Kenneth Kuttler

16 CHAPTER 1. SOME PREREQUISITE TOPICS

for some value of r and t. First here is a graph of this parametric function of t for t ∈ [0,2π]on the left, when r = 4. Note how the graph misses the origin 0+ i0. In fact, the closedcurve is in the exterior of a circle which has the point 0+ i0 on its inside.

-50 0 50

-50

-2 0 2

-2

-4 -2 0 2 4 6

-2

r too big r too small r just right

Next is the graph when r = .5. Note how the closed curve is included in a circle whichhas 0+ i0 on its outside. As you shrink r you get closed curves. At first, these closedcurves enclose 0+ i0 and later, they exclude 0+ i0. Thus one of them should pass throughthis point. In fact, consider the curve which results when r = 1.386 which is the graph onthe right. Note how for this value of r the curve passes through the point 0+ i0. Thus forsome t, 1.386(cos t + isin t) is a solution of the equation p(z) = 0 or very close to one.

Now here is a short rigorous proof for those who have studied analysis. The neededanalysis will be presented later in the book. You need the extreme value theorem for exam-ple.

Proof: Suppose the nonconstant polynomial p(z) = a0 + a1z+ · · ·+ anzn,an ̸= 0, hasno zero in C. Since lim|z|→∞ |p(z)|= ∞, there is a z0 with

|p(z0)|= minz∈C|p(z)|> 0

Then let q(z) = p(z+z0)p(z0)

. This is also a polynomial which has no zeros and the minimum of

|q(z)| is 1 and occurs at z = 0. Since q(0) = 1, it follows q(z) = 1+akzk +r (z) where r (z)is of the form

r (z) = amzm +am+1zm+1 + ...+anzn for m > k.

Choose a sequence, zn → 0, such that akzkn < 0. For example, let −akzk

n = (1/n) so zn =

(−ak)1/k ( 1

)1/kand Then

|q(zn)| =∣∣∣1+akzk + r (z)

∣∣∣≤ 1−1/n+ |r (zn)|

≤ 1− 1n+

∑j=m

∣∣a j∣∣ |ak|1/k

(1n

)( j−k)/k

< 1

for all n large enough because the sum is smaller than 1 whenever n is large enough, show-ing |q(zn)|< 1 whenever n is large enough. This is a contradiction to |q(z)| ≥ 1. ■

1.11 Ordered FieldsTo do linear algebra, you need a field which is something satisfying the axioms listed inTheorem 1.5.1. This is generally all that is needed to do linear algebra but for the sake ofcompleteness, the concept of an ordered field is considered here. The real numbers alsohave an order defined on them. This order may be defined by reference to the positive real

16 CHAPTER 1. SOME PREREQUISITE TOPICSfor some value of r and t. First here is a graph of this parametric function of t for t € [0, 27]on the left, when r = 4. Note how the graph misses the origin 0+ i0. In fact, the closedcurve is in the exterior of a circle which has the point 0+ i0 on its inside.r too small r just right50 > fo 4, O 2Vo vo} | : | V\ / 0-50 ~e NY 22 O 2 4 2 0 2 4 6Xx x XxNext is the graph when r = .5. Note how the closed curve is included in a circle whichhas 0+ i0 on its outside. As you shrink r you get closed curves. At first, these closedcurves enclose 0 + i0 and later, they exclude 0 + i0. Thus one of them should pass throughthis point. In fact, consider the curve which results when r = 1.386 which is the graph onthe right. Note how for this value of r the curve passes through the point 0+ 10. Thus forsome ft, 1.386 (cost + isint) is a solution of the equation p(z) = 0 or very close to one.Now here is a short rigorous proof for those who have studied analysis. The neededanalysis will be presented later in the book. You need the extreme value theorem for exam-ple.Proof: Suppose the nonconstant polynomial p(z) = ag + a1Z+ +++ +4nZ",an # 0, hasno zero in C. Since lim),_,.. |p (z)| =, there is a zo with|p (zo)| = min|p(z)| >0zECThen let g(z) =? see . This is also a polynomial which has no zeros and the minimum of|q(z)| is 1 and occurs at z= 0. Since q(0) = 1, it follows g(z) = 1+ agz* +r(z) where r(z)is of the form1(z) = ane" Fam ert +. bane" for m > k.Choose a sequence, z, —> 0, such that ayz < 0. For example, let —agz« = (1/n) so zm =(—a,)!/* (4) /K and Thenla(zn)| = |L tah +r(2)| <1 = 1/n+ |r(e)|1 1 L/k 1 (i-k)/k< 1--+4+- = 1eafor all n large enough because the sum is smaller than 1 whenever n is large enough, show-ing |g (Zn)| < 1 whenever n is large enough. This is a contradiction to |g(z)| > 1.1.11 Ordered FieldsTo do linear algebra, you need a field which is something satisfying the axioms listed inTheorem 1.5.1. This is generally all that is needed to do linear algebra but for the sake ofcompleteness, the concept of an ordered field is considered here. The real numbers alsohave an order defined on them. This order may be defined by reference to the positive real