tag:blogger.com,1999:blog-9004452969563620551.post6136951424478101352..comments2017-11-23T14:25:22.867+05:30Comments on Paramanand's Math Notes: Fundamental Theorem of Algebra: Two Proofstesternoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-9004452969563620551.post-51100808487669173392013-11-01T14:55:03.353+05:302013-11-01T14:55:03.353+05:30@adnoctem,
I am aware of this proof via Liouville&...@adnoctem,<br />I am aware of this proof via Liouville's theorem, but to comprehend it you need to be aware of Liouville's theorem first. The proof of Liouville's theorem itself hinges on the Cauchy's integral theorem and bounds of the first derivative of an analytic function. I believe such a proof of fundamental theorem of algebra is difficult and looks too much "complex analytic" compared to the more algebraical proof which I have presented.<br /><br />Anyway thanks for sharing your views on this post.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-38375314321575302322013-10-31T21:38:08.074+05:302013-10-31T21:38:08.074+05:30There is a proof in the complex variable theory. T...There is a proof in the complex variable theory. This follows from Liouville's Theorem: A bounded function that is holomorphic everywhere in the complex plane, must necessarily be a constant function. The Fundamental Theorem of Algebra is now proven as follows. Let $f(z)$ be a polynomial function without any roots in $\mathbb C$. Then, $1/f$ satisfies all the criteria for Liouville's theorem and is therefore constant; therefore $f$ must be constant. <br />adnoctemhttp://adnoctem.wordpress.com/noreply@blogger.com