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Complex Polynomials and Minimal Values

  1. Mar 19, 2007 #1
    Hello, I'm trying to prove the fundamental thereom of algebra using the minimal modulous principal. There is one step that I cannot make sense out of logically and I was wondering if one of you could explain it to me.

    If the limit as |z| approaches infinity implies |P(z)| approaches infinity for some non-constant polynomial then there exists an a in C such that |P(a)| <or= |P(z)| for all z. Why does this follow?
  2. jcsd
  3. Mar 19, 2007 #2


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    I don't quite remember exactly how it goes, but I remember it using 1/|P(z)|.

    I'd have to wait until I get home to find out how it went.
  4. Mar 19, 2007 #3
    So, after reading wikipedia for awhile. It follows since you can find a closed disk around 0 with radius r where |f(z)|>or=|f(0)| where |z|>r since the limit goes to infinity. And since the disk is a compact set it must contain a minimum value of z_0. Thus |f(z_0)| <or= |f(x)| for all x in C.

    Note: z_0 is on interior of the disk not the boundary.
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