Subject: Re: [HM] Fundamental Theorem of Algebra
From: John Conway (conway@math.Princeton.EDU)
Date: Sat Apr 08 2000 - 14:05:31 EDT
The simplest proof of the "fundamental theorem" has always
been this:
1) prove that for any real C we can find an R such that
|f(z)| > C for |z| > R.
{This is easy - if the polynomial is
z^n + az^n-1 + ... + k
then if |z| > C( 1 + |1| + ... + |k|) I think we're OK.
2) So the infimum of |f(z)| over the whole plane equals
that over some closed disc |z| =< R.
3) Prove or quote that the infimum of a continuous function
over a closed and bounded set is attained. [The most
easily-comprehended proof is by the bisection method
of the Bolzano-Weierstrass theorem.]
4) So we can suppose that there's a particular value Z
of z at which |f(z)| is minimal. Shift to make Z = 0.
We now want to prove f(0) = 0.
5) If not, we can suppose f(0) = 1, so f looks like
1 + az + bz^2 + ... .
Now if only |z| is sufficiently small we have
|bz^2 + ... | < |az|/2.
6) Now by chose of arg(z) we can make az be negative real,
and then if it's small enough, it's easy using 5) to
see that | 1 + az + bz^2 + ... | < 1, contradicting
our assumption.
John Conway
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