Kenneth Kuttler

12 CHAPTER 1. SOME PREREQUISITE TOPICS

-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, 1.3862(cos t + isin t) is a solution of the equation p(z) = 0.

Now here is a rigorous proof for those who have studied analysis. It depends on the ex-treme value theorem from calculus applied to the continuous function f (x,y)≡ |p(x+ iy)|.

Proof: This will make sense for those who have had an advanced calsulus course.Suppose the nonconstant polynomial p(z) = a0 + a1z+ · · ·+ anzn,an ̸= 0, has no zero inC. 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)| is1 and occurs at z = 0. Since q(0) = 1, it follows q(z) = 1+ akzk + r (z) where r (z) is ofthe form

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

Choose a sequence, z j → 0, such that akzkj < 0. For example, you could let z j be such that

akzkj = −1/ j. The terms in r (z) are all of higher order than k and so for n large enough.

Then ∣∣q(z j)∣∣= ∣∣∣1+akzk

j + r (z j)∣∣∣≤ 1−1/ j+

∣∣r (z j)∣∣< 1

for all j large enough, showing∣∣q(z j)

∣∣ < 1 whenever j is large enough. This is a contra-diction to |q(z)| ≥ 1. ■

At this point, you could skip Chapters 2 and 3 and go directly to Chapter 4 if desired.

1.9 Exercises

1. Prove by induction that ∑nk=1 k3 =

n4 +12

n3 +14

n2.

2. Prove by induction that whenever n≥ 2,∑nk=1

1√k>√

3. Prove by induction that 1+∑ni=1 i(i!) = (n+1)!.

4. The binomial theorem states (x+ y)n = ∑nk=0(n

)xn−kyk where(

n+1k

(nk

k−1

)if k ∈ [1,n] ,

(n0

)≡ 1≡

(nn

)

12 CHAPTER 1. SOME PREREQUISITE TOPICSr too big r too small r just right50 D / 4/ © 2V0 vy ot | : | V\ } 0-50 ; 2 \ / 2-50 0 50 2 O 2 4 2 0 2 4 6Xx Xx XxNext is the graph when r = .5. Note how the closed curve is included in a circle whichhas 0+ 70 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, 1.3862 (cost + isint) is a solution of the equation p(z) = 0.Now here is a rigorous proof for those who have studied analysis. It depends on the ex-treme value theorem from calculus applied to the continuous function f (x,y) = |p (x+iy)|.Proof: This will make sense for those who have had an advanced calsulus course.Suppose the nonconstant polynomial p(z) = ap + a,z+ ++: +nz",a, #0, has no zero inC. Since lim),-,.. |p (z)| = 99, there is a zo with |p(zo)| = minzec|p(z)| > 0. Then let— +0)4(2) = “pe1 and occurs at z= 0. Since q(0) = 1, it follows q(z) = 1 +ayz* + r(z) where r(z) is ofthe form. This is also a polynomial which has no zeros and the minimum of |q (z)| isr(z) =Gmnz" + ami"! +...+4,z" form > k.Choose a sequence, z; —> 0, such that az <0. For example, you could let z; be such thatay = —1/j. The terms in r(z) are all of higher order than k and so for n large enough.Thenla(z))| =| +aneh +r (@)| 1-1/7 |r@)| <1for all j large enough, showing la(< i)| < 1 whenever j is large enough. This is a contra-diction to |q(z)|>1.™At this point, you could skip Chapters 2 and 3 and go directly to Chapter 4 if desired.1.9 Exercises1 1 11. Prove by induction that )°7_ | P= rue + 5” + ra12. Prove by induction that whenever n > 2,)°7_, 74 > Vn.3. Prove by induction that 1+)7_,i(i!) =(n+1)!.4. The binomial theorem states (x + y)” = Yio (7)x" *y* whereMr) = (ection (et ()