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THE FUNDAMENTAL THEOREM OF ALGEBRA

DAVID GLYNN

1. Introduction

Theorem 1.1. Every non-constant polynomial with complex coefficients has at least one root in the field of complex numbers.

The above theorem is known as the Fundamental Theorem of Algebra. It is a theorem which has garnered the contributions of some of the greatest names in mathematics such as Gauss, Cauchy, Liouville and Laplace. With all this interest, it is unsurprising that numerous proofs of the theorem have been discovered. The proof in this report is an elegant and comparably short argument which appears in [1, Chapter 19]. There is also a discussion of a topological approach to proving the theorem, as given by Peter Phelan during his talk.

2. Important Preliminary Facts

First we shall establish some important facts necessary to complete the proof.

They are generally covered in a first-year analysis course and so the proofs are omitted.

(1) Polynomials are continuous functions.

(2) Let z ∈ C have absolute value 1. Then for any positive integer m there existsζ∈Cwithζm=z. ζ is called them-th root of unity ofz.

(3) A continuous function f on a compact set S attains a minimum inS.

Now we proceed to the first step of our argument, a proof of what is commonly known as d’Alembert’s Lemma.

3. d’Alembert’s Lemma Letp(z) =Pn

k=0ckzk be a complex polynomial of degreen≥1.

Lemma 3.1 (d’Alembert). If p(a) 6= 0 then every disk D around a contains an interior point b with|p(b)|<|p(a)|.

Proof. Suppose the diskD has radius R. Therefore all points in the interior ofD are of the forma+wwith|w|< R. First we shall demonstrate the following (1) p(a+w) =p(a) +cwm(1 +r(w)),

where c is a nonzero complex number, 1 ≤ m ≤ n, and r(w) is a polynomial of degreen−mwithout constant term.

1

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2 DAVID GLYNN

Using the binomial theorem, we get p(a+w) =

n

X

k=0

ck(a+w)k =

n

X

k=0

ck

k

X

i=0

k i

ak−iwi

=

n

X

i=0

(

n

X

k=i

k i

ckak−i)wi (Rewriting the double summation)

=

n

X

k=0

ckak+

n

X

i=1

(

n

X

k=i

k i

ckak−i)wi (Taking out thei= 0 part)

=p(a) +

n

X

i=1

diwi

Letm= min{i≥1 :di6= 0}, setc=dm and factorcwm out to get p(a+w) =p(a) +cwm(1 +r(w))

The next step is to bound|cwm| and|r(w)|from above.

Letρ1:= mp

|p(a)/c, so if |w|< ρ1then|cwm|<|p(a)|.

Also sincer(w) is continuous andr(0) = 0, we know that there existsρ2such that

|r(w)|<1 whenever|w|< ρ2.

Hence for|w|less thanρ= min(ρ1, ρ2) we have

(2) |cwm|<|p(a)|

⇒ |cwm|

|p(a)| <1

(3) |r(w)|<1

Now consider the quantity−|p(a)/c|p(a)/c . Clearly it has absolute value 1, so there exists ζ∈Cwithζm=−|p(a)/c|p(a)/c.

Let∈Rsatisfy 0< <min(ρ, R). Settingw0=ζ, consider the pointb=a+w0. The pointbis inD since|w0|= < R, and we claim that it satisfies|p(b)|<|p(a)|.

By (1) we have,

(4) |p(b)|=|p(a+w0)|=|p(a) +cwm0 (1 +r(w0))|, Now define a factorδ by

cwm0 =cmζm=− m

|p(a)/c|p(a) =−δp(a), by (2) ourδ satisfies

0< δ=m |c|

|p(a)| <1 , since < ρ

Using the triangle inequality we get for the right-hand term of (4)

|p(a) +cwm0(1 +r(w0))|=|p(a)−δp(a)(1 +r(w0))|

=|(1−δ)p(a)−δp(a)r(w0)|

≤(1−δ)|p(a)|+δ|p(a)||r(w0)|

<(1−δ)|p(a)|+δ|p(a)|=|p(a)|

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THE FUNDAMENTAL THEOREM OF ALGEBRA 3

where we have used (3) to get the strict inequality. Thus we have arrived at the desired result

|p(b)|<|p(a)|

4. Proof of the Fundamental Theorem

Now that we have established the validity of d’Alembert’s Lemma, the proof of the theorem follows quite quickly. First we note that|p(z)|goes to

infinity as |z| → ∞. Clearly, p(z)z−n = Pn

k=0ckzk−n approaches the leading coefficientcn as |z| → ∞. Therefore|p(z)|must go to infinity as|z| → ∞.

Consequently, there exists R >0 such that |p(z)|> |p(0)| for all points z on the circle {z : |z| =R}. Since p(z) is continuous and the disk D = {z : |z| ≤R} is compact,|p(z)| attains a minimum value at somez0∈D. Because |p(z)|>|p(0)|

for all z on the boundary of D, z0 must lie in the interior. But by d’Alembert’s Lemma we know that in any disk around z0 ∈ Do we can find a point z1 with

|p(z1)| < |p(z0)|, so long as |p(z0)| 6= 0. This would be a contradiction to the minimality of|p(z0)|, so we must have thatp(z0) = 0.

5. A topological argument

The following argument was outlined at the end of Peter Phelan’s talk, and draws on elements from topology.

First, we need to introduce the concept of the winding number of a closed curve.

The winding number of a closed curve around a given point is an integer representing the total number of times that curve travels anti-clockwise around the point. The sign of the number depends on the orientation of the curve, it is negative if the curve travels clockwise around the point. For example, the winding number of z(t) =eit,0 ≤t ≤2πis 1 while the winding number of z(t) =e−it,0≤t≤4π is -2.

Now suppose that the polynomialp(z) has no roots,p(z)6= 0 for allz∈C. We will think ofp(z) as a map from the complex plane to the complex plane. It maps any circleC={z∈C:|z|=R}to a closed curveγR.

Figure 1. A circle mapped to a closed curve inC

What happens to the winding number of γR when R is very large and when R= 0? WhenRis sufficiently large, the leading termzn dominates all other terms in p(z). The curve z(θ) =Re (0 ≤θ ≤2π) revolves once anti-clockwise around

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4 DAVID GLYNN

the origin, so it has a winding number of 1 around (0,0). Thereforez(θ)n=Reinθ revolvesntimes anti-clockwise around the origin, so it has a winding number ofn around (0,0). For sufficiently largeR,γR also windsntimes around the origin, as the leading term dominates all other terms. If|z|= 0, thenγ0 is simply the point p(0), which we assumed is nonzero. Therefore the winding number ofγ0around the origin is 0. If we changeRcontinuously thenγRwill deform continuously. However we know that at someR the winding number must change, which only happens if the curveγR includes the origin (0,0). But then for somez0 on the circle|z|=R we have p(z0) = 0. This contradicts our assumption thatp(z) has no roots, so it must have at least one.

References

[1] Martin Aigner and Gunter M. Ziegler. Proofs from The Book. Springer-Verlag, Berlin, fifth edition, 2014. Including illustrations by Karl H. Hofmann.

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