Kenneth Kuttler

16 CHAPTER 1. PRELIMINARIES

Example 1.5.7 Factor the polynomial x3 − 27.

First find the cube roots of 27. By the above procedure using De Moivre’s theorem,

these cube roots are 3, 3(

−12 + i

√32

), and 3

(−12 − i

√32

). Therefore, x3 + 27 =

(x− 3)

(x− 3

(−1

2+ i

√3

))(x− 3

(−1

2− i

√3

)).

Note also(x− 3

(−12 + i

√32

))(x− 3

(−12 − i

√32

))= x2 + 3x+ 9 and so

x3 − 27 = (x− 3)(x2 + 3x+ 9

)where the quadratic polynomial, x2 + 3x + 9 cannot be factored without using complexnumbers.

The real and complex numbers both are fields satisfying the axioms on Page 10 and it isusually one of these two fields which is used in linear algebra. The numbers are often calledscalars. However, it turns out that all algebraic notions work for any field and there aremany others. For this reason, I will often refer to the field of scalars as F although F willusually be either the real or complex numbers. If there is any doubt, assume it is the fieldof complex numbers which is meant.

1.6 The Fundamental Theorem of Algebra

The reason the complex numbers are so significant in linear algebra is that they are alge-braically complete. This means that every polynomial

∑nk=0 akz

k, n ≥ 1, an ̸= 0, havingcoefficients ak in C has a root in in C. I will give next a simple explanation of why it isreasonable to believe in this theorem followed by a legitimate proof. The first completelycorrect proof of this theorem was given in 1806 by Argand although Gauss is often creditedwith proving it earlier and many others worked on it in the 1700’s.

Theorem 1.6.1 Let p (z) = anzn + an−1z

n−1 + · · ·+ a1z + a0 where each ak is a complexnumber and an ̸= 0, n ≥ 1. Then there exists w ∈ C such that p (w) = 0.

To begin with, here is the informal explanation. Dividing by the leading coefficient an,there is no loss of generality in assuming that the polynomial is of the form

p (z) = zn + an−1zn−1 + · · ·+ a1z + a0

If a0 = 0, there is nothing to prove because p (0) = 0. Therefore, assume a0 ̸= 0. Fromthe polar form of a complex number z, it can be written as |z| (cos θ + i sin θ). Thus, byDeMoivre’s theorem,

zn = |z|n (cos (nθ) + i sin (nθ))

It follows that zn is some point on the circle of radius |z|nDenote by Cr the circle of radius r in the complex plane which is centered at 0. Then

if r is sufficiently large and |z| = r, the term zn is far larger than the rest of the polyno-mial. It is on the circle of radius |z|n while the other terms are on circles of fixed mul-

tiples of |z|k for k ≤ n − 1. Thus, for r large enough, Ar = {p (z) : z ∈ Cr} describesa closed curve which misses the inside of some circle having 0 as its center. It won’tbe as simple as suggested in the following picture, but it will be a closed curve thanksto De Moivre’s theorem and the observation that the cosine and sine are periodic. Nowshrink r. Eventually, for r small enough, the non constant terms are negligible and so Ar

is a curve which is contained in some circle centered at a0 which has 0 on the outside.

16 CHAPTER 1. PRELIMINARIESExample 1.5.7 Factor the polynomial x — 27.First find the cube roots of 27. By the above procedure using De Moivre’s theorem,these cube roots are 3,3 (+ + ig?) , and 3 (+ — i). Therefore, x? + 27 =eafea(sea) (Cia)Note also (~-3 (= +18) (x-3(=- i@)) =a? + 32+ 9 and sox® — 27 = (a —3) (a? +3249)where the quadratic polynomial, 2? + 32 +9 cannot be factored without using complexnumbers.The real and complex numbers both are fields satisfying the axioms on Page 10 and it isusually one of these two fields which is used in linear algebra. The numbers are often calledscalars. However, it turns out that all algebraic notions work for any field and there aremany others. For this reason, I will often refer to the field of scalars as F although F willusually be either the real or complex numbers. If there is any doubt, assume it is the fieldof complex numbers which is meant.1.6 The Fundamental Theorem of AlgebraThe reason the complex numbers are so significant in linear algebra is that they are alge-braically complete. This means that every polynomial 577, anz*, n > 1,a, 4 0, havingcoefficients a, in C has a root in in C. I will give next a simple explanation of why it isreasonable to believe in this theorem followed by a legitimate proof. The first completelycorrect proof of this theorem was given in 1806 by Argand although Gauss is often creditedwith proving it earlier and many others worked on it in the 1700’s.Theorem 1.6.1 Let p(z) = anz" + @n_12”"-' + +++ +412 +49 where each ay, is a complexnumber and a, #0,n > 1. Then there exists w € C such that p(w) = 0.To begin with, here is the informal explanation. Dividing by the leading coefficient ay,there is no loss of generality in assuming that the polynomial is of the formp(z)= 2" +an 12") +s +aiz+a9If ao = 0, there is nothing to prove because p(0) = 0. Therefore, assume ap # 0. Fromthe polar form of a complex number z, it can be written as |z|(cos@+isin@). Thus, byDeMoivre’s theorem,= |z|" (cos (nO) + isin (n@))It follows that 2” is some point on the circle of radius |z|”Denote by C; the circle of radius r in the complex plane which is centered at 0. Thenif r is sufficiently large and |z| = r, the term z” is far larger than the rest of the polyno-mial. It is on the circle of radius |z|" while the other terms are on circles of fixed mul-tiples of |z|* for k < n—1. Thus, for r large enough, A, = {p(z): 2 € C;,} describesa closed curve which misses the inside of some circle having 0 as its center. It won’tbe as simple as suggested in the following picture, but it will be a closed curve thanksto De Moivre’s theorem and the observation that the cosine and sine are periodic. Nowshrink r. Eventually, for r small enough, the non constant terms are negligible and so A,is a curve which is contained in some circle centered at ag which has 0 on the outside.