra is also true: a polynomial of degree can be factored completely into a product of linear polynomials:
A startlingly simple proof is based on Liouville's theorem: If is a polynomial function of a complex variable then both and will be holomorphic in any domain where ,[url=http://lunettes-de-soleil.webnode.fr]lunette chanel[/url]. By Liouville's theorem,[url=http://lunettes-de-soleil.webnode.fr]chanel en ligne[/url], it must be constant,2013 the style One important case of the Fundament, so must also be constant.
Using algebra (and a bit of real analysis)
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Using the fundamental group of the punctured plane
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Using the second homotopy group of the Riemann sphere
Without loss of generality the leading coefficient of is 1. We consider as a map from the Riemann sphere to itself (taking infinity to infinity). By considering the homotopy , where , this map is homotopic to the map . Hence it suffices to show that the map is not null homotopic. However, in the homotopy group we have , and so it suffices to show that is not null homotpic, which is equivalent to the fact that the sphere is not contractible.
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