From ca7d72c0f8b072a8768fdff641c3e87f52d84b7c Mon Sep 17 00:00:00 2001
From: Juan Carlos Ponce Campuzano <37394697+jcponce@users.noreply.github.com>
Date: Sat, 30 Nov 2024 17:55:40 +1000
Subject: [PATCH] Update fundamental-theorem-of-algebra.html
---
content/fundamental-theorem-of-algebra.html | 40 ++++++++++++++++-----
1 file changed, 32 insertions(+), 8 deletions(-)
diff --git a/content/fundamental-theorem-of-algebra.html b/content/fundamental-theorem-of-algebra.html
index 6f28c8a..c9c3770 100644
--- a/content/fundamental-theorem-of-algebra.html
+++ b/content/fundamental-theorem-of-algebra.html
@@ -211,7 +211,8 @@ Proof & historical notes
\]
By making $r$ arbitrarily large, we can make $\abs{f'(z)}$
as small as we wish. This means that $f'(z_0)= 0$ for all points
- $z_0$ in the complex plane. Hence, by Theorem 3
+ $z_0$ in the complex plane. Hence, by
+ Theorem 3
in the Complex Differentiation section, $f$ must be constant.
@@ -234,16 +235,39 @@ Proof & historical notes
\[
f(z) = \frac{1}{p(z)}
\]
- is an entire function. For $\abs{z}\to \infty$ we see that
- $\abs{f(z)}\to 0.$
+ is an entire function.
+ Since $\abs{p(z)}\to \infty$ as
+ $\abs{z}\to \infty,$ there exists $M,$ $R\gt 0$ such that
+ $\abs{p(z)}\gt M$ if $\abs{z}\gt R.$
+ This implies that for $\abs{z}\gt R,$
+ \[
+ \abs{f(z)} = \abs{\frac{1}{p(z)}} \lt \frac{1}{M}.
+ \]
+ Now, $f$ is continuous on the region $\abs{z}\leq R,$ which is closed and bounded.
+ By Theorem 3, in the Continuity section,
+ $f$ is also bounded for $\abs{z}\leq R.$
Hence $f(z)$ is bounded in the entire complex plane.
- Liouville's Theorem then implies that $f(z)$ is a constant, and consequently
- $p(z)$ is a contant as well.
- But this is a contradiction to our underlying assumption that
- $p(z)$ was not a constant polynomial. We conclude that there must
+ Liouville's Theorem then implies that $f(z)$ is a constant,
+ and consequently $p(z)$ is a constant as well.
+ But this is a contradiction. We conclude that there must
exist at least one number $z$ for which $p(z) = 0.$
+
+
+ It follows in the usual way that any polynomial $p(z)$ of degree
+ $n,$ with complex coefficients, can be expressed as a
+ product of linear terms:
+ \begin{eqnarray}\label{factorized}
+ p(z) = c\left(z-\alpha_1\right)\left(z-\alpha_2\right)\cdots \left(z-\alpha_n\right),
+ \end{eqnarray}
+ where $c$ and $\alpha_k$ $(k=1,2,\ldots, n)$ are complex constants.
+ Of course, some of the constants $\alpha_k$ in expression (\ref{factorized}) may
+ appear more than once, and it is clear that $p(z)$ cannot have
+ more than $n$ distinct zeros.
+
+
+
According to B. Fine & G. Rosenberger
[7, Chapter 1],
@@ -326,7 +350,7 @@
References
Boas, Jr., R. P. (1964). Yet another proof of the fundamental theorem of algebra, Amer. Math. Monthly, 71, 180.
Byl, J. (1999). A simple proof of the fundamental theorem of algebra, Internat. J. Math. Ed. Sci. Tech. 30, no. 4, 602-603.
Fefferman, C. (1967). An easy proof of the fundamental theorem of algebra, Amer. Math. Monthly, 74, 854-855.
- Fine, B. & Rosengerger. G. (1997). The Fundamental Theorem of Algebra. New York: Springer-Verlag Inc.
+ Fine, B. & Rosenberger. G. (1997). The Fundamental Theorem of Algebra. New York: Springer-Verlag Inc.
Gauss, C. F. (1799). Demonstratio nova theorematis omnem functionem algebraicum rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse, Helmstedt dissertation, reprinted in Werke, Vol. 3, 1-30.
Gauss, C. F. (1850). Beiträge zur Theorie der algebraischen Gleichungen". Abhandlungen der Königlichen Gesellschaft der Wissenschaften zu Göttingen., 4: 34-35.
Redheffer, R. M. (1964). What! Another note just on the fundamental theorem of algebra?, Amer. Math. Monthly, 71, 180-185.