Mathematical Communication, Assignment 2
Fundamental Theorem of Algebra
Alice Hedenlund
Oscar Mickelin
[email protected]
[email protected]
October 29, 2014
Mathematics students in an informal setting, for instance during a colloquium
or lunch seminar.
Target group:
The fundamental theorem of algebra is probably one of the most famous theorems in mathematics. Simply put, it states that the eld of complex numbers is algebraically closed, or that
every non-constant polynomial with complex coecients in one variable has at least one root.
Repeated polynomial division shows that this is equivalent to saying that every n-degree complex
polynomial has exactly n roots counted with multiplicity.
The rst proof of the theorem was published by d'Alembert in 1746, but it was not complete.
Other attempts were made by famous mathematicians such as Euler, Laplace and Lagrange. The
credit for producing the rst correct proof is often given to Gauss, for work he did in his doctoral
thesis in 1799. However, even his proof had gaps. Today, many widely dierent proofs exist;
some use complex analysis, other are more geometric in nature, and there are even constructive
ones. However, common among all of them is that they involve some analysis, or at least some
topological concept like continuity. Any proof must in some way use the completeness of the real
numbers, so it is not possible to construct an entirely algebraic proof. Hence, it could be argued
that the theorem is neither fundamental nor algebraic at all, at least not in the sense that we
view algebra today.
The proof below is one that employs only a small amount of analysis and no major theorems
from complex analysis.
Fundamental Theorem of Algebra.
If f (z) = a0 + a1 z + . . . + an z n ∈ C[z] is non-constant, then f (z0 ) = 0 for some z0 ∈ C.
For the sake of contradiction, assume that f (z) 6= 0 for all z ∈ C. To begin with, we
observe that |f (z)| → ∞ when |z| → ∞, since the reverse triangle inequality gives that
Proof.
|f (z)| ≥ |z|n − |an−1 ||z|n−1 − . . . − |a0 | → ∞,
(1)
as |z| tends to innity. Hence, there is some R > 0 such that |f (z)| ≥ |f (0)| if |z| > R. Now, the
disc |z| ≤ R is closed and bounded, and z 7→ |f (z)| is a continuous function, so it must attain a
minumum on this disc. Let z0 be the point on the disc such that |f (z)| is minimal, that is
|f (z0 )| ≤ |f (z)| for all |z| ≤ R.
(2)
Next, note that the point z = 0 has absolute value less than R, so we certainly have |f (z0 )| ≤
|f (0)|, since |f (z0 )| was minimal. If we now take any z with |z| ≥ R, this implies that |f (z0 )| ≤
1
|f (0)| ≤ |f (z)|, by the choice of R. Combining this with Eq. (2) and the assumption that f has
no roots results in
0 < |f (z0 )| ≤ |f (z)|,
(3)
for all z ∈ C.
Now, without loss of generality, we may assume that z0 = 0 and f (0) = 1, possibly after
replacing f (z) by (f (z0 ))−1 f (z + z0 ). We can then write
f (z) = 1 + ak z k + · · · + an z n ,
(4)
where k ≥ 1 and ak 6= 0. Choose ω ∈ C such that |ω| = 1 and such that ak ω k = −|ak |, as shown
in Fig. 1.
Im(z)
ak
ω
θ
θ/k
Re(z)
1
|ak|
Figure 1: Figure showing the choice of ω as a function of ak .
Choose also ε < 1 such that εk |ak | < 1. It then follows from Eq. 3 that
k
k+1 k+1
n n
ε
+ . . . + an ω ε |f (z0 )| = 1 ≤ |f (εω)| = 1 − ε |ak | + ak+1 ω
k
≤ 1 − ε |ak | + εk+1 (|ω|k+1 |ak+1 | + . . . + |ω|n |an |)
= 1 − εk |ak | + εk+1 (|ak+1 | + · · · + |an |).
This implies that
|ak | ≤ ε(|ak+1 | + · · · + |an |).
(5)
(6)
(7)
(8)
If we nally let ε → 0 in Eq. (8), we obtain ak = 0, which is a contradiction. The assumption
that f (z) 6= 0 for all z ∈ C must then be wrong, which nishes the proof.
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